Structure of block quantum dynamical semigroups and their product systems

TL;DR

This paper investigates the structure of block quantum dynamical semigroups on $C^*$-algebras and von Neumann algebras, establishing a connection between their inclusion systems, morphisms, and product systems, with implications for dilations.

Contribution

It introduces a contractive morphism linking inclusion systems of von Neumann modules and demonstrates how such morphisms can be lifted to product systems, advancing the understanding of block quantum semigroups.

Findings

Existence of a contractive morphism T between inclusion systems.

Representation of $\psi_t$ via the morphism T and generating vectors.

The $E_0$-dilation of block quantum Markov semigroups remains a block map.

Abstract

W. Paschke's version of Stinespring's theorem associates a Hilbert $C^*$-module along with a generating vector to every completely positive map. Building on this, to every quantum dynamical semigroup (QDS) on a $C^*$-algebra $\mathcal A$ one may associate an inclusion system $E=(E_t)$ of Hilbert $\mathcal A$-$\mathcal A$-modules with a generating unit $\xi =(\xi_t)$. Suppose $\mathcal B$ is a von Neumann algebra, consider $M_2(\mathcal B)$, the von Neumann algebra of $2\times 2$ matrices with entries from $\mathcal B$. Suppose $(\Phi_t)_{t\ge 0}$ with $\Phi_t=\begin{pmatrix} \phi_t^1& \psi_t \psi_t^*&\phi_t^2 \end{pmatrix},$ is a QDS on $M_2(B)$ which acts block-wise and let $(E^i_t)_{t\ge 0}$ be the inclusion system associated to the diagonal QDS $(\phi^i_t)_{t\ge 0}$ with the generating unit $(\xi_t^i)_{t\ge 0}, i=1,2.$ It is shown that there is a contractive (bilinear) morphism $T=(T_t)_{t\ge0}$ from $(E^2_t)_{t\ge 0}$ to $(E^1_t)_{t\ge 0}$ such that $\psi_t(a)=\langle \xi^1_t, T_t a\xi^2_t\rangle $ for all $a\in\mathcal B.$ We also prove that any contractive morphism between inclusion systems of von Neumann $\mathcal B$-$\mathcal B$-modules can be lifted as a morphism between the product systems generated by them. We observe that the $E_0$-dilation of a block quantum Markov semigroup (QMS) on a unital $C^*$-algebra is again a semigroup of block maps.

Equations

$$\varphi (a)=⟨\xi ,a\xi ⟩=L_{\xi }^{\ast }{L}_{a\xi }=L_{\xi }^{\ast }\rho (a)L_{\xi }.$$

$$⟨\xi \odot \zeta ,a\xi \odot \zeta ⟩=⟨\zeta ,⟨\xi ,a\xi ⟩\zeta ⟩=⟨\zeta ,\varphi (a)\zeta ⟩=\psi \circ \varphi (a)\text{for all }a\in \mathcal{A}.$$

$$\kappa \mapsto \xi \odot \zeta ,$$

$$({\beta }_{r,s}\odot {\mathrm{id}}_{E_t}){\beta }_{r+s,t}=({\mathrm{id}}_{E_r}\odot {\beta }_{s,t}){\beta }_{r,s+t}.$$

$${\varphi }_t(b)=⟨{\xi }_t,b{\xi }_t⟩\text{ for }b\in \mathcal{B}.$$

$${\varphi }_t\circ {\varphi }_s(b)={\varphi }_t(⟨{\xi }_s,b{\xi }_s⟩)=⟨{\xi }_t,⟨{\xi }_s,b{\xi }_s⟩{\xi }_t⟩=⟨{\xi }_s\odot {\xi }_t,b({\xi }_s\odot {\xi }_t)⟩=⟨{\xi }_{t+s},b{\xi }_{t+s}⟩={\varphi }_{t+s}(b).$$

$${\xi }_{t+s}\mapsto {\xi }_s\odot {\xi }_t.$$

$$({\beta }_{r,s}\odot I_{E_t}){\beta }_{r+s,t}({\xi }_{r+s+t})$$

$$={\xi }_r\odot ({\xi }_s\odot {\xi }_t)=(I_{E_r}\odot {\beta }_{s,t})({\xi }_r\odot {\xi }_{s+t})$$

$$=(I_{E_r}\odot {\beta }_{s,t}){\beta }_{r,s+t}({\xi }_{r+s+t})$$

$$T_{s+t}={\gamma }_{s,t}^{\ast }(T_s\odot T_t){\beta }_{s,t}\text{ for all }s,t\in \mathbb{T}.$$

$${\gamma }_{s,t}{T}_{s+t}=(T_s\odot T_t){\beta }_{s,t}\text{ for all }s,t\in \mathbb{T}.$$

$$x=\left(\begin{array}{cc}pxp& px{p}^{\prime }\\ p^{\prime }xp& p^{\prime }x{p}^{\prime }\end{array}\right)\in \left(\begin{array}{cc}p\mathcal{A}p& p\mathcal{A}{p}^{\prime }\\ p^{\prime }\mathcal{A}p& p^{\prime }\mathcal{A}{p}^{\prime }\end{array}\right).$$

$$\Phi (x)=\left(\begin{array}{cc}\Phi (pxp)& \Phi (px{p}^{\prime })\\ \Phi (p^{\prime }xp)& \Phi (p^{\prime }x{p}^{\prime })\end{array}\right)\in \left(\begin{array}{cc}q\mathcal{B}q& q\mathcal{B}{q}^{\prime }\\ q^{\prime }\mathcal{B}q& q^{\prime }\mathcal{B}{q}^{\prime }\end{array}\right).$$

$$\Phi =\left(\begin{array}{cc}{\varphi }_{11}& {\varphi }_{12}\\ {\varphi }_{21}& {\varphi }_{22}\end{array}\right).$$

$$\Phi \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}⟨x_1,a{x}_1⟩& ⟨x_1,by⟩\\ ⟨y,c{x}_1⟩& ⟨y,dy⟩\end{array}\right)+\left(\begin{array}{cc}0& 0\\ 0& ⟨x_2,d({\mathrm{id}}_{E_2}-T^{\ast }T)x_2⟩\end{array}\right).$$

$$xb:=x\left(\begin{array}{cc}b& 0\\ 0& b\end{array}\right)\text{ and }⟨x,y{⟩}_{\Sigma }:=\sum _{i,j=1}^2⟨x,y{⟩}_{i,j}\text{for }x,y\in F,b\in \mathcal{B}.$$

$$[x]b=[x\left(\begin{array}{cc}b& 0\\ 0& b\end{array}\right)]\text{ and }⟨[x],[y]⟩=⟨x,y{⟩}_{\Sigma }=\sum _{i,j=1}^2⟨x,y{⟩}_{i,j}\text{for }x,y\in F,b\in \mathcal{B}.$$

$$\text{norm}[x]=\text{norm}{\sum _{i,j=1}^2⟨x,x{⟩}_{i,j}}^{\frac{1}{2}}=\text{norm}x\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right).$$

$$\text{norm}[x]\odot (g_1+g_2)={⟨g_1+g_2,\sum _{i,j=1}^2⟨x,x{⟩}_{i,j}(g_1+g_2)⟩}^{\frac{1}{2}}=\text{norm}x\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\odot \left(\begin{array}{c}{g}_1\\ g_2\end{array}\right).$$

$$a[x]:=[ax]\text{for }a\in \mathcal{A},x\in F$$

$$\text{norm}[x]\odot g=\text{norm}x\odot \left(\begin{array}{c}g\\ g\end{array}\right).$$

$$ax:=\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right)x\text{for }x\in F,a\in \mathcal{A}.$$

$$E^{(\mathcal{B})}\simeq \{[x{]}_F:x\in E\}\subseteq F^{(\mathcal{B})}.$$

$$\text{norm}{x_i-{\mathbb{E}}_{ii}x}^2=\text{norm}{{\mathbb{E}}_{ii}x{\mathbb{E}}_{jj}}^2=\text{norm}⟨{\mathbb{E}}_{ii}x{\mathbb{E}}_{jj},{\mathbb{E}}_{ii}x{\mathbb{E}}_{jj}⟩=\text{norm}{\mathbb{E}}_{jj}\Phi ({\mathbb{E}}_{ii}){\mathbb{E}}_{jj}=0,$$

$$\text{norm}{x_i-x{\mathbb{E}}_{ii}}^2=\text{norm}{{\mathbb{E}}_{jj}x{\mathbb{E}}_{ii}}^2=\text{norm}⟨{\mathbb{E}}_{jj}x{\mathbb{E}}_{ii},{\mathbb{E}}_{jj}x{\mathbb{E}}_{ii}⟩=\text{norm}{\mathbb{E}}_{ii}\Phi ({\mathbb{E}}_{jj}){\mathbb{E}}_{ii}=0.$$

$$x_i={\mathbb{E}}_{ii}x=x{\mathbb{E}}_{ii},i=1,2\text{ and hence }x=({\mathbb{E}}_{11}+{\mathbb{E}}_{22})x=x_1+x_2.$$

$$\Phi (A)=⟨x,Ax⟩=\sum _{i,j=1}^2⟨x_i,A{x}_j⟩=\left(\begin{array}{cc}⟨x_1,A{x}_1{⟩}_{11}& ⟨x_1,A{x}_2{⟩}_{12}\\ ⟨x_2,A{x}_1{⟩}_{21}& ⟨x_2,A{x}_2{⟩}_{22}\end{array}\right),$$

$$ax:=\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right)x\text{for }a\in \mathcal{A},x\in {\hat{E}}_i.$$

$$⟨[x_i],a[x_i]⟩=\sum _{r,s=1}^2{⟨{\mathbb{E}}_{ii}x,\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right){\mathbb{E}}_{ii}x⟩}_{r,s}=\sum _{r,s=1}^2\Phi {\left({\mathbb{E}}_{ii}\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right){\mathbb{E}}_{ii}\right)}_{r,s}={\varphi }_i(a).$$

$$⟨U[z],U[w]⟩=\sum _{i,j=1}^2⟨{\mathbb{E}}_{12}z,{\mathbb{E}}_{12}w{⟩}_{i,j}=\sum _{i,j=1}^2⟨z,{\mathbb{E}}_{21}{\mathbb{E}}_{12}w{⟩}_{i,j}=\sum _{i,j=1}^2⟨z,w{⟩}_{i,j}=⟨[z],[w]⟩,$$

Full text

1. Introduction

It is well-known that a block matrix $\begin{pmatrix}A&B\\ B^{*}&D\end{pmatrix}$ of operators on a direct sum of Hilbert spaces $\mathcal{H}\oplus\mathcal{K}$ is positive if and only if $A,D$ are positive and there exists a contraction $K:\mathcal{K}\to\mathcal{H}$ such that $B=A^{\frac{1}{2}}KD^{\frac{1}{2}}.$ This says that the positivity of a block matrix is determined up to a contraction by the positive diagonals. We want to look at the structure of block completely positive (CP) maps, that is, completely positive maps which send $2\times 2$ block operators as above to $2\times 2$ block operators. Such maps have already appeared in many different contexts. For example, Paulsen uses the block CP maps in [15] to prove that every completely polynomially bounded operator is similar to a contraction. The structure of completely bounded (CB) maps are understood using the $2\times 2$ block CP maps (See [15, 16, 21],[14, Chapter 8]). The usual way to study the structure of CP maps into $\mathscr{B}(\mathcal{H})$ is via Stinespring dilation theorem ([20]), which says that if $\phi:\mathcal{A}\to\mathscr{B}(\mathcal{H})$ is a CP map then there is a triple $(\mathcal{K},\pi,V)$ of a Hilbert space $\mathcal{K}$, a representation $\pi:\mathcal{A}\to\mathscr{B}(\mathcal{K})$ and a bounded operator $V\in\mathscr{B}(\mathcal{H},\mathcal{K})$ such that $\phi(a)=V^{*}\pi(a)V$ for all $a\in\mathcal{A}.$ If $\Phi=\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix}:M_{2}(\mathcal{A})\to M_{2}(\mathscr{B}(\mathcal{H}))$ is a block CP map, then the diagonals $\phi_{i},i=1,2$ are also CP maps on $\mathcal{A}.$ Also the Stinespring representation of $\Phi$ gives us natural Stinespring representations for $\phi_{i}$ by the appropriate compressions. In [16, Corollary 2.7] Paulsen and Suen proved that: if $\Phi=\begin{pmatrix}\phi&\psi\\ \psi^{*}&\phi\end{pmatrix}:M_{2}(\mathcal{A})\to M_{2}(\mathscr{B}(\mathcal{H}))$ is CP and if $\phi$ has the minimal Stinespring representation $(\mathcal{K},\pi,V)$ then there exists a contraction $T\in\pi(\mathcal{A})^{\prime}$ such that $\psi(\cdot)=V^{*}\pi(\cdot)TV.$ In [9] Furuta studied the completion problems of partial matrices of block completely positive maps (see Remark 3.10). These results show the importance of studying block CP maps. In this article, we want to study one parameter semigroups of block CP maps.

To begin with, the Stinespring’s theorem approach is not so convenient to study compositions of CP maps between general $C^{*}$-algebras. In [13] Paschke proved a structure theorem for CP maps between $C^{*}$-algebras, which is a generalization of Stinespring theorem, we call it as GNS-construction for CP maps, in view of its close connection with the familiar GNS-construction for positive linear functionals on $C^{*}$-algebras. This theorem says that given a CP map $\phi:\mathcal{A}\to\mathcal{B}$ there is a pair $(E,\xi)$ of a Hilbert $\mathcal{A}$-$\mathcal{B}$-module $E$ (see below for definitions) and a cyclic vector $\xi$ in $E$ such that $\phi(a)=\langle\xi,a\xi\rangle$ for all $a\in\mathcal{A}.$ The advantage of GNS-construction is that we can write the GNS-construction of the composition of two CP maps as a submodule of the tensor product of their GNS-constructions (see Remark 2.10). Employing this scheme we can associate an inclusion system (synonymous with subproduct system) $(E_{t})$ of Hilbert $\mathcal{B}$-$\mathcal{B}$-modules with a generating unit $(\xi_{t})$ to any one parameter semigroup of completely positive maps on $\mathcal{B}.$ It may be recalled that semigroups of completely positive maps on a $C^{*}$-algebra are known as quantum dynamical semigroups (QDS), semigroups of unital completely positive maps are known as quantum Markov semigroups (QMS) and semigroups of unital endomorphisms are known as $E_{0}$-semigroups. In [4] Bhat proved that any QMS on $\mathscr{B}(\mathcal{H})$ admits a unique $E_{0}$-dilation, and in [5] extended the result to QMS on unital $C^{*}$-subalgebras of $\mathscr{B}(\mathcal{H})$. Later in [8] Bhat and Skeide constructed the $E_{0}$-dilation for arbitrary quantum Markov semigroups (QMS) on abstract unital $C^{*}$-algebras, using the technology of Hilbert $C^{*}$-modules. Here one sees for the first time subproduct systems and product systems of Hilbert $C^{*}$-modules. Muhly and Solel (in [12]) took a dual approach to achieve this, where they have called these Hilbert $C^{*}$-modules as $C^{*}$-correspondences.

While studying units of $E_{0}$-semigroups of $\mathscr{B}(\mathcal{H})$ Powers was led into considering block CP semigroups (See [17] and [6], [19]). In [7] Bhat and Mukherjee proved a structure theorem for block QMS on $\mathscr{B}(\mathcal{H}\oplus\mathcal{K})$. The main point is that when we have a block QMS, there is a contractive morphism between inclusion systems of diagonal CP semigroups. Moreover, this morphism lifts to associated product systems. The main goal of this paper is to explore the structure of block quantum dynamical semigroups on general von Neumann algebras. The extension of the theory from $\mathscr{B}(\mathcal{H})$ case is not straightforward for the following reason. In the case of $\mathscr{B}(\mathcal{H})$, we need only to consider product systems of Hilbert spaces, whereas now we need to deal with both product systems of Hilbert $\mathcal{B}$-modules and also product systems of Hilbert $M_{2}(\mathcal{B})$-modules (see Theorem 3.7) and their inter-dependences. But a careful analysis of these modules does lead us to a morphism between inclusion systems as in the $\mathscr{B}(\mathcal{H})$ case and this morphism can also be lifted to a morphism at the level of associated product systems (Theorem 5.3). At various steps we consider adjoints of maps between our modules and so it is convenient to have von Neumann modules. The picture is unclear for Hilbert $C^{*}$-modules.

In Section 2 we recall the preliminaries. In Section 3 we prove a structure theorem for block CP maps from $M_{2}(\mathcal{A})$ to $M_{2}(\mathcal{B})$ when $\mathcal{B}$ is a von Neumann algebra also we give an example to indicate that we can not replace $\mathcal{B}$ by an arbitrary $C^{*}$-algebra. We extend this result to semigroups of block CP maps on $M_{2}(\mathcal{B})$ in Section 4.1. In Section 4.2 we show that the $E_{0}$-dilation of a block QMS is again block semigroup. In the final section we prove that any morphism between inclusion systems of von Neumann $\mathcal{B}$-$\mathcal{B}$-modules can be lifted as a morphism between the product systems generated by them. Subproduct systems and inclusion systems are synonyms. The word ‘subproduct systems’ seems to be better established now. Since we are mostly following the ideas and notations of [7], we will continue to call these objects as inclusion systems.

Given a linear map $\psi:\mathcal{A}\to\mathcal{B}$ between $C^{*}$-algebras, $\psi^{*}$ denotes the linear map $\psi^{*}:\mathcal{A}\to\mathcal{B}$ defined by $\psi^{*}(a)=\psi(a^{*})^{*}$ for all $a\in\mathcal{A}.$ For $h\in\mathcal{H},k\in\mathcal{K},$ $\outerproduct{k}{h}$ denotes the bounded linear operator $\outerproduct{k}{h}:\mathcal{H}\to\mathcal{K}$ defined by $\outerproduct{k}{h}(h^{\prime})=\langle h,h^{\prime}\rangle k$ for all $h^{\prime}\in\mathcal{H}.$ All our Hilbert spaces are taken as complex and separable, with scalar products linear in the second variable.

2. Preliminaries

For the basic theory about Hilbert $C^{*}$-modules we refer to [10, 8, 3, 13, 18]. Here we recall only the most required definitions and results about Hilbert $C^{*}$-modules. A pre-Hilbert $\mathcal{B}$-module $E$ is a vector space, which is a right $\mathcal{B}$-module (compatible with scalar multiplication) together with a $\mathcal{B}$-valued inner product. It is said to be a Hilbert $\mathcal{B}$-module if it is complete in the norm given by $\norm{x}=\sqrt{\norm{\langle x,x\rangle}}$ for $x\in E.$ If we have just semi-inner product on a $\mathcal{B}$-module $E,$ using the Cauchy-Schwarz inequality $\langle x,y\rangle\langle y,x\rangle\leq\norm{\langle y,y\rangle}\langle x,x\rangle,$ we can see that $N=\{x\in E:\langle x,x\rangle=0\}$ is a $\mathcal{B}$-submodule and hence $E/N$ is a pre-Hilbert $\mathcal{B}$-module equipped with the natural inner product. A Hilbert $\mathcal{A}$-$\mathcal{B}$-module $E$ is a Hilbert $\mathcal{B}$-module with a nondegenerate action of $\mathcal{A}$ on $E.$ Two Hilbert $\mathcal{A}$-$\mathcal{B}$-modules $E$ and $F$ are said to be isomorphic if there is a bilinear unitary between them and in this case, we write $E\simeq F.$

Definition 2.1**.**

Let $E$ be a Hilbert $\mathcal{A}$-$\mathcal{B}$-module and $F$ be a Hilbert $\mathcal{B}$-$\mathcal{C}$-module. Then $\langle x\otimes y,x^{\prime}\otimes y^{\prime}\rangle=\langle y,\langle x,x^{\prime}\rangle y^{\prime}\rangle$ defines a semi-inner product on (the algebraic tensor product) $E\otimes F$ with the natural right $\mathcal{C}$-action. Let $N=\{w\in E\otimes F:\langle w,w\rangle=0\}.$ The interior tensor product of $E$ and $F$ is defined as the completion of ${E\otimes F}/N$ and it is denoted by $E\odot F.$

Observe that $E\odot F$ in previous definition is a Hilbert $\mathcal{A}$-$\mathcal{C}$-module with the natural left action of $\mathcal{A}.$ We denote the equivalence class of $x\otimes y$ in $E\odot F$ by $x\odot y.$ It may be noted that for $b\in\mathcal{B}$, $xb\odot y=x\odot by.$ Let $E,E^{\prime}$ be Hilbert $\mathcal{A}$-$\mathcal{B}$-modules and $F,F^{\prime}$ be Hilbert $\mathcal{B}$-$\mathcal{C}$-modules. If $T:E\to E^{\prime}$ and $S:F\to F^{\prime}$ are bounded bilinear maps then, $T\odot S:E\odot F\to E^{\prime}\odot F^{\prime}$ is a bounded bilinear map defined by $(T\odot S)(x\odot y)=Tx\odot Sy$ for $x\in E,y\in F.$

Let $\mathcal{A}$ and $\mathcal{B}$ be $C^{*}$-algebras and let $\mathcal{G}$ be a Hilbert space on which $\mathcal{B}$ is represented nondegenerately ($\mathcal{G}$ can be viewed as a Hilbert $\mathcal{B}$-$\mathbb{C}$-module). Let $E$ be a Hilbert $\mathcal{A}$-$\mathcal{B}$-module. Consider the tensor product $\mathcal{H}=E\odot\mathcal{G},$ which is a Hilbert $\mathcal{A}$-$\mathbb{C}$-module. That is, $\mathcal{H}$ is a Hilbert space with a representation $\rho:\mathcal{A}\to\mathscr{B}(\mathcal{H}).$ The map $\rho$ is called the Stinespring representation of $\mathcal{A}$ associated with $E$ and $\mathcal{G}$ (see Remark 2.6 below). For $x\in E$ let $L_{x}:\mathcal{G}\to\mathcal{H}$ be defined by $L_{x}(g)=x\odot g,$ then $L_{x}\in\mathscr{B}(\mathcal{G},\mathcal{H})$ with $L_{x}^{*}:x^{\prime}\odot g\mapsto\langle x,x^{\prime}\rangle g.$ Define $\eta:E\to\mathscr{B}(\mathcal{G},\mathcal{H})$ by $\eta(x)=L_{x}.$ Then we have $L_{x}^{*}L_{y}=\langle x,y\rangle\in\mathcal{B}\subseteq\mathscr{B}(\mathcal{G}),$ hence, if the representation of $\mathcal{B}$ on $\mathcal{G}$ is faithful then so is $\eta.$ Also we have $L_{axb}=\rho(a)L_{x}b$ so that we may identify $E$ as a concrete subset of $\mathscr{B}(\mathcal{G},\mathcal{H}).$ The map $\eta$ is called the Stinespring representation of $E$ (associated with $\mathcal{G}$).

In particular, if $\mathcal{B}$ is a von Neumann algebra on a Hilbert space $\mathcal{G},$ we always consider $E$ as a concrete subset of $\mathscr{B}(\mathcal{G},E\odot\mathcal{G}).$

Definition 2.2**.**

Let $\mathcal{B}$ be a von Neumann algebra on a Hilbert space $\mathcal{G}.$ A Hilbert $\mathcal{B}$-module $E$ is said to be a von Neumann $\mathcal{B}$-module if $E$ is strongly closed in $\mathscr{B}(\mathcal{G},E\odot\mathcal{G}).$ Further, if $\mathcal{A}$ is a von Neumann algebra, a von Neumann $\mathcal{B}$-module $E$ is said to be a von Neumann $\mathcal{A}$-$\mathcal{B}$-module if it is a Hilbert $\mathcal{A}$-$\mathcal{B}$-module such that the Stinespring representation $\rho:\mathcal{A}\to\mathscr{B}(E\odot\mathcal{G})$ is normal.

Remark 2.3**.**

Let $\mathcal{A}$ be a $C^{*}$-algebra and $\mathcal{B}$ be a von Neumann algebra on a Hilbert space $\mathcal{G}$. Let $E$ be a Hilbert $\mathcal{A}$-$\mathcal{B}$-module. Then $E$ can be completed in strong operator topology to get, $\overline{E}^{s}$ which is a Hilbert $\mathcal{A}$-$\mathcal{B}$-module and is a von Neumann $\mathcal{B}$-module. Here the left action by $\mathcal{A}$ need not be normal.

Remark 2.4**.**

If $E$ is a von Neumann $\mathcal{B}$-module, then $\mathscr{B}^{a}(E)$ is a von Neumann subalgebra of $\mathscr{B}(E\odot\mathcal{G}).$ von Neumann modules are self-dual and hence any bounded right linear map between von Neumann modules is adjointable. If $F$ is a von Neumann submodule of $E,$ then there exists a projection $p$ $(p=p^{2}=p^{*})$ in $\mathscr{B}^{a}(E)$ onto $F$, that is, $p(E)=F$ and $E$ decomposes as $E=F\oplus F^{\perp}.$

Let $\mathcal{A}$ and $\mathcal{B}$ be unital $C^{*}$-algebras and let $\phi:\mathcal{A}\to\mathcal{B}$ be a CP map. Consider the algebraic tensor product $\mathcal{A}\otimes\mathcal{B}.$ For $a,a^{\prime}\in\mathcal{A},b,b^{\prime}\in\mathcal{B},$ define $\langle a\otimes b,a^{\prime}\otimes b^{\prime}\rangle=b^{*}\phi(a^{*}a^{\prime})b.$ Then $\langle\cdot,\cdot\rangle$ is a semi-inner product on $\mathcal{A}\otimes\mathcal{B}.$ Let $N_{\mathcal{A}\otimes\mathcal{B}}=\{w\in\mathcal{A}\otimes\mathcal{B}:\langle w,w\rangle=0\}.$ Let $E$ be the completion of ${\mathcal{A}\otimes\mathcal{B}}/N_{\mathcal{A}\otimes\mathcal{B}}.$ Then $E$ is a Hilbert $\mathcal{A}$-$\mathcal{B}$-module in a natural way. Let $\xi=\mathbf{1}\otimes\mathbf{1}+N_{\mathcal{A}\otimes\mathcal{B}},$ then we have $\phi(a)=\langle\xi,a\xi\rangle.$ Moreover, $\xi$ is cyclic (i.e., $E=\overline{\operatorname{span}}(\mathcal{A}\xi\mathcal{B})$). The pair $(E,\xi)$ is called the GNS-construction of $\phi$ and $E$ is called the GNS-module for $\phi.$ Obviously, $\phi$ is unital if and only if $\langle\xi,\xi\rangle=\mathbf{1}.$

Definition 2.5**.**

Let $\phi:\mathcal{A}\to\mathcal{B}$ be a CP map. Let $E$ be a Hilbert $\mathcal{A}$-$\mathcal{B}$-module and $\xi\in E,$ We call $(E,\xi)$ as a GNS-representation for $\phi$ if $\phi(a)=\langle\xi,a\xi\rangle$ for all $a\in\mathcal{A}.$ It is said to be minimal if $E=\overline{\operatorname{span}}(\mathcal{A}\xi\mathcal{B}).$

Note that the GNS-module in the GNS-construction is minimal. If $(E,\xi)$ and $(F,\zeta)$ are two minimal GNS-representations for $\phi$ then the map $\xi\mapsto\zeta$ extends as a bilinear unitary from $E$ to $F.$ Hence the GNS-representation is unique up to (unitary) isomorphism.

Remark 2.6**.**

Let $\phi:\mathcal{A}\to\mathscr{B}(\mathcal{G})$ be a CP map.

Suppose $(E,\xi)$ is the GNS-construction for $\phi.$ Let $\eta:E\to\mathscr{B}(\mathcal{G},\mathcal{H})$ be the Stinespring representation of $E$ as defined above, then

[TABLE]

Note that $L_{\xi}$ is an isometry in $\mathscr{B}(\mathcal{G},\mathcal{H})$ if and only if $\phi$ is unital. Note also that $\overline{\operatorname{span}}\{\rho(a)L_{\xi}g:a\in\mathcal{A},g\in\mathcal{G}\}=\overline{\operatorname{span}}\{a\xi\odot g:a\in\mathcal{A},g\in\mathcal{G}\}=E\odot\mathcal{G}=\mathcal{H}.$ So we obtain the usual minimal Stinespring representation $(\mathcal{H},\rho,L_{\xi})$ of $\phi.$

Conversely, if $(\mathcal{H},\pi,V)$ is the minimal Stinespring representation for $\phi.$ Consider $\mathscr{B}(\mathcal{G},\mathcal{H})$ as a Hilbert $\mathcal{A}$-$\mathscr{B}(\mathcal{G})$-module, where the left action of $\mathcal{A}$ is given by the representation $\pi.$ Let $E=\overline{\operatorname{span}}\leavevmode\nobreak\ \mathcal{A}V\mathscr{B}(\mathcal{G})\subseteq\mathscr{B}(\mathcal{G},\mathcal{H}).$ Then $(E,V)$ is a minimal GNS-representation for $\phi.$

Proposition 2.7**.**

If $E$ is the GNS-module of a normal completely positive map $\phi:\mathcal{A}\to\mathcal{B}$ between von Neumann algebras, then $\overline{E}^{s}$ is a von Neumann $\mathcal{A}$-$\mathcal{B}$-module.

Proposition 2.8**.**

Let $E$ be a von Neumann $\mathcal{A}$-$\mathcal{B}$-module and let $F$ be a von Neumann $\mathcal{B}$-$\mathcal{C}$-module where $\mathcal{C}$ acts on a Hilbert space $\mathcal{G}.$ Then the strong closure $\overline{E\odot F}^{s}$ of the tensor product $E\odot F$ in $\mathscr{B}(\mathcal{G},E\odot F\odot\mathcal{G}),$ is a von Neumann $\mathcal{A}$-$\mathcal{C}$-module.

Definition 2.9**.**

Due to Propositions 2.7, and 2.8 we make the following conventions:

1. (1)

Whenever $\mathcal{B}$ is a von Neumann algebra and $\phi:\mathcal{A}\to\mathcal{B}$ is a CP map, by GNS-module we always mean $\overline{E}^{s},$ where $E$ is the GNS-module, constructed above. 2. (2)

If $E$ and $F$ are von Neumann modules, by tensor product of $E$ and $F$ we mean the strong closure $\overline{E\odot F}^{s}$ of $E\odot F$ and we still write $E\odot F.$

Remark 2.10**.**

Let $\phi:\mathcal{A}\to\mathcal{B}$ and $\psi:\mathcal{B}\to\mathcal{C}$ be CP maps with GNS-representations $(E,\xi)$ and $(F,\zeta)$ respectively. Let $(K,\kappa)$ be the GNS-construction of $\psi\circ\phi.$ Note that

[TABLE]

This says that $(E\odot F,\xi\odot\zeta)$ is a GNS-representation (not necessarily minimal) for $\psi\circ\phi.$ Thus the the mapping

[TABLE]

extends as a unique bilinear isometry from $K$ to $E\odot F.$ Hence we may identify $K$ as the submodule $\overline{\operatorname{span}}(\mathcal{A}\xi\odot\zeta\mathcal{C})$ of $E\odot F.$

Note that $E\odot F=\overline{\operatorname{span}}(\mathcal{A}\xi\mathcal{B}\odot\mathcal{B}\zeta\mathcal{C})=\overline{\operatorname{span}}(\mathcal{A}\xi\odot\mathcal{B}\zeta\mathcal{C})=\overline{\operatorname{span}}(\mathcal{A}\xi\mathcal{B}\odot\zeta\mathcal{C}).$

In the following we define quantum dynamical semigroups and see their connection with inclusion systems. We may take $\mathbb{T}$ as either the semigroup of non-negative integers $\mathbb{Z}_{+}$ or as the semigroup of non-negative reals $\mathbb{R}_{+}$ under addition, but our real interest lies in $\mathbb{T}=\mathbb{R}_{+}$, in view of quantum theory of open systems. For more details on this theory look at [2, 1, 8].

Definition 2.11**.**

Let $\mathcal{A}$ be a unital $C^{*}$-algebra. A family $\phi=(\phi_{t})_{t\in\mathbb{T}}$ of CP maps on $\mathcal{A}$ is said to be a quantum dynamical semigroup (QDS) or one-parameter CP-semigroup if

1. (1)

$\phi_{s+t}=\phi_{s}\circ\phi_{t}$ for all $t\in\mathbb{T},$ 2. (2)

$\phi_{0}(a)=a$ for all $a\in\mathcal{A},$ 3. (3)

$\phi_{t}(\mathbf{1})\leq\mathbf{1}$ for all $t\in\mathbb{T},$ (contractivity)

It is said to be conservative QDS or quantum Markov semigroup (QMS) if $\phi_{t}$ is unital for all $t\in\mathbb{T}.$ In practice, in addition to (1)-(3) we may assume continuity of $t\to\phi_{t}(a)$ in different topologies, depending upon the context.

Definition 2.12**.**

Let $\mathcal{B}$ be a $C^{*}$-algebra. An inclusion system $(E,\beta)$ is a family $E=(E_{t})_{t\in\mathbb{T}}$ of Hilbert $\mathcal{B}$-$\mathcal{B}$-modules with $E_{0}=\mathcal{B}$ and a family $\beta=(\beta_{s,t})_{s,t\in\mathbb{T}}$ of bilinear isometries $\beta_{s,t}:E_{s+t}\to E_{s}\odot E_{t}$ such that, for all $r,s,t\in\mathbb{T},$

[TABLE]

It is said to be a product system if every $\beta_{s,t}$ is unitary.

Remark 2.13**.**

If $\mathcal{B}$ is von Neumann algebra in Definition 2.12, then we consider inclusion system of von Neumann $\mathcal{B}$-$\mathcal{B}$-modules.

Definition 2.14**.**

Let $(E,\beta)$ be an inclusion system. A family $\xi^{\odot}=(\xi_{t})_{t\in\mathbb{T}}$ of vectors $\xi_{t}\in E_{t}$ is called a unit for the inclusion system, if $\beta_{s,t}(\xi_{s+t})=\xi_{s}\odot\xi_{t}.$ It is said to be unital if $\langle\xi_{t},\xi_{t}\rangle=1$ for all $t\in\mathbb{T},$ and generating if $\xi_{t}$ is cyclic in $E_{t}$ for all $t\in\mathbb{T}.$ Suppose $(E,\beta)$ is a product system, a unit $\xi^{\odot}=(\xi_{t})_{t\in\mathbb{T}}$ is said to be a generating unit for the product system $(E,\beta)$ if $E_{t}$ is spanned by images of elements $b_{n}\xi_{t_{n}}\odot\dots\odot b_{1}\xi_{t_{1}}b_{0}$ ($t_{i}\in\mathbb{T},\sum t_{i}=t,b_{i}\in\mathcal{B}$) under successive applications of appropriate mappings $\operatorname{id}\odot\beta_{s,s^{\prime}}^{*}\odot\operatorname{id}.$

Suppose $(E,\beta)$ is an inclusion system with a unit $\xi^{\odot}.$ Consider $\phi_{t}:\mathcal{B}\to\mathcal{B}$ defined by

[TABLE]

Then as $\beta_{s,t}$’s are bilinear isometries and $\xi^{\odot}$ is a unit, for $b\in\mathcal{B}$ we have

[TABLE]

That is, $(\phi_{t})_{t\in\mathbb{T}}$ is a QDS. Clearly $(\phi_{t})_{t\in\mathbb{T}}$ is a QMS if $\xi^{\odot}$ is unital. In the converse direction we have the following remark:

Remark 2.15**.**

Let $\phi=(\phi_{t})_{t\in\mathbb{T}}$ be a QDS on a unital $C^{*}$-algebra $\mathcal{B}$ and let $(E_{t},\xi_{t})$ be the (minimal) GNS-construction for $\phi_{t}.$ (Recall that $\xi_{t}$ is a cyclic vector in $E_{t}$ such that $\phi_{t}(b)=\langle\xi_{t},b\xi_{t}\rangle$ for all $b\in\mathcal{B}$). Note that $E_{0}=\mathcal{B}$ and $\xi_{0}=\mathbf{1}.$ Define $\beta_{s,t}:E_{s+t}\to E_{s}\odot E_{t}$ by

[TABLE]

Then by Remark 2.10 $\beta_{s,t}$’s are bilinear isometries. Now

[TABLE]

shows that $(E=(E_{t}),\beta=(\beta_{s,t}))$ is an inclusion system of Hilbert $\mathcal{B}$-$\mathcal{B}$-module. It is obvious that $\xi^{\odot}=(\xi_{t})$ is a generating unit for $(E,\beta).$

Suppose $\mathcal{B}$ is a von Neumann algebra and each $\phi_{t}$ is a normal CP map on $\mathcal{B},$ then recall from Proposition 2.7 (see also Definition 2.9) that the GNS-module $E_{t}=\overline{E}^{s}_{t}$ is a von Neumann $\mathcal{B}$-$\mathcal{B}$-module for all $t\in\mathbb{T}.$ In this case, $(E,\beta)$ is an inclusion system of von Neumann $\mathcal{B}$-$\mathcal{B}$-modules with the generating unit $\xi^{\odot}.$

Definition 2.16**.**

For a QDS $\phi=(\phi_{t})_{t\geq 0}$ on $\mathcal{B},$ the inclusion system with the generating unit $(E,\beta,\xi^{\odot})$ as given in Remark 2.15 is called the inclusion system associated to $\phi.$ Sometimes we will just write $(E,\xi^{\odot})$ instead of $(E,\beta,\xi^{\odot}).$

Definition 2.17**.**

Let $(E,\beta)$ and $(F,\gamma)$ be two inclusion systems. Let $T=(T_{t})_{t\in\mathbb{T}}$ be a family of adjointable bilinear maps $T_{t}:E_{t}\to F_{t},$ satisfying $\norm{T_{t}}\leq e^{tk}$ for some $k\in\mathbb{R}.$ Then $T$ is said to be a morphism or a weak morphism from $(E,\beta)$ to $(F,\gamma)$ if every $\gamma_{s,t}$ is adjointable and

[TABLE]

It is said to be a strong morphism if

[TABLE]

3. Block CP maps

Let $\mathcal{A}$ be a unital $C^{*}$-algebra. Let $p\in\mathcal{A}$ be a projection. Set $p^{\prime}=\mathbf{1}-p.$ Then for every $x\in\mathcal{A}$ we have the following block decomposition:

[TABLE]

Definition 3.1**.**

Let $\mathcal{A}$ and $\mathcal{B}$ be unital $C^{*}$-algebras. Let $p\in\mathcal{A}$ and $q\in\mathcal{B}$ be projections. We say that a map $\Phi:\mathcal{A}\to\mathcal{B}$ is a block map (with respect to $p$ and $q$) if $\Phi$ respects the above block decomposition. i.e., for all $x\in\mathcal{A}$ we have

[TABLE]

If $\Phi:\mathcal{A}\to\mathcal{B}$ is a block map, then we get the following four maps: $\phi_{11}:p\mathcal{A}p\to q\mathcal{B}q,$ $\phi_{12}:p\mathcal{A}p^{\prime}\to q\mathcal{B}q^{\prime},\phi_{21}:p^{\prime}\mathcal{A}p\to q^{\prime}\mathcal{B}q,$ and $\phi_{22}:p^{\prime}\mathcal{A}p^{\prime}\to q^{\prime}\mathcal{B}q^{\prime}.$ So we write $\Phi$ as

[TABLE]

Lemma 3.2**.**

Let $\mathcal{A}$ and $\mathcal{B}$ be unital $C^{*}$-algebras. For $i=1,2,$ let $\phi_{i}:\mathcal{A}\to\mathcal{B}$ be a CP map with GNS-representation $(E_{i},x_{i}).$ Suppose $T:E_{2}\to E_{1}$ is an adjointable bilinear contraction and $\psi:\mathcal{A}\to\mathcal{B}$ is given by $\psi(a)=\langle x_{1},Tax_{2}\rangle.$ Then the block map $\Phi=\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix}:M_{2}(\mathcal{A})\to M_{2}(\mathcal{B})$ is CP.

Proof.

Set $y=Tx_{2}\in E_{1}.$ Then

[TABLE]

Clearly $\begin{pmatrix}a&b\\ c&d\end{pmatrix}\mapsto\begin{pmatrix}\langle x_{1},ax_{1}\rangle&\langle x_{1},by\rangle\\ \langle y,cx_{1}\rangle&\langle y,dy\rangle\end{pmatrix}$ is CP. Since $T$ is an adjointable bilinear contraction, $(\operatorname{id}_{E_{2}}-T^{*}T)$ is bilinear and positive. Hence $\begin{pmatrix}a&b\\ c&d\end{pmatrix}\mapsto\begin{pmatrix}0&0\\ 0&\langle x_{2},d(\operatorname{id}_{E_{2}}-T^{*}T)x_{2}\rangle\end{pmatrix}$ is CP. Therefore $\Phi$ is CP. ∎

Let $F$ be a Hilbert $M_{2}(\mathcal{B})$-module. Define a right $\mathcal{B}$-module action and a $\mathcal{B}$-valued semi-inner product $\langle\cdot,\cdot\rangle_{\Sigma}$ on $F$ by

[TABLE]

where $\langle x,y\rangle_{i,j}$ denotes the $(i,j)$th entry of $\langle x,y\rangle\in M_{2}(\mathcal{B}).$

Let $F^{(\mathcal{B})}$ denote the quotient space $F/N$ where $N=\{x:\langle x,x\rangle_{\Sigma}=0\}.$ (We denote the coset $x+N$ of $x\in F$ by $[x]_{F}$ or just by $[x]$). Then $F^{(\mathcal{B})}$ is a pre-Hilbert $\mathcal{B}$-module with right $\mathcal{B}$-action and inner product given by

[TABLE]

Proposition 3.3**.**

If $F$ is a Hilbert (von Neumann) $M_{2}(\mathcal{B})$-module, then $F^{(\mathcal{B})}$ is a Hilbert (von Neumann) $\mathcal{B}$-module.

Proof.

Let $F$ be a Hilbert $M_{2}(\mathcal{B})$-module. For each $x\in F,$ we have $[x]=[x\begin{pmatrix}1/2&1/2\\ 1/2&1/2\end{pmatrix}]$ and

[TABLE]

Consider a Cauchy sequence $([x_{n}])_{n\geq 1}$ in $F^{(\mathcal{B})}.$ Set $y_{n}=x_{n}\begin{pmatrix}1&1\\ 1&1\end{pmatrix}\in F.$ Then by (10), $(y_{n})_{n\geq 1}$ is a Cauchy sequence in $F.$ Let $y=\lim_{n\to\infty}y_{n}$ in $F.$ Then $y=y\begin{pmatrix}1/2&1/2\\ 1/2&1/2\end{pmatrix}.$ Take $x=\frac{y}{2}.$ Then, again by using (10), we see that $([x_{n}])_{n\geq 1}$ converges to $[x]$ in $F^{(\mathcal{B})}.$ Thus $F^{(\mathcal{B})}$ is complete.

Now assume that $F$ is von Neumann $M_{2}(\mathcal{B})$-module. Let $\mathcal{B}\subseteq\mathscr{B}(\mathcal{G}).$ So $F^{(\mathcal{B})}\subseteq\mathscr{B}(\mathcal{G},F^{(\mathcal{B})}\odot\mathcal{G})$ and $F\subseteq\mathscr{B}(\mathcal{G}^{2},F\odot\mathcal{G}^{2})$ where $\mathcal{G}^{2}=\mathcal{G}\oplus\mathcal{G}.$ We have for $x\in F,g_{1},g_{2}\in\mathcal{G},$

[TABLE]

Using (11), we can prove as in the above case, that $F^{(\mathcal{B})}$ is SOT closed in $\mathscr{B}(\mathcal{G},F^{(\mathcal{B})}\odot\mathcal{G})$ and hence $F^{(\mathcal{B})}$ is a von Neumann $\mathcal{B}$-module. ∎

Let $F$ be a Hilbert $M_{2}(\mathcal{B})$-module. Suppose $F$ has a nondegenerate left action of $\mathcal{A},$ then (10) implies that the natural left action of $\mathcal{A}$ on $F^{(\mathcal{B})}$ given by

[TABLE]

is a well defined nondegenerate action.

Proposition 3.4**.**

If $F$ is a Hilbert (von Neumann) $\mathcal{A}$-$M_{2}(\mathcal{B})$-module, then $F^{(\mathcal{B})}$ is a Hilbert (von Neumann) $\mathcal{A}$-$\mathcal{B}$-module with the left action defined in (12).

Proof.

If $F$ is a Hilbert $\mathcal{A}$-$M_{2}(\mathcal{B})$-module, then clearly $F^{(\mathcal{B})}$ is a Hilbert $\mathcal{A}$-$\mathcal{B}$-module. We shall prove that if $F$ is a von Neumann $\mathcal{A}$-$M_{2}(\mathcal{B})$-module, then $F^{(\mathcal{B})}$ is a von Neumann $\mathcal{A}$-$\mathcal{B}$-module. Let $\mathcal{B}\subseteq\mathscr{B}(\mathcal{G}).$ So $F^{(\mathcal{B})}\subseteq\mathscr{B}(\mathcal{G},F^{(\mathcal{B})}\odot\mathcal{G})$ and $F\subseteq\mathscr{B}(\mathcal{G}^{2},F\odot\mathcal{G}^{2})$ where $\mathcal{G}^{2}=\mathcal{G}\oplus\mathcal{G}.$ We must show that the Stinespring representation $\rho:\mathcal{A}\to\mathscr{B}(F^{(\mathcal{B})}\odot\mathcal{G})$ of $\mathcal{A}$ given by $\rho(a)([x]\odot g)=a[x]\odot g$ is normal. For any $x\in F,g\in\mathcal{G},$ a computation similar to (11) implies that

[TABLE]

As the Stinespring representation $\hat{\rho}:\mathcal{A}\to\mathscr{B}(F\odot\mathcal{G}^{2})$ given by $\tilde{\rho}(a)(x\odot\underline{g})=ax\odot\underline{g}$ for $a\in\mathcal{A},\underline{g}\in\mathcal{G}^{2}$ is normal, using (13), we can see that $\rho$ is normal. ∎

Remark 3.5**.**

Suppose $F$ is a Hilbert (von Neumann) $M_{2}(\mathcal{A})$-$M_{2}(\mathcal{B})$-module, then we can consider $F$ as a Hilbert (von Neumann) $\mathcal{A}$-$M_{2}(\mathcal{B})$-module by considering the left action of $\mathcal{A}$ given by

[TABLE]

Therefore, Proposition 3.4 shows that, if $F$ is a Hilbert (von Neumann) $M_{2}(\mathcal{A})$-$M_{2}(\mathcal{B})$-module, then $F^{(\mathcal{B})}$ is a Hilbert (von Neumann) $\mathcal{A}$-$\mathcal{B}$-module.

Remark 3.6**.**

Let $E\subseteq F$ be a $M_{2}(\mathcal{B})$-submodule of a $M_{2}(\mathcal{B})$-module $F.$ Then

[TABLE]

Theorem 3.7**.**

Let $\mathcal{A}$ be a unital $C^{*}$-algebra and $\mathcal{B}$ be a von Neumann algebra on a Hilbert space $\mathcal{G}.$ For $i=1,2,$ let $\phi_{i}:\mathcal{A}\to\mathcal{B}$ be a CP map with a GNS-representation $(F_{i},y_{i}).$ Suppose $\Phi=\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix}:M_{2}(\mathcal{A})\to M_{2}(\mathcal{B})$ is a block CP map for some CB map $\psi:\mathcal{A}\to\mathcal{B}$ then, there is an adjointable bilinear contraction $T:F_{2}\to F_{1}$ such that $\psi(a)=\langle y_{1},Tay_{2}\rangle$ for all $a\in\mathcal{A}$.

Proof.

Let $(E,x)$ be the (minimal) GNS-construction for $\Phi.$ So, $E$ is a von Neumann $M_{2}(\mathcal{B})$-module and Hilbert $M_{2}(\mathcal{A})$-$M_{2}(\mathcal{B})$-module. Let $\mathbb{E}_{ij}:=\mathbf{1}\otimes E_{ij}$ in $\mathcal{A}\otimes M_{2},$ or $\mathcal{B}\otimes M_{2}$, depending upon the context, where $\{E_{ij}\}$’s are the matrix units in $M_{2}.$ Set $\hat{E}_{i}:=\mathbb{E}_{ii}E\subseteq E,i=1,2.$ Then $\hat{E}_{i}$’s are SOT closed (as $\mathbb{E}_{ii}$’s are projections) $M_{2}(\mathcal{B})$-submodules of $E$ such that $E=\hat{E}_{1}\oplus\hat{E}_{2}.$

Let $x_{i}:=\mathbb{E}_{ii}x\mathbb{E}_{ii}\in\hat{E}_{i},i=1,2.$ Clearly $\langle x_{1},x_{2}\rangle=0.$ Also for $i,j=1,2$ and $i\neq j,$

[TABLE]

and

[TABLE]

Thus

[TABLE]

As $\Phi$ is a block map, for $A\in M_{2}(\mathcal{A}),$ using (15) we have

[TABLE]

where $\langle a,b\rangle_{ij}$ denotes the $(i,j)$th entry of $\langle a,b\rangle\in M_{2}(\mathcal{B}).$

Consider the Hilbert $\mathcal{A}$-$\mathcal{B}$-module and von Neumann $\mathcal{B}$-module $E^{(\mathcal{B})}$ (as described in Remark 3.5), and consider the von Neumann $\mathcal{B}$-modules $\hat{E}_{i}^{(\mathcal{B})},i=1,2.$ Observe that $\hat{E}_{i}$ has a non-degenerate left action of $\mathcal{A}$ given by

[TABLE]

Therefore, Proposition 3.4 shows that $\hat{E}_{i}^{(\mathcal{B})}$ is also a Hilbert $\mathcal{A}$-$\mathcal{B}$-module for $i=1,2.$

We have $E^{(\mathcal{B})}\simeq\hat{E}_{1}^{(\mathcal{B})}\oplus\hat{E}_{2}^{(\mathcal{B})}$ (via $[y]_{E}\mapsto[\mathbb{E}_{11}y]_{\hat{E}_{1}}+[\mathbb{E}_{22}y]_{\hat{E}_{2}}$ for $y\in E$). For $a\in\mathcal{B}$ and $i=1,2$ see that,

[TABLE]

This shows that $(\hat{E}_{i}^{(\mathcal{B})},[x_{i}])$ is a GNS-representation (not necessarily minimal) for $\phi_{i},i=1,2.$ Define $U:\hat{E}_{2}^{(\mathcal{B})}\to\hat{E}_{1}^{(\mathcal{B})}$ by $U[w]=[\mathbb{E}_{12}w]$ for all $w\in\hat{E}_{2}.$ Then, for all $z,w\in\hat{E}_{2},$

[TABLE]

also for $y\in\hat{E}_{1}$ we have $\mathbb{E}_{21}y\in\hat{E}_{2}$ such that

[TABLE]

Therefore $U$ is a unitary from the von Neumann $\mathcal{B}$-module $\hat{E}_{2}^{(\mathcal{B})}$ to the von Neumann $\mathcal{B}$-module $\hat{E}_{1}^{(\mathcal{B})}.$ Now for $a\in\mathcal{A},w\in\hat{E}_{2},$

[TABLE]

Thus $U:\hat{E}_{2}^{(\mathcal{B})}\to\hat{E}_{1}^{(\mathcal{B})}$ is a bilinear (adjointable) unitary between the Hilbert $\mathcal{A}$-$\mathcal{B}$ modules.

Let $\tilde{F_{i}}:=\overline{\operatorname{span}}^{s}\mathcal{A}y_{i}\mathcal{B}\subseteq F_{i}$ and $\tilde{E_{i}}=\overline{\operatorname{span}}^{s}\mathcal{A}[x_{i}]\mathcal{B}\subseteq\hat{E}_{1}^{(\mathcal{B})},$ so that $(\tilde{F_{i}},y_{i})$ and $(\tilde{E_{i}},[x_{i}])$ are minimal GNS-representations for $\phi_{i},i=1,2.$ Therefore, $\tilde{V}_{i}:\tilde{F}_{i}\to\tilde{E}_{i}$ given by

[TABLE]

extends to a bilinear (adjointable) unitary. Let $V_{i}:F_{i}\to\hat{E}_{i}^{(\mathcal{B})}$ be the extension of $\tilde{V_{i}},$ by defining it to be zero on the complement $\tilde{F_{i}}^{\perp}$ of $\tilde{F_{i}}.$ Note that $V_{i}$ is a bilinear partial isometry with initial space $\tilde{F_{i}}$ and final space $\tilde{E_{i}}$ for $i=1,2.$ Take $T:=V_{1}^{*}UV_{2}.$

Now consider, for $a\in\mathcal{A},$

[TABLE]

This completes the proof. ∎

Remark 3.8** (Uniqueness).**

With the hypothesis and notations of Theorem 3.7 let $T,T^{\prime}:F_{2}\to F_{1}$ be any two adjointable bilinear contractions such that $\psi(a)=\langle y_{1},Tay_{2}\rangle=\langle y_{1},T^{\prime}ay_{2}\rangle$ for all $a\in\mathcal{A},$ then

[TABLE]

for $a_{1},a_{2}\in\mathcal{A},b_{1},b_{2}\in\mathcal{B}$ and hence $P_{\tilde{F}_{1}}TP_{\tilde{F}_{2}}=P_{\tilde{F}_{1}}T^{\prime}P_{\tilde{F}_{2}}$ where $P_{\tilde{F}_{i}}:F_{i}\to F_{i}$ is the projection onto $\tilde{F}_{i}.$ This in particular shows that the contraction $T$ in Theorem 3.7 is unique if $F_{i}$’s are minimal GNS-modules.

Corollary 3.9**.**

Let $\mathcal{A}$ be a unital $C^{*}$-algebra. For $i=1,2,$ let $\varphi_{i}:\mathcal{A}\to\mathscr{B}(\mathcal{H})$ be a completely positive map with the minimal Stinespring representation $(\mathcal{K}_{i},\pi_{i},V_{i}).$ Suppose $\Phi:M_{2}(\mathcal{A})\to M_{2}(\mathscr{B}(\mathcal{H})),$ defined by $\Phi=\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix}$ is block CP for some CB map $\psi:\mathcal{A}\to\mathscr{B}(\mathcal{H}),$ then there is a unique contraction $T:\mathcal{K}_{2}\to\mathcal{K}_{1}$ with $\pi_{1}(a)T=T\pi_{2}(a)$ for all $a\in\mathcal{A}$ such that $\psi(a)=V_{1}^{*}T\pi_{2}(a)V_{2}$ for all $a\in\mathcal{A}$.

Proof.

Given that $(\mathcal{K}_{i},\pi_{i},V_{i})$ is a minimal Stinespring representation for $\phi_{i},i=1,2.$ Let $(E_{i},V_{i})$ be the minimal GNS-representation for $\phi_{i},i=1,2$ as explained in Remark 2.6, where $E_{i}=\overline{\operatorname{span}}^{s}\pi_{i}(\mathcal{A})V_{i}\mathscr{B}(\mathcal{H})\subseteq\mathscr{B}(\mathcal{H},\mathcal{K}_{i}).$ By Theorem 3.7, there exists an adjointable bilinear contraction $\hat{T}:E_{2}\to E_{1}$ such that

[TABLE]

As $(\mathcal{K}_{i},\pi_{i},V_{i})$ is the minimal Stinespring representation for $\phi,$ we have $\mathcal{K}_{i}=\overline{\pi_{i}(\mathcal{A})V_{i}\mathcal{H}}.$ Define $T:\mathcal{K}_{2}\to\mathcal{K}_{1}$ by

[TABLE]

Let $h$ be a non-zero vector in $\mathcal{H}.$ As $\hat{T}$ is right $\mathscr{B}(\mathcal{H})$-linear and contraction, we have for $a\in\mathcal{A},h\in\mathcal{H},$

[TABLE]

This implies

[TABLE]

Therefore $T$ is a well-defined contraction. Now as $\hat{T}$ is left $\mathcal{A}$-linear, for all $a,b\in\mathcal{A}$ and $h\in\mathcal{H},$ we have

[TABLE]

Thus $T\pi_{2}(a)=\pi_{1}(a)T,$ for all $a\in\mathcal{A}.$ Now (17) shows that $\psi(a)h=V_{1}^{*}T\pi_{2}(a)V_{2}h$ for all $h\in\mathcal{H}.$ For the uniqueness of $T,$ let $T^{\prime}$ be another contraction such that $T^{\prime}\pi_{2}(a)=\pi_{1}(a)T^{\prime}$ and $\psi(a)=V_{1}^{*}T^{\prime}\pi_{2}(a)V_{2}$ for all $a\in\mathcal{A}.$ Consider for $a,b\in\mathcal{A}$ and $h,g\in\mathcal{H},$

[TABLE]

This proves the uniqueness of $T.$ ∎

Remark 3.10**.**

(i). Corollary 3.9 can be proved directly (without deducing from Theorem 3.7).

(ii). Given two CP maps $\phi_{i}:\mathcal{A}\to\mathscr{B}(\mathcal{H}),i=1,2.$ Let $(\mathcal{K}_{i},\pi_{i},V_{i})$ be the minimal Stinespring representation for $\phi_{i},i=1,2.$ Suppose the block map $\Phi=\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix}$ is CP for some CB map $\psi:\mathcal{A}\to\mathscr{B}(\mathcal{H}).$ Then Furuta in [9, Proposition 6.1] proved that: $\psi$ is non-trivial (non-zero) if and only if there exists a non-zero operator $T:\mathcal{K}_{2}\to\mathcal{K}_{1}$ such that $T\pi_{1}(a)=\pi_{2}(a)T$ for all $a\in\mathcal{A}.$ On the other hand, Corollary 3.9 explicitly tells us the structure of $\psi$ from the minimal Stinespring representations of $\phi_{i}$’s.

(iii). Corollary 3.9 is a generalization of [16, Corollary 2.7] (namely, when $\phi_{1}=\phi_{2}$ in Corollary 3.9, we get the result of Paulsen and Suen [16, Corollary 2.7]).

The following example shows that we cannot replace the von Neumann algebra $\mathcal{B}$ in Theorem 3.7 by an arbitrary $C^{*}$-algebra.

Example 3.11**.**

Let $\mathcal{A}=\mathcal{B}=C([0,1]),$ the commutative unital $C^{*}$-algebra of continuous functions on $[0,1].$ Let $E=C([0,1]).$ It is a Hilbert $\mathcal{A}$-$\mathcal{B}$-module with the natural actions and standard inner product: $\langle f,g\rangle=f^{*}g.$ Let

[TABLE]

Consider the CP map $\Phi:M_{2}(\mathcal{A})\to M_{2}(\mathcal{B})$ defined by

[TABLE]

Note that $\Phi$ is the block CP map $\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix},$ where $\phi_{i},\psi:\mathcal{A}\to\mathcal{B}$ are given by

[TABLE]

Therefore, $(E,h_{i})$ is a GNS-representation for $\phi_{i},i=1,2.$ Let $E_{i}=\overline{\operatorname{span}}\leavevmode\nobreak\ \mathcal{A}h_{i}\mathcal{B}\subseteq E.$ Then $(E_{i},h_{i})$ is the minimal GNS-representation for $\phi_{i},i=1,2.$ Note that

[TABLE]

Now suppose that there exists a bilinear contraction $T:E_{2}\to E_{1}$ such that $\psi(f)=\langle h_{1},Tfh_{2}\rangle$ for all $f\in\mathcal{A}.$ Then $\psi(h_{2})=h_{1}=h_{1}T(h_{2}).$ That is, $t=tT(h_{2})(t)$ for all $t\in[0,1].$ This implies that $T(h_{2})(t)=1$ for all $t\neq 0.$ This is a contradiction to $T(h_{2})\in E_{1}.$

Remark 3.12**.**

Let $\mathcal{A}$ and $\mathcal{B}$ be unital $C^{*}$-algebras and let $\Phi:M_{2}(\mathcal{A})\to M_{2}(\mathcal{B})$ be a block CP map $\Phi=\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix}.$ Suppose $\mathcal{B}$ is a unital subalgebra of $\mathscr{B}(\mathcal{H})$ for some Hilbert space $\mathcal{H}.$ Let $\mathcal{C}$ be the von Neumann algebra $\overline{\mathcal{B}}^{s}.$ Now enlarge the codomain of $\Phi$ to $M_{2}(\mathcal{C}).$ That is, consider the block CP map $\tilde{\Phi}:M_{2}(\mathcal{A})\to M_{2}(\mathcal{C}),$ such that $\tilde{\Phi}(A)=\Phi(A).$

Let $\tilde{\Phi}=\begin{pmatrix}\tilde{\phi}_{1}&\tilde{\psi}\\ \tilde{\psi}^{*}&\tilde{\phi}_{2}\end{pmatrix}.$ Then, by Theorem 3.7 we get a bilinear contraction $\tilde{T}:\tilde{E}_{2}\to\tilde{E}_{1}$ such that $\psi(a)=\tilde{\psi}(a)=\langle x_{1},\tilde{T}ax_{2}\rangle$ for all $a\in\mathcal{A},$ where $(\tilde{E}_{i},x_{i})$ is the GNS-construction for $\tilde{\phi}_{i},i=1,2.$ Note that $(\tilde{E}_{i},x_{i})$ is not a GNS-representation for $\phi_{i},i=1,2$ as $\tilde{E}_{i}$ is an Hilbert $\mathcal{A}$-$\mathcal{C}$-module which need not be an Hilbert $\mathcal{A}$-$\mathcal{B}$ module.

In particular, in Example 3.11 if we enlarge the codomain $\mathcal{B}$ to $\mathcal{C}=\overline{\mathcal{B}}^{s}=L^{\infty}([0,1]),$ then with the above notations, we have $\tilde{E}_{i}=\overline{\operatorname{span}}\leavevmode\nobreak\ \mathcal{A}h_{i}\mathcal{C}=L^{\infty}([0,1]),i=1,2.$ Note also that there exists a bilinear contraction $\tilde{T}:\tilde{E}_{2}\to\tilde{E}_{1}$ given by $\tilde{T}f=f,f\in E_{2}$ such that $\psi(f)=\tilde{\psi}(f)=\langle h_{1},Tfh_{2}\rangle$ for all $f\in\mathcal{A}.$

The following example is a modification of Example 3.11 to get an example of a unital block CP map $\Phi:M_{2}(\mathcal{B})\to\mathcal{M}_{2}(\mathcal{B}).$

Example 3.13**.**

Let $\mathcal{A}$ be the unital $C^{*}$-algebra $C([0,1]).$ Let $\mathcal{B}=\mathcal{A}\oplus\mathcal{A}$ and let $F=\mathcal{A}\oplus\mathcal{A}$ be the Hilbert $\mathcal{B}$-$\mathcal{A}$-module with the module actions and inner product given by

[TABLE]

for $k\in\mathcal{A},\begin{pmatrix}k_{1}\\ k_{2}\end{pmatrix}\in\mathcal{B},\begin{pmatrix}f_{1}\\ f_{2}\end{pmatrix},\begin{pmatrix}g_{1}\\ g_{2}\end{pmatrix}\in F.$ Consider $E=F\oplus F$ as a Hilbert $\mathcal{B}$-$\mathcal{B}$-module with right action

[TABLE]

inner product

[TABLE]

and the left action

[TABLE]

Let $h_{11}(t)=t,h_{12}(t)=\sqrt{1-t^{2}},h_{21}(t)=1,h_{22}(t)=0$ for $t\in[0,1].$ Let

[TABLE]

Let $\Phi:M_{2}(\mathcal{B})\rightarrow M_{2}(\mathcal{B})$ be the block CP map $\Phi=\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix},$ where $\phi_{i},\psi:\mathcal{B}\to\mathcal{B}$ are defined by

[TABLE]

Let $E_{i}=\overline{\operatorname{span}}\mathcal{B}h_{i}\mathcal{B}\subseteq E,i=1,2.$ Then $E_{1}=F_{1}\oplus F_{1}$ with $F_{1}=C_{0}([0,1])\oplus C_{1}([0,1])$ where $C_{j}([0,1])=\{f\in C([0,1]):f(j)=0\}$ for $j=0,1,$ and $E_{2}=F_{2}\oplus F_{2}$ with $F_{2}=C([0,1])\oplus 0.$ Now suppose there exists a bilinear contraction $T:E_{2}\rightarrow E_{1}$ such that $\psi(f)=\langle h_{1},Tfh_{2}\rangle$ for all $f\in\mathcal{B}.$ Then for $f=\begin{pmatrix}h_{21}\\ h_{22}\end{pmatrix}\in\mathcal{B},$

[TABLE]

where $Th_{2}=\begin{pmatrix}l_{11}\\ l_{12}\end{pmatrix}\oplus\begin{pmatrix}l_{21}\\ l_{22}\end{pmatrix}\in E_{1}.$ Therefore $h_{11}=h_{11}l_{11}+h_{12}h_{22}.$ Hence $t=tl_{11}(t)$ for all $t\in[0,1].$ Hence $l_{11}(t)=1$ for $t\neq 0.$ This is a contradiction to the assumption that $Th_{2}\in E_{1}.$ So no such $T$ exists.

We could not get any reasonable answer to the following question.

Problem 3.14**.**

Let $\mathcal{A},\mathcal{B}$ be unital $C^{*}$-algebras and let $p\in\mathcal{A},q\in\mathcal{B}$ be projections. Let $\Phi=\begin{pmatrix}\phi_{1}&\psi\\ \psi^{*}&\phi_{2}\end{pmatrix}$ be a block CP map from $\mathcal{A}$ to $\mathcal{B}$ with respect to $p$ and $q.$ Let $(E_{i},\xi_{i})$ be GNS-representation of $\phi_{i},i=1,2.$ Can we prove a theorem similar to Theorem 3.7? In other words what is the structure of $\psi$ in terms of $(E_{i},\xi_{i})?$

4. Semigroups of block CP maps

4.1. Structure of block quantum dynamical semigroups

In this section, we shall prove a structure theorem similar to (or using) Theorem 3.7 for semigroups of block CP maps. We shall start with a few basic examples of semigroups of block CP maps, which are of interest.

Example 4.1**.**

Let $\mathcal{H}$ be a Hilbert space. Let $(\theta_{t})_{t\geq 0}$ be an $E_{0}$-semigroup on $\mathscr{B}(\mathcal{H}).$ Let $(U_{t})_{t\geq 0}$ be a family of unitaries in $\mathscr{B}(\mathcal{H})$ forming left cocycle for $\theta$, that is, $U_{0}=I,U_{s+t}=U_{s}\theta_{s}(U_{t})$, $t\mapsto U_{t}$ continuous in SOT. Let $\psi_{t}(X)=U_{t}\theta_{t}(X)U_{t}^{*}$ for $X\in\mathscr{B}(\mathcal{H}).$ Then $(\psi_{t})_{t\geq 0}$ is an $E_{0}$-semigroup, cocycle conjugate to $(\theta_{t})_{t\geq 0}.$ Define $\tau_{t}:\mathscr{B}(\mathcal{H}\oplus\mathcal{H})\to\mathscr{B}(\mathcal{H}\oplus\mathcal{H})$ by

[TABLE]

Then clearly $(\tau_{t})_{t\geq 0}$ is a block $E_{0}$-semigroup.

Example 4.2**.**

Let $(a_{t})_{t\geq 0}$ and $(b_{t})_{t\geq 0}$ be semigroups on a $C^{*}$-algebra $\mathcal{B}$ and let $(\phi_{t}^{i})_{t\geq 0},i=1,2,$ be two QDS on $\mathcal{B}$ such that $\phi_{t}^{1}(\cdot)-a_{t}(\cdot)a_{t}^{*}$ and $\phi_{t}^{2}(\cdot)-b_{t}(\cdot)b_{t}^{*}$ are CP maps for all $t\geq 0.$ Define $\tau_{t}:M_{2}(\mathcal{B})\to M_{2}(\mathcal{B})$ by

[TABLE]

Then $\tau_{t}$ is CP, for all $t\geq 0,$ as

[TABLE]

Clearly $(\tau_{t})_{t\geq 0}$ is a block QDS.

Lemma 4.3**.**

Let $\mathcal{B}$ be a unital $C^{*}$-algebra. Given two inclusion systems $(E^{i},\beta^{i},\xi^{\odot i})$ associated to a pair of CP semigroups $\phi^{i}=(\phi_{t}^{i})_{t\geq 0},i=1,2$ on $\mathcal{B}$ and a contractive morphism $T=(T_{t}):E^{2}\to E^{1},$ there is a block CP semigroup $\Phi=(\Phi_{t})_{t\geq 0}$ on $M_{2}(\mathcal{B})$ such that $\Phi_{t}=\begin{pmatrix}\phi_{t}^{1}&\psi_{t}\\ \psi_{t}^{*}&\phi_{t}^{2}\end{pmatrix}$ and $\psi_{t}(a)=\langle\xi_{t}^{1},T_{t}(a\xi_{t}^{2})\rangle.$

Proof.

Define $\Phi_{t}:M_{2}(\mathcal{B})\to M_{2}(\mathcal{B})$ as the block maps $\Phi_{t}=\begin{pmatrix}\phi_{t}^{1}&\psi_{t}\\ \psi_{t}^{*}&\phi_{t}^{2}\end{pmatrix},$ where $\psi_{t}(b)=\langle\xi_{t}^{1},T_{t}(b\xi_{t}^{2})\rangle.$ Then, as $T_{t}:E_{t}^{2}\to E_{t}^{1}$ is an adjointable bilinear contraction, $\Phi_{t}$ is CP for all $t\geq 0$ (see the proof of Lemma 3.2). Now we shall show that $\Phi=(\Phi_{t})_{t\geq 0}$ is a semigroup on $M_{2}(\mathcal{B}).$ We have

[TABLE]

It is clear from this, that to show $\Phi$ is a semigroup, it is enough to show that $(\psi_{t})_{t\geq 0}$ is a semigroup. Now as $T$ is a morphism it is easy to see that $(\psi_{t})_{t\geq 0}$ is a semigroup. ∎

In the following we have the converse of Lemma 4.3, when $\mathcal{B}$ is a von Neumann algebra. Example 3.11 says that we cannot take $\mathcal{B}$ as an arbitrary $C^{*}$-algebra.

Theorem 4.4**.**

Let $\mathcal{B}$ be a von Neumann algebra. Let $\Phi=(\Phi_{t})_{t\geq 0}$ be a semigroup of block normal CP maps on $M_{2}(\mathcal{B})$ with $\Phi_{t}=\begin{pmatrix}\phi_{t}^{1}&\psi_{t}\\ \psi_{t}^{*}&\phi_{t}^{2}\end{pmatrix}.$ Then, there are inclusion systems $(E^{i},\beta^{i},\xi^{\odot i}),i=1,2$ associated to $\phi^{i}$ (canonically arising from the inclusion system associated to $\Phi$) and a unique contractive (weak) morphism $T=(T_{t}):E^{2}\to E^{1}$ such that $\psi_{t}(a)=\langle\xi_{t}^{1},T_{t}a\xi_{t}^{2}\rangle$ for all $a\in\mathcal{B},t\geq 0.$

Proof.

We shall prove this extending the same ideas of the proof of Theorem 3.7 to the semigroup level. Let $(E=(E_{t}),\beta=(\beta_{t,s}),\eta^{\odot}=(\eta_{t}))$ be the inclusion system associated to $\Phi.$ Note that $E_{t}$’s are von Neumann $M_{2}(\mathcal{B})$-$M_{2}(\mathcal{B})$-modules. Let $\mathbb{E}_{ij}:=\mathbf{1}\otimes E_{ij}\in\mathcal{B}\otimes M_{2},$ where $E_{ij}$’s are the matrix units in $M_{2}.$ Let $\hat{E}_{t}^{i}:=\mathbb{E}_{ii}E_{t}\subseteq E_{t},i=1,2.$ Then $\hat{E}_{t}^{i}$’s are SOT closed $M_{2}(\mathcal{B})$-submodules of $E_{t}$ such that $E_{t}=\hat{E}_{t}^{1}\oplus\hat{E}_{t}^{2}$ for all $t\geq 0.$ Let $\eta_{t}^{i}:=\mathbb{E}_{ii}\eta_{t}\mathbb{E}_{ii}\in\hat{E}_{t}^{i},$ $i=1,2.$ Then we have (as in the proof of Theorem 3.7)

[TABLE]

As $\beta_{t,s}:E_{t+s}\to E_{t}\odot E_{s}$ are the canonical maps $\eta_{t+s}\mapsto\eta_{t}\odot\eta_{s},$ using (21) we have,

[TABLE]

Consider the von Neumann $\mathcal{B}$-$\mathcal{B}$-modules $E_{t}^{(\mathcal{B})}$ (as described in Remark 3.5) and the von Neumann $\mathcal{B}$-modules $\hat{E}_{t}^{i(\mathcal{B})}$ (see Proposition 3.3). Notice that $\hat{E}_{t}^{i(\mathcal{B})}$ is also a von Neumann $\mathcal{B}$-$\mathcal{B}$-module for $i=1,2$ with the left action of $\mathcal{B}$ given by

[TABLE]

Then, we have $E_{t}^{(\mathcal{B})}\simeq\hat{E}_{t}^{1(\mathcal{B})}\oplus\hat{E}_{t}^{2(\mathcal{B})}$ (as two-sided von Neumann modules) for all $t\geq 0$ (as in the proof of Theorem 3.7). Let $\xi_{t}^{i}=[\eta_{t}^{i}]\in\hat{E}_{t}^{i(\mathcal{B})},i=1,2.$ Then for $a\in\mathcal{B},i=1,2,$ we have

[TABLE]

Therefore, $(\hat{E}_{t}^{i(\mathcal{B})},\xi_{t}^{i})$ is a GNS-representation (not necessarily minimal) for $\phi_{t}^{i},i=1,2.$ Let $E_{t}^{i}=\overline{\operatorname{span}}^{s}\mathcal{B}\xi_{t}^{i}\mathcal{B}\subseteq\hat{E}_{t}^{i(\mathcal{B})}$ be the minimal GNS-module for $\phi_{t}^{i}$ for $i=1,2.$ Let $\beta_{t,s}^{i}:E_{t+s}^{i}\to E_{t}^{i}\odot E_{s}^{i}$ be the canonical maps (as in Remark 2.15) given by

[TABLE]

so that $(E^{i}=(E_{t}^{i}),\beta^{i}=(\beta_{t,s}^{i}),\xi^{\odot i}=(\xi^{i}_{t}))$ is the inclusion system associated to $\phi^{i},i=1,2.$ (Equation (22) shows that, we get the inclusion systems associated to $\phi^{i}$’s in a canonical way from the inclusion system associated to $\Phi$.)

Let $V_{t}^{i}:E_{t}^{i}\to\hat{E}_{t}^{i(\mathcal{B})}$ be the inclusion maps and let $U_{t}:\hat{E}_{t}^{2(\mathcal{B})}\to\hat{E}_{t}^{1(\mathcal{B})}$ be defined by

[TABLE]

Then $V_{t}^{i}$’s are adjointable, bilinear isometries and $U_{t}$’s are bilinear unitaries (as in the proof of Theorem 3.7). Take $T_{t}:=V_{t}^{1*}U_{t}V_{t}^{2}.$ Then $T_{t}:E_{t}^{2}\to E_{t}^{1}$ is an adjointable, bilinear contraction such that for $a\in\mathcal{B},$

[TABLE]

For $a,b,c,d\in\mathcal{B},$

[TABLE]

shows that $T:=(T_{t})_{t\geq 0}$ is a morphism of inclusion systems from $(E^{2},\beta^{2})$ to $(E^{1},\beta^{1}).$

To prove the uniqueness of $T$, let $T^{\prime}=(T^{\prime}_{t})_{t\geq 0}$ be another morphism of inclusion systems from $(E^{2},\beta^{2},\xi^{\odot 2})$ to $(E^{1},\beta^{1},\xi^{\odot 1})$ such that $\psi_{t}(a)=\langle\xi_{t}^{1},T^{\prime}_{t}(a\xi_{t}^{2})\rangle$ for all $a\in\mathcal{B},t\geq 0,$ then

[TABLE]

for $a_{1},a_{2},b_{1},b_{2}\in\mathcal{B}$ and hence $T_{t}=T^{\prime}_{t}$ for all $t\geq 0.$ ∎

Example 4.5**.**

Let $\mathcal{B}$ be a von Neumann algebra. Let $E$ be a von Neumann $M_{2}(\mathcal{B})$-$M_{2}(\mathcal{B})$-module. Take $\beta=\begin{pmatrix}\beta_{1}&0\\ 0&\beta_{2}\end{pmatrix}$ in $M_{2}(\mathcal{B})$ and $\zeta\in E$ such that $\zeta=\mathbb{E}_{11}\zeta\mathbb{E}_{11}+\mathbb{E}_{22}\zeta\mathbb{E}_{22},$ where $\mathbb{E}_{ij}=\mathbf{1}\otimes E_{ij}\in\mathcal{B}\otimes M_{2}$ and $\{E_{ij}\}_{i,j=1}^{2}$ are the matrix units in $M_{2}.$

Let $\xi^{\odot}(\beta,\zeta)=(\xi_{t}{(\beta,\zeta)})_{t\in\mathbb{R}_{+}}\in\mathbb{I}\!\!\Gamma^{\odot}(E),$ the product system of time ordered Fock module over $E,$ where the component $\xi_{t}^{n}$ of $\xi_{t}{(\beta,\zeta)}\in\mathbb{I}\!\!\Gamma_{t}(E)$ in the $n$-particle $(n>0)$ sector is defined as

[TABLE]

and $\xi_{t}^{0}=e^{t\beta}.$ Then it follows from [11, Theorem 3] that, $\xi^{\odot}(\beta,\zeta)$ is a unit for the product system $\mathbb{I}\!\!\Gamma^{\odot}(E).$ Further if $\Phi_{t}^{(\beta,\zeta)}:M_{2}(\mathcal{B})\to M_{2}(\mathcal{B})$ is defined by

[TABLE]

then $\Phi:=(\Phi_{t})_{t\geq 0}$ is a uniformly continuous CP-semigroup on $M_{2}(\mathcal{B}),$ with bounded generator

[TABLE]

Let $\zeta_{i}=\mathbb{E}_{ii}\zeta\mathbb{E}_{ii},i=1,2,$ then $\zeta=\zeta_{1}+\zeta_{2},\langle\zeta_{1},\zeta_{2}\rangle=0.$ Let $\tau:M_{2}(\mathcal{B})\to M_{2}(\mathcal{B})$ be defined by $\tau(A)=\langle\zeta,A\zeta\rangle,A\in M_{2}(\mathcal{B}),$ then $\tau$ is a block CP map, say $\tau=\begin{pmatrix}\tau_{11}&\tau_{12}\\ \tau_{12}^{*}&\tau_{22}\end{pmatrix}.$

Note that $(E^{(\mathcal{B})},[\zeta_{i}])$ is a GNS-representation for $\tau_{ii},i=1,2,$ where $E^{(\mathcal{B})}$ is the von Neumann $\mathcal{B}$-$\mathcal{B}$-module as described in Remark 3.5.

Let $E_{i}=\overline{\operatorname{span}}^{s}\mathcal{B}[\zeta_{i}]\mathcal{B}\subseteq E^{(\mathcal{B})}$ be the minimal GNS-representation for $\tau_{ii},i=1,2$ and let $T:E_{2}\to E_{1}$ be the unique bilinear, adjointable contraction such that $\tau_{12}(a)=\langle[\zeta_{1}],Ta[\zeta_{2}]\rangle$ as given in Theorem 3.7. Therefore, we have

[TABLE]

and hence

[TABLE]

where

[TABLE]

and

[TABLE]

[TABLE]

for $a\in\mathcal{B}.$ Therefore, for $A=\begin{pmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}\in M_{2}(\mathcal{B}),$

[TABLE]

Now note that the inclusion system $(E^{i}=(E_{t}^{i}),\xi^{i}=(\xi_{t}^{i}(\beta_{i},[\zeta_{i}])))$ associated to $\phi^{i}=(e^{tL_{ii}^{(\beta_{i},[\zeta_{i}])}})_{t\geq 0}$ is a subsystem of the product system of time-ordered Fock module $\mathbb{I}\!\!\Gamma^{\odot}(E_{i})$ over $E_{i},i=1,2.$

Let $w=(w_{t})_{t\geq 0}$ be the contractive morphism from $(E^{2},\xi^{2})$ to $(E^{1},\xi^{1})$ such that

[TABLE]

As any morphism maps a unit to a unit we have

[TABLE]

for some $\gamma_{w},\eta_{w}:\mathcal{B}\times E_{2}\to\mathcal{B}\times E_{1}.$ Hence from (28) and (29) we have

[TABLE]

Now by differentiating (28), we get

[TABLE]

Therefore as (27)=(30) we have $\gamma_{w}(\beta_{2},[\zeta_{2}])=\beta_{2}$ and $\eta_{w}(\beta_{2},[\zeta_{2}])=T[\zeta_{2}].$ Thus, the unique morphism $(w_{t})$ is given by

[TABLE]

4.2. $E_{0}$-dilation of block quantum Markov semigroups

In this subsection we shall prove that if we have a block QMS on a unital $C^{*}$-algebra then the $E_{0}$-dilation constructed in [8] is also a semigroup of block maps.

Let $\mathcal{B}$ be a unital $C^{*}$-algebra. Let $p\in\mathcal{B}$ be a projection. Denote $p^{\prime}=\mathbf{1}-p.$ Let $\Phi=(\Phi_{t})_{t\geq 0}$ be a block QMS on $\mathcal{B}$ with respect to $p.$

(We have some changes in the notations from [8]: $\mathcal{E}_{t}\leadsto E_{t},E_{t}\leadsto\mathcal{E}_{t},E\leadsto\mathcal{E}$)

Let $(E=(E_{t}),\xi^{\odot}=(\xi_{t}))$ be the inclusion system associated to $\Phi.$ Recall from [8, Sections 4, 5] that

[TABLE]

That is, we have a $\mathcal{B}$-module $\mathcal{E}$ with $\mathcal{E}\simeq\mathcal{E}\odot\mathcal{E}_{t},$ a representation $j_{0}:\mathcal{B}\to\mathscr{B}^{a}(\mathcal{E})$ ($b\mapsto\ket{\xi}b\bra{\xi}$) and endomorphisms $\vartheta_{t}:\mathcal{B}^{a}(\mathcal{E})\to\mathcal{B}^{a}(\mathcal{E})$ defined by $\vartheta_{t}(a)=a\odot\operatorname{id}_{\mathcal{E}_{t}}$ such that $(\vartheta_{t})_{t\geq 0}$ is an $E_{0}$-dilation of $(\Phi_{t})_{t\geq 0}.$ Moreover, we have the Markov property

[TABLE]

This implies that $j_{0}(\mathbf{1})\vartheta_{t}(j_{0}(\mathbf{1}))j_{0}(\mathbf{1})=j_{0}(\Phi_{t}(\mathbf{1}))=j_{0}(\mathbf{1}).$ Since $j_{0}(\mathbf{1})$ is a projection, we have $j_{0}(\mathbf{1})\leq\vartheta_{t}(j_{0}(\mathbf{1}))$ and hence $(\vartheta_{t}(j_{0}(\mathbf{1})))_{t\geq 0}$ is an increasing family of projections. Hence it converges in SOT. Now if $k_{s}:\mathcal{E}_{s}\to\mathcal{E}$ are the canonical maps ($x_{s}\mapsto\xi\odot x_{s}$) then

[TABLE]

Hence $\vartheta_{t}(j_{0}(\mathbf{1}))(\mathcal{E}_{t})=\vartheta_{t}(j_{0}(\mathbf{1}))(\xi\odot\mathcal{E}_{t})=(\outerproduct{\xi}{\xi}\odot\operatorname{id}_{\mathcal{E}_{t}})(\xi\odot\mathcal{E}_{t})=\xi\odot\mathcal{E}_{t}$ shows that $\vartheta_{t}(j_{0}(\mathbf{1}))_{t\geq 0}$ is converging in SOT to $\operatorname{id}_{\mathcal{E}},$ the identity on $\mathcal{E}.$

Now for $q=p$ or $p^{\prime},$ consider $\vartheta_{t}(j_{0}(q))=\vartheta_{t}(|\xi\rangle q\langle\xi|).$ Note that since $\Phi$ is a unital block semigroup $\Phi_{t}(q)=q$ for $q=p,p^{\prime}.$ Hence by the Markov property (31) we have

[TABLE]

Note that $j_{0}(\mathbf{1})=j_{0}(p)+j_{0}(p^{\prime})$ and $j_{0}(p)j_{0}(p^{\prime})=j_{0}(p^{\prime})j_{0}(p)=0.$ Hence multiplying by $j_{0}(q)$ on both sides of Equation (33) we get

[TABLE]

Since $j_{0}(q)$ is a projection, we have $j_{0}(q)\leq\vartheta_{t}(j_{0}(q))$ for all $t,$ hence $\vartheta_{s}(j_{0}(q))\leq\vartheta_{t}(j_{0}(q))$ for $s\leq t.$ Therefore $(\vartheta_{t}(j_{0}(q)))_{t\geq 0}$ is an increasing family of projections in $\mathcal{B}^{a}(\mathcal{E}).$ Say $(\vartheta_{t}(j_{0}(p)))_{t\geq 0}$ converges to $P.$ Then as $(\vartheta_{t}(j_{0}(\mathbf{1}))_{t\geq 0}$ converges to $\operatorname{id}_{\mathcal{E}},$ $(\vartheta_{t}(j_{0}(p^{\prime})))_{t\geq 0}$ will converge to $P^{\prime}=\operatorname{id}_{\mathcal{E}}-P.$ Note that we have $PP^{\prime}=0$ and

[TABLE]

Thus, we have $\mathcal{E}=\mathcal{E}^{(1)}\oplus\mathcal{E}^{(2)}$ where $\mathcal{E}^{(1)}=P(\mathcal{E})$ and $\mathcal{E}^{(2)}=P^{\prime}(\mathcal{E}).$

Lemma 4.6**.**

$P(\mathcal{E}_{t})=\vartheta_{t}(j_{0}(p))(\mathcal{E}_{t})$* and $P^{\prime}(\mathcal{E}_{t})=\vartheta_{t}(j_{0}(p^{\prime}))(\mathcal{E}_{t})$ for all $t\geq 0.$ *

Proof.

Fix $t\geq 0.$ It is enough to prove for $q=p,p^{\prime}$ that $\vartheta_{s}(j_{0}(q))(x)=\vartheta_{t}(j_{0}(q))(x)$ if $x\in\mathcal{E}_{t},s\geq t.$ Note that (since $\langle\xi^{t},q\xi^{t}\rangle=\Phi_{t}(q)=q$ for $q=p$ or $p^{\prime}$) we have

[TABLE]

This implies that $p\xi^{t}=\xi^{t}p=p\xi^{t}p.$ Similarly we have $p^{\prime}\xi^{t}=\xi^{t}p^{\prime}=p^{\prime}\xi^{t}p^{\prime}.$

Let $q=p$ or $p^{\prime}$ and let $s\geq t.$ If $x\in\mathcal{E}_{t},$ then $\xi^{s-t}\odot x\in\mathcal{E}_{s}$ and

[TABLE]

∎

We have from [8, Theorem 5.4] that

[TABLE]

Now we shall prove a similar result for $\mathcal{E}^{(i)}$’s by recalling the proof of this result. It is important to note that we are not getting something like $\mathcal{E}^{(i)}=\mathcal{E}^{(i)}\odot\mathcal{E}_{t}^{(i)}$, and we have not even bothered to define $\mathcal{E}_{t}^{(i)}$.

Lemma 4.7**.**

$\mathcal{E}^{(i)}\simeq\mathcal{E}^{(i)}\odot\mathcal{E}_{t},$* for $i=1,2,t\geq 0.$*

Proof.

Let $k_{t}:\mathcal{E}_{t}\to\mathcal{E}$ be the canonical maps (isometries). Then $u_{t}:\mathcal{E}\odot\mathcal{E}_{t}\to\mathcal{E}$ defined by

[TABLE]

for $x_{s}\in\mathcal{E}_{s},y_{t}\in\mathcal{E}_{t},$ is a unitary ([8, Theorem 5.4]). Hence, we have $\mathcal{E}\simeq\mathcal{E}\odot\mathcal{E}_{t}.$ Since $\mathcal{E}=\mathcal{E}^{(1)}\oplus\mathcal{E}^{(2)},$ we have, $\mathcal{E}\simeq\mathcal{E}^{(1)}\odot\mathcal{E}_{t}\oplus\mathcal{E}^{(2)}\odot\mathcal{E}_{t}.$

We shall prove that, the restriction of this unitary $u_{t}$ to $\mathcal{E}^{(i)}\odot\mathcal{E}_{t}$ is a unitary from $\mathcal{E}^{(i)}\odot\mathcal{E}_{t}$ onto $\mathcal{E}^{(i)}.$ It is enough to prove that $u_{t}(\mathcal{E}^{(i)}\odot\mathcal{E}_{t})\subseteq\mathcal{E}^{(i)}.$ To prove this, (from (32), (34) and Lemma 4.6) it is sufficient to prove that $u_{t}(\vartheta_{s}(j_{0}(p))k_{s}(\mathcal{E}_{s})\odot\mathcal{E}_{t})\subseteq\mathcal{E}^{(1)}$ and $u_{t}(\vartheta_{s}(j_{0}(p^{\prime}))k_{s}(\mathcal{E}_{s})\odot\mathcal{E}_{t})\subseteq\mathcal{E}^{(2)}.$ To prove this consider for $q=p$ or $p^{\prime}$ and $x_{s}\in\mathcal{E}_{s}$

[TABLE]

which is in $\mathcal{E}^{(1)}$ if $q=p$ and is in $\mathcal{E}^{(2)}$ if $q=p^{\prime}.$ ∎

Theorem 4.8**.**

The $E_{0}$-dilation $\vartheta=(\vartheta_{t})_{t\geq 0}$ of $\Phi$ is a semigroup of block maps with respect to the projection $P$ defined above.

Proof.

As $\mathcal{E}=\mathcal{E}^{(1)}\oplus\mathcal{E}^{(2)},$ we have

[TABLE]

For any $i,j\in\{1,2\},$ let $a\in\mathscr{B}^{a}(\mathcal{E}^{(i)},\mathcal{E}^{(j)}),$ then

[TABLE]

Therefore $\vartheta_{t}$ acts block-wise. ∎

5. Lifting of morphisms

In this section we will show that any (weak) morphism between two inclusion systems of von Neumann $\mathcal{B}$-$\mathcal{B}$-modules can be always lifted as a morphism between the product systems generated by them.

We shall introduce some notations and results from [8] and [7]. For all $t>0$ we define

[TABLE]

and for ${\mathfrak{s}}=(s_{m},\dots,s_{1})\in\mathbb{J}_{s}$ and ${\mathfrak{t}}=(t_{n},\dots,t_{1})\in\mathbb{J}_{t}$ we define the joint tuple ${\mathfrak{s}}\smile{\mathfrak{t}}\in\mathbb{J}_{s+t}$ by

[TABLE]

We have a partial order $``\geq"$ on $\mathbb{J}_{t}$ as follows: ${\mathfrak{t}}\geq{\mathfrak{s}}=(s_{m},\dots,s_{1}),$ if for each $j$ $(1\leq j\leq m)$ there are (unique) ${\mathfrak{s}}_{j}\in\mathbb{J}_{s_{j}}$ such that ${\mathfrak{t}}={\mathfrak{s}}_{m}\smile\dots\smile{\mathfrak{s}}_{1}$ (In this case we also write ${\mathfrak{s}}\leq{\mathfrak{t}},$ to mean ${\mathfrak{t}}\geq{\mathfrak{s}}$).

For $t=0$ we extend the definition of $\mathbb{J}_{t}$ as $\mathbb{J}_{0}=\{()\},$ where $()$ is the empty tuple. Also for ${\mathfrak{t}}\in\mathbb{J}_{t}$ we put ${\mathfrak{t}}\smile()={\mathfrak{t}}=()\smile{\mathfrak{t}}.$

Now we will describe the construction of product system generated by an inclusion system of von Neumann $\mathcal{B}$-$\mathcal{B}$-modules using the inductive limits. (This construction holds also for Hilbert $\mathcal{B}$-$\mathcal{B}$-modules along the same lines, but as we are going to prove the lifting theorem only for von Neumann $\mathcal{B}$-$\mathcal{B}$-modules, we confine ourselves to von Neumann modules).

Let $(E=(E_{t})_{t\in\mathbb{T}},\beta=(\beta_{s,t})_{s,t\in\mathbb{T}})$ be an inclusion system of von Neumann $\mathcal{B}$-$\mathcal{B}$-modules. Fix $t\in\mathbb{T}.$ Let $E_{\mathfrak{t}}:=E_{t_{n}}\odot\cdots\odot E_{t_{1}}$ for ${\mathfrak{t}}=(t_{n},\dots,t_{1})\in\mathbb{J}_{t}.$ For all ${\mathfrak{t}}=(t_{n},\dots,t_{1})\in\mathbb{J}_{t}$ we define $\beta_{{\mathfrak{t}}(t)}:E_{t}\to E_{\mathfrak{t}}$ by

[TABLE]

and for ${\mathfrak{t}}=(t_{n},\dots,t_{1})={\mathfrak{s}}_{m}\smile\dots\smile{\mathfrak{s}}_{1}\geq{\mathfrak{s}}=(s_{m},\dots,s_{1})$ with $|{\mathfrak{s}}_{j}|=s_{j},$ we define $\beta_{{\mathfrak{t}}{\mathfrak{s}}}:E_{\mathfrak{s}}\to E_{\mathfrak{t}}$ by

[TABLE]

Then it is clear from the definitions that $\beta_{{\mathfrak{t}}{\mathfrak{s}}},{\mathfrak{t}}\geq{\mathfrak{s}}$ are bilinear isometries and $\beta_{{\mathfrak{t}}{\mathfrak{s}}}\beta_{{\mathfrak{s}}\mathfrak{r}}=\beta_{{\mathfrak{t}}\mathfrak{r}}$ for ${\mathfrak{t}}\geq{\mathfrak{s}}\geq\mathfrak{r}.$ That is, the family $(E_{\mathfrak{t}})_{{\mathfrak{t}}\in\mathbb{J}_{t}}$ with $(\beta_{{\mathfrak{t}}{\mathfrak{s}}})_{{\mathfrak{s}}\leq{\mathfrak{t}}}$ is an inductive system of von Neumann $\mathcal{B}$-$\mathcal{B}$-modules. Hence the inductive limit $\mathcal{E}_{t}=\underset{{\mathfrak{t}}\in\mathbb{J}_{t}}{\lim\operatorname{ind}}\leavevmode\nobreak\ E_{\mathfrak{t}}$ is also a von Neumann $\mathcal{B}$-$\mathcal{B}$-module and the canonical mappings $i_{\mathfrak{t}}:E_{\mathfrak{t}}\to\mathcal{E}_{t}$ are bilinear isometries (cf. [8, Proposition 4.3]).

For ${\mathfrak{s}}\in\mathbb{J}_{s},{\mathfrak{t}}\in\mathbb{J}_{t}$ it is clear that $E_{\mathfrak{s}}\odot E_{\mathfrak{t}}=E_{{\mathfrak{s}}\smile{\mathfrak{t}}}.$ Using this observation we define $B_{st}:\mathcal{E}_{s}\odot\mathcal{E}_{t}\to\mathcal{E}_{s+t}$ by

[TABLE]

Then $(\mathcal{E}=(\mathcal{E}_{t})_{t\in\mathbb{T}},B=(B_{st})_{s,t\in\mathbb{T}})$ forms a product system (cf. Bhat and Skeide [8, Theorem 4.8 and page 41]).

Definition 5.1**.**

Given an inclusion system $(E,\beta),$ the product system $(\mathcal{E},B)$ described above is called the product system generated by the inclusion system $(E,\beta).$

We recall the following: Let $\mathcal{B}$ be a von Neumann algebra on a Hilbert space $\mathcal{G}.$ Let $E$ be a von Neumann $\mathcal{B}$-module. Then $\mathcal{H}=E\odot\mathcal{G}$ is a Hilbert space such that $E\subseteq\mathscr{B}(\mathcal{G},\mathcal{H})$ via $E\ni x\mapsto L_{x}\in\mathscr{B}(\mathcal{G},\mathcal{H}),$ where $L_{x}:\mathcal{G}\to\mathcal{H}$ is defined by $L_{x}(g)=x\odot g$ for $g\in\mathcal{G}.$ Note that $E$ is strongly closed in $\mathscr{B}(\mathcal{G},\mathcal{H}).$ Sometimes we write $xg$ instead of $x\odot g$ with the above identification in mind.

Remark 5.2**.**

Let $(\mathcal{E},B)$ be the product system generated by the inclusion system $(E,\beta)$ on a von Neumann algebra $\mathcal{B}\subseteq\mathscr{B}(\mathcal{G}).$ Let $i_{\mathfrak{t}}:E_{\mathfrak{t}}\to\mathcal{E}_{t},{\mathfrak{t}}\in\mathbb{J}_{t}$ be the canonical bilinear isometries. Then $i_{\mathfrak{t}}i_{\mathfrak{t}}^{*}$ increases to identity in strong operator topology, that is, for all $x\in\mathcal{E}_{t}$ and $g\in\mathcal{G},$ we have

[TABLE]

Now we shall prove the lifting theorem almost along the same lines of the proof of [7, Theorem 11]

Theorem 5.3**.**

Let $\mathcal{B}$ be a von Neumann algebra on a Hilbert space $\mathcal{G}.$ Let $(E,\beta)$ and $(F,\gamma)$ be two inclusion systems of von Neumann $\mathcal{B}$-$\mathcal{B}$-modules generating two product systems $(\mathcal{E},B),(\mathcal{F},C)$ respectively. Let $i,j$ be their respective inclusion maps. Suppose $T:(E,\beta)\to(F,\gamma)$ is a (weak) morphism then there exists a unique morphism $\hat{T}:(\mathcal{E},B)\to(\mathcal{F},C)$ such that $T_{s}=j_{s}^{*}\hat{T}_{s}i_{s}$ for all $s\in\mathbb{T}.$

Proof.

Given that $T:(E,\beta)\to(F,\gamma)$ is a morphism. Let $k$ be such that $\norm{T_{s}}\leq e^{ks}$ for all $s\in\mathbb{T}.$ For ${\mathfrak{s}}=(s_{n},...,s_{1})\in\mathbb{J}_{s},$ define $T_{\mathfrak{s}}:E_{\mathfrak{s}}\to F_{\mathfrak{s}}$ by $T_{\mathfrak{s}}=T_{s_{n}}\odot\cdots\odot T_{s_{1}}.$ Let $i_{\mathfrak{s}}:E_{\mathfrak{s}}\to\mathcal{E}_{s}$ and $j_{\mathfrak{s}}:F_{\mathfrak{s}}\to\mathcal{F}_{s}$ be the canonical bilinear isometries. Then for ${\mathfrak{s}}\leq{\mathfrak{t}}$ in $\mathbb{J}_{s}$ we have

[TABLE]

Consider for ${\mathfrak{s}}\in\mathbb{J}_{s},\Phi_{\mathfrak{s}}:=j_{\mathfrak{s}}T_{\mathfrak{s}}i_{\mathfrak{s}}^{*}.$ Set $P_{\mathfrak{s}}=j_{\mathfrak{s}}j_{\mathfrak{s}}^{*}$ and $Q_{\mathfrak{s}}=i_{\mathfrak{s}}i_{\mathfrak{s}}^{*}.$ Then by Remark 5.2 $(P_{\mathfrak{s}})_{{\mathfrak{s}}\in\mathbb{J}_{s}}$ and $(Q_{\mathfrak{s}})_{{\mathfrak{s}}\in\mathbb{J}_{s}}$ are families of increasing projections. Now for $\mathfrak{r}\leq{\mathfrak{s}},$ $i_{\mathfrak{r}}=i_{\mathfrak{s}}\beta_{{\mathfrak{s}}\mathfrak{r}},$ $j_{\mathfrak{r}}=j_{\mathfrak{s}}\gamma_{{\mathfrak{s}}\mathfrak{r}}$ implies that $\beta_{{\mathfrak{s}}\mathfrak{r}}=i_{\mathfrak{s}}^{*}i_{\mathfrak{r}},\gamma_{{\mathfrak{s}}\mathfrak{r}}=j_{\mathfrak{s}}^{*}j_{\mathfrak{r}},$ hence it follows from (39) that $P_{\mathfrak{r}}\Phi_{\mathfrak{s}}Q_{\mathfrak{r}}=\Phi_{\mathfrak{r}}.$

For all $s\in\mathbb{T},$ $\mathcal{E}_{s}\subseteq\mathscr{B}(\mathcal{G},\mathcal{E}_{s}\odot\mathcal{G})$ and $\mathcal{F}_{s}\subseteq\mathscr{B}(\mathcal{G},\mathcal{F}_{s}\odot\mathcal{G}).$ Fix $s\in\mathbb{T}.$ Let $x\in\mathcal{E}_{s},g\in\mathcal{G}$ and let $\epsilon>0.$ Using (38) choose $\mathfrak{r}_{0}\in\mathbb{J}_{s}$ such that

[TABLE]

Then, for any ${\mathfrak{s}}\in\mathbb{J}_{s},$ we have

[TABLE]

Let ${\mathfrak{t}}\geq{\mathfrak{s}}\geq\mathfrak{r}_{0}\in\mathbb{J}_{s}.$ As $(P_{\mathfrak{s}})_{{\mathfrak{s}}\in\mathbb{J}_{s}}$ and $(Q_{\mathfrak{s}})_{{\mathfrak{s}}\in\mathbb{J}_{s}}$ are increasing families of projections, we have

[TABLE]

Hence for ${\mathfrak{t}}\geq{\mathfrak{s}}\geq\mathfrak{r}_{0}\in\mathbb{J}_{s},$ we have $\norm{\Phi_{\mathfrak{t}}Q_{\mathfrak{r}_{0}}(x)g}^{2}\geq\norm{\Phi_{\mathfrak{s}}Q_{\mathfrak{r}_{0}}(x)g}^{2}.$ Also

[TABLE]

for all ${\mathfrak{s}}\in\mathbb{J}_{s}.$ Thus $(\norm{\Phi_{\mathfrak{s}}Q_{\mathfrak{r}_{0}}(x)g}^{2})_{{\mathfrak{s}}\in\mathbb{J}_{s}}$ is a Cauchy net, hence choose $\mathfrak{r}_{1}\in\mathbb{J}_{s},\mathfrak{r}_{1}\geq\mathfrak{r}_{0}$ such that

[TABLE]

Therefore for ${\mathfrak{t}}\geq{\mathfrak{s}}\geq\mathfrak{r}_{1}$ in $\mathbb{J}_{s},$ from (42) and (43) we have

[TABLE]

Now for ${\mathfrak{t}}\geq{\mathfrak{s}}\geq\mathfrak{r}_{1}$ in $\mathbb{J}_{s},$ from (41) and (44) we have

[TABLE]

Thus $\lim\limits_{{\mathfrak{s}}\in\mathbb{J}_{s}}\Phi_{\mathfrak{s}}(x)g$ exists. Define $\hat{T}_{s}(x)g:=\lim\limits_{{\mathfrak{s}}\in\mathbb{J}_{s}}\Phi_{\mathfrak{s}}(x)g$ for $s>0.$ This defines a bounded bilinear map $\hat{T}_{s}:\mathcal{E}_{s}\to\mathcal{F}_{s}$ for all $s\in\mathbb{T}$.

Now for ${\mathfrak{s}}\in\mathbb{J}_{s}$ and for all $x_{\mathfrak{s}}\in E_{\mathfrak{s}},g\in\mathcal{G},$ we have

[TABLE]

Thus $T_{\mathfrak{s}}=j_{\mathfrak{s}}^{*}\hat{T}_{s}i_{\mathfrak{s}}$ for all ${\mathfrak{s}}\in\mathbb{J}_{s}$ and $s\in\mathbb{T}.$ In particular $T_{s}=j_{s}^{*}\hat{T}_{s}i_{s}$ for all $s\in\mathbb{T}.$

Now we shall prove that $(\hat{T}_{t})_{t\in\mathbb{T}}$ is a morphism of product systems. For ${\mathfrak{t}}\in\mathbb{J}_{t},{\mathfrak{s}}\in\mathbb{J}_{s}$ and $x_{{\mathfrak{t}}}\in E_{\mathfrak{t}},x_{\mathfrak{s}}\in E_{\mathfrak{s}},y_{{\mathfrak{t}}}\in F_{\mathfrak{t}},y_{\mathfrak{s}}\in F_{\mathfrak{s}}$ consider,

[TABLE]

Thus $\hat{T}_{s+t}=C_{s,t}^{*}(\hat{T}_{s}\odot\hat{T}_{t})B_{s,t}$ for all $s,t\in\mathbb{T}.$ ∎

Acknowledgments. The first author thanks J C Bose Fellowship and the second author thanks NBHM, and the Indian Statistical Institute for research funding.