On stable maps of operator algebras

TL;DR

This paper introduces a new Morita-type equivalence for operator algebras, characterizes stable isomorphism, and explores relations between $C^*$-algebras using operator algebra techniques.

Contribution

It defines a strong Morita-type equivalence for operator algebras and characterizes stable isomorphism through this relation, extending the understanding of operator algebra equivalences.

Findings

Equivalence $A extasciitilde _{\sigma riangle } B$ iff $A$ and $B$ are stably isomorphic.

Characterization of the relation $ extasciitilde _{\sigma riangle }$ for $C^*$-algebras via onto $*$-homomorphisms.

Decomposition of $C^*$-algebras under mutual $ extasciitilde _{\sigma riangle }$ relations using projections in their biduals.

Abstract

We define a strong Morita-type equivalence $\sim _{\sigma \Delta }$ for operator algebras. We prove that $A\sim _{\sigma \Delta }B$ if and only if $A$ and $B$ are stably isomorphic. We also define a relation $\subset _{\sigma \Delta }$ for operator algebras. We prove that if $A$ and $B$ are $C^*$-algebras, then $A\subset _{\sigma \Delta } B$ if and only if there exists an onto $*$-homomorphism $\theta :B\otimes \mathcal K \rightarrow A\otimes \mathcal K,$ where $\mathcal K$ is the set of compact operators acting on an infinite dimensional separable Hilbert space. Furthermore, we prove that if $A$ and $B$ are $C^*$-algebras such that $A\subset _{\sigma \Delta } B$ and $B\subset _{\sigma \Delta } A $, then there exist projections $r, \hat r$ in the centers of $A^{**}$ and $B^{**}$, respectively, such that $Ar\sim _{\sigma \Delta }B\hat r$ and $A (id_{A^{**}}-r) \sim _{\sigma \Delta }B(id_{B^{**}}-\hat r). $

Equations

$$\alpha (A)={\overline{[M^{\ast }\beta (B)M]}}^{\Vert \cdot \Vert },\beta (B)={\overline{[M\alpha (A)M^{\ast }]}}^{\Vert \cdot \Vert }.$$

$$\{x\in B(H):(I_H-n)xn=0,\forall n\in \mathcal{N}\}.$$

$$\underset{l}{\lim }\sum _{i=1}^l{m}_i{m}_i^{\ast }m=m,\underset{l}{\lim }\sum _{i=1}^{l}m{n}_i^{\ast }{n}_i=m,\forall m\in M$$

$$\Vert \sum _{i=1}^l{m}_i{m}_i^{\ast }\Vert \le 1,\Vert \sum _{i=1}^l{n}_i^{\ast }{n}_i\Vert \le 1,\forall l.$$

$$B={\overline{[M^{\ast }AM]}}^{\Vert \cdot \Vert },MB{M}^{\ast }\subseteq A.$$

$$b=\sum _{l=1}^{\infty }\sum _{n=1}^{\infty }\sum _{i,j=1}^l\frac{m_i^{\ast }{a}_n{m}_i{m}_j^{\ast }{a}_n^{\ast }{m}_j}{2^n{2}^l\Vert a_n{\Vert }^2}.$$

$$\Vert \sum _{i,j=1}^l{m}_i^{\ast }{a}_n{m}_i{m}_j^{\ast }{a}_n^{\ast }{m}_j\Vert ={\Vert \sum _{i=1}^l{m}_i^{\ast }{a}_n^{\ast }{m}_i\Vert }^2\le \Vert a_n{\Vert }^2,$$

$$\sum _{l=1}^{\infty }\sum _{n=1}^{\infty }\Vert \sum _{i,j=1}^l\frac{m_i^{\ast }{a}_n{m}_i{m}_j^{\ast }{a}_n^{\ast }{m}_j}{2^n{2}^l\Vert a_n{\Vert }^2}\Vert <+\infty .$$

$$\varphi (b)=0\Rightarrow \varphi (\sum _{i,j=1}^l{m}_i^{\ast }{a}_n{m}_i{m}_j^{\ast }{a}_n^{\ast }{m}_j)=0,\forall n,l.$$

$$\varphi (\sum _{i=1}^l{m}_i^{\ast }{a}_n{m}_i{m}^{\ast }as{t}^{\ast }r)=0\forall n,l.$$

$$m_i{m}^{\ast }as{t}^{\ast }r\in M{M}^{\ast }AM{M}^{\ast }M\subseteq AM,$$

$$\underset{n}{\lim }{a}_n{m}_i{m}^{\ast }s{t}^{\ast }r=m_i{m}^{\ast }s{t}^{\ast }r.$$

$$\underset{l}{\lim }\sum _{i=1}^l{m}_i^{\ast }{m}_i{m}^{\ast }=m^{\ast },$$

$$E={\overline{[M_2^{\ast }F{M}_1]}}^{\Vert \cdot \Vert },F={\overline{[M_{2}E{M}_1^{\ast }]}}^{\Vert \cdot \Vert }.$$

$${\overline{[F{F}^{\ast }]}}^{\Vert \cdot \Vert }={\overline{[M_{2}E{E}^{\ast }{M}_2^{\ast }]}}^{\Vert \cdot \Vert }.$$

$$\Vert \sum _{i=1}^l{m}_i{m}_i^{\ast }\Vert \le 1\forall l,$$

$$b=\sum _{l=1}^{\infty }\sum _{n=1}^{\infty }\sum _{i,j=1}^l\frac{m_i{a}_n{m}_i^{\ast }{m}_j{a}_n^{\ast }{m}_j^{\ast }}{2^n{2}^l\Vert a_n{\Vert }^2}$$

$$E={\overline{[M_{2}F{M}_1^{\ast }]}}^{\Vert \cdot \Vert },M_2^{\ast }E{M}_1\subseteq F.$$

$$\Pi(E)=\left(\begin{array}[]{clr}\overline{[E^{*}E]}^{\|\cdot\|}&E^{*}\\ E&\overline{[EE^{*}]}^{\|\cdot\|}\end{array}\right),\;\;\;\Pi(F)=\left(\begin{array}[]{clr}\overline{[F^{*}F]}^{\|\cdot\|}&F^{*}\\ F&\overline{[FF^{*}]}^{\|\cdot\|}\end{array}\right).$$

$$(M_1\oplus M_2)\Pi (F)(M_1\oplus M_2{)}^{\ast }=\Pi (E)$$

$$(M_1\oplus M_2{)}^{\ast }\Pi (E)(M_1\oplus M_2{)}^{\ast }\subseteq \Pi (F).$$

$$NDM{M}^{\ast }D{N}^{\ast }NDM\subseteq NDM.$$

$$Z=\left(\begin{array}[]{clr}\overline{[M^{*}D]}^{\|\cdot\|}\\ \overline{[ND]}^{\|\cdot\|}\end{array}\right).$$

$$ZZ^{*}=\left(\begin{array}[]{clr}M^{*}DM&M^{*}DN^{*}\\ NDM&NDN^{*}\end{array}\right).$$

$$\{m_i:i\in \mathbb{N}\}\subseteq M,\{n_i:i\in \mathbb{N}\}\subseteq N$$

$$\Vert \sum _{i=1}^l{m}_i^{\ast }{m}_i\Vert \le 1,\Vert \sum _{i=1}^l{n}_i{n}_i^{\ast }\Vert \le 1,\forall l$$

$$\underset{l}{\lim }\sum _{i=1}^l{m}_i^{\ast }{m}_i{m}^{\ast }=m^{\ast }\forall m\in M,\underset{l}{\lim }\sum _{i=1}^l{n}_i{n}_i^{\ast }n=n\forall n\in N.$$

$$a_{l}=\left(\begin{array}[]{clr}\sum_{i=1}^{l}m_{i}^{*}m_{i}&0\\ 0&\sum_{i=1}^{l}n_{i}n_{i}^{*}\end{array}\right),l\in\mathbb{N}$$

$$\left(\begin{array}[]{clr}M^{*}M&0\\ 0&NN^{*}\end{array}\right)\subseteq ZZ^{*},$$

$${\overline{[Z^{\ast }Z]}}^{\Vert \cdot \Vert }={\overline{DM{M}^{\ast }D+DN{N}^{\ast }D}}^{\Vert \cdot \Vert }.$$

Full text

2010 Mathematics Subject Classification. 47L30 (primary), 46L05, 47L05, 47L25, 47L35, 16D90 (secondary)

1. Introduction

Two operator algebras $A$ and $B$ are called stably isomorphic if the algebras $A\otimes\mathcal{K}$ and $B\otimes\mathcal{K}$ are isomorphic as operator algebras. Here, $\mathcal{K}$ is the set of compact operators acting on $l^{2}(\mathbb{N}).$ Stably isomorphic $C^{*}$-algebras are strongly Morita equivalent in the sense of Rieffel. The same is true of non-self-adjoint operator algebras if we consider the strong Morita equivalence that was introduced by Blecher, Muhly, and Paulsen in [2]. Meanwhile, the converse is not true, even in the case of $C^{*}$-algebras [4].

We introduce a new Morita type equivalence between operator algebras: Let $A$ and $B$ be operator algebras that are possibly non-self-adjoint. We say that $A$ and $B$ are $\sigma-$strongly $\Delta$-equivalent, and write $A\sim_{\sigma\Delta}B$, if there exist completely isometric homomorphisms $\alpha:A\rightarrow\alpha(A),\;\;\beta:B\rightarrow\beta(B)$ and a $\sigma$-ternary ring of operators $M$ such that

[TABLE]

See the definition of the $\sigma-$ternary ring of operators in Definition 2.1.

In the proof of [9, Theorem 3.2], see also [9, Lemma 3.4], we noticed that if $A,B$ are operator algebras possessing countable approximate identities, $M$ is a ternary ring of operators and the triple $(A,B,M)$ satisfies (1.1) then $M$ is necessarily a $\sigma-$ternary ring of operators. We used this fact in order to prove that $A$ and $B$ are stably isomorphic. Subsequently in [11, Theorem 4.6] we extended the proof in the case of operator spaces. In the present paper, we prove that $\sim_{\sigma\Delta}$ is an equivalence relation in the class of operator algebras and we use this fact to prove that $A\sim_{\sigma\Delta}B$ if and only if $A$ and $B$ are stably isomorphic.

In [9], we studied the relationship between $A$ and $B$ when (1.1) holds for a ternary ring (TRO) of operators $M$ that is not necessarily a $\sigma$-TRO. This relation is not equivalent to the existence of an operator algebra isomorphism between $A\otimes\mathcal{K}$ and $B\otimes\mathcal{K}.$

We also consider a weaker relation $\subset_{\sigma\Delta}$ between operator algebras: We say that $A,$ $\sigma\Delta$-embeds into $B$ if there exists a projection $p$ in the center of $\Delta(B^{**}),$ where $\Delta(B^{**})$ is the diagonal of the second dual operator algebra of $B,$ such that $pBp$ is an operator algebra and $A\sim_{\sigma\Delta}pBp.$ In this case, we write $A\subset_{\sigma\Delta}B.$ We prove that $\subset_{\sigma\Delta}$ is transitive. For the case of $C^{*}$-algebras, we prove that $A\subset_{\sigma\Delta}B$ if and only if there exists an onto $*$-homomorphism from $B\otimes\mathcal{K}$ onto $A\otimes\mathcal{K},$ which is true if and only if there exists an ideal $I$ of $B$ such that $A\sim_{\sigma\Delta}B/I.$

We investigate whether it is true that $A\sim_{\sigma\Delta}B$ if $A\subset_{\sigma\Delta}B$ and $B\subset_{\sigma\Delta}A.$ In general, this is not true (see Section 5). It is also not true even in the case of $C^{*}$- algebras (see Example 4.9). However, we prove that if $A$ and $B$ are $C^{*}$-algebras such that $A\subset_{\sigma\Delta}B$ and $B\subset_{\sigma\Delta}A$, then there exist projections $r,\hat{r}$ in the centers of $A^{**}$ and $B^{**}$, respectively, such that $Ar\sim_{\sigma\Delta}B\hat{r}$ and $A(id_{A^{**}}-r)\sim_{\sigma\Delta}B(id_{B^{**}}-\hat{r}).$ A dual version of the results obtained in this article can be found in [10].

In the following we describe the notations and symbols used in this paper. If $H,K$ are Hilbert spaces, then $B(H,K)$ is the space of bounded operators from $H$ to $K.$ We write $B(H)$ for $B(H,H).$ A ternary ring of operators (TRO) is a subspace of some $B(H,K)$ satisfying $MM^{*}M\subseteq M$ (see the definition of a $\sigma$-TRO in Definition 2.1). An operator algebra is an operator space and Banach algebra for which there exists a completely isometric homomorphism $\alpha:A\rightarrow B(H).$ In this article, when we consider an operator algebra, we mean an operator algebra with a contractive approximate identity. We note that $C^{*}$-algebras possess contractive approximate identities automatically. If $X$ is an operator space, then $M_{\infty}(X)$ is the set of $\infty\times\infty$ matrices whose finite submatrices have uniformly bounded norm. The space $M_{\infty}(X)$ is an operator space. In addition, $M_{\infty}^{fin}(X)$ will denote the subspace of $M_{\infty}(X)$ consisting of ”finitely supported matrices.” We write $K_{\infty}(X)$ for the norm closure in $M_{\infty}(X)$ of $M_{\infty}^{fin}(X).$ It is well-known that the space $K_{\infty}(X)$ is completely isometric isomorphic with $X\otimes\mathcal{K},$ where $\otimes$ is the minimal tensor product [1]. For further details on the operator space theory that is used in this paper, we refer the reader to the books by [1], [6], [12], and [13].

A nest $\mathcal{N}\subseteq B(H)$ is a totally ordered set of orthogonal projections containing the zero and identity operators that are closed under arbitrary suprema and infima. Given a nest $\mathcal{N}\subseteq B(H),$ by $\mathrm{Alg}(\mathcal{N})$ we denote the corresponding nest algebra:

[TABLE]

Given an operator algebra $A,$ we denote its center by $Z(A)$ and its diagonal $A\cap A^{*}$ by $\Delta(A).$ If $S$ is a subset of a vector space, then we denote the linear span of the elements of $S$ by $[S]$.

2. Preliminaries

The purpose of this section is to prove Lemma 2.5, which is required to prove that $\sim_{\sigma\Delta}$ is an equivalence relation in Section 3.

Definition 2.1**.**

Let $H,K$ be Hilbert spaces, and $M\subseteq B(H,K)$ be a norm closed TRO. We call $M$ $\sigma$-TRO if there exist sequences $\{m_{i},\;n_{i},\;i\in\mathbb{N}\}\subseteq M$ such that

[TABLE]

and

[TABLE]

Remark 2.1**.**

A norm-closed TRO $M$ is a $\sigma$- TRO if and only if the $C^{*}$ algebras $\overline{[M^{*}M]}^{\|\cdot\|},\overline{[MM^{*}]}^{\|\cdot\|}$ are $\sigma$-unital. A proof of this fact can be found in Theorem 2.1 in [3].

Lemma 2.2**.**

Let $A\subseteq B(H),B\subseteq B(K)$ be $C^{*}$ algebras and $M\subseteq B(H,K)$ be a $\sigma$-TRO such that

[TABLE]

If $A$ is $\sigma$-unital, then $B$ is $\sigma$-unital.

Proof.

Suppose that $(a_{n})_{n\in\mathbb{N}}\subseteq A$ such that $\lim_{n}a_{n}a=a\;\forall\;\;a\in\;A.$ In addition, let $\{m_{i}:i\in\mathbb{N}\}\subseteq M$ be such that $\lim_{l}\sum_{i=1}^{l}m_{i}^{*}m_{i}m^{*}=m^{*}\forall\;m\;\in M$ and $\left\|\sum_{i=1}^{l}m_{i}^{*}m_{i}\right\|\leq 1\;\forall\;l.$ It suffices to prove that $B$ contains a strictly positive element. Define

[TABLE]

Because

[TABLE]

we have that

[TABLE]

Thus, the element $b$ is well defined. Observe that $b\geq 0$ if $\phi$ is a state of $B$ such that

[TABLE]

If $a\in A,m,s,t,r,\in M,$ then the Cauchy-Schwartz inequality for $\phi$ implies that

[TABLE]

Because

[TABLE]

we have that

[TABLE]

Thus, $\phi(\sum_{i=1}^{l}m_{i}^{*}m_{i}m^{*}astr^{*})=0,\;\forall\;l.$ Because

[TABLE]

we have that $\phi(m^{*}ast^{*}r)=0$ for all $m,s,t,r\in M,a\in A.$ Because $B=\overline{[M^{*}AMM^{*}M]}^{\|\cdot\|}$, we conclude that $\phi=0$. This contradiction shows that $b$ is strictly positive.

∎

Lemma 2.3**.**

Let $E,F,M_{1},M_{2}$ be TROs such that the algebra $\overline{[M_{2}^{*}M_{2}]}^{\|\cdot\|}$ is $\sigma$ - unital and

[TABLE]

If it also holds that the algebra $\overline{[EE^{*}]}^{\|\cdot\|}$ is $\sigma$-unital, then the algebra $\overline{[FF^{*}]}^{\|\cdot\|}$ is also $\sigma$-unital.

Proof.

Observe that

[TABLE]

Let $\{m_{i}:i\in\mathbb{N}\}\subseteq M_{2}$ be such that $\sum_{i=1}^{l}m_{i}m_{i}^{*}m=m\forall\;m\;\in\;M_{2},$

[TABLE]

and let $(a_{n})_{n}\subseteq\overline{[EE^{*}]}^{\|\cdot\|}$ be a $\sigma$-unit. As in Lemma 2.2, we can prove that the element

[TABLE]

is strictly positive in $\overline{[FF^{*}]}^{\|\cdot\|}.$ Thus, $\overline{[FF^{*}]}^{\|\cdot\|}$ is $\sigma$-unital. ∎

Lemma 2.4**.**

Let $E,F,M_{1},M_{2}$ be TROs such that $M_{1},M_{2},F$ are $\sigma$-TROs and

[TABLE]

Then, $E$ is a $\sigma$-TRO.

Proof.

It suffices to prove that the $C^{*}$-algebras $\overline{[EE^{*}]}^{\|\cdot\|},\overline{[E^{*}E]}^{\|\cdot\|}$ are $\sigma$-unital. Define the $C^{*}$-algebras

[TABLE]

Because $F$ is a $\sigma$-TRO, the algebra $\Pi(F)$ is $\sigma$-unital. Furthermore, it easy to see that

[TABLE]

and

[TABLE]

Lemma 2.2 implies that $\Pi(E)$ is $\sigma$-unital. Thus, the $C^{*}$-algebras $\overline{[EE^{*}]}^{\|\cdot\|},\overline{[E^{*}E]}^{\|\cdot\|}$ are $\sigma$-unital. ∎

Lemma 2.5**.**

Let $H,K,L$ be Hilbert spaces, $M\subseteq B(H,K),N\subseteq B(K,L)$ be $\sigma$-TROs, and $D$ be the $C^{*}$ algebra generated by the sets $MM^{*},N^{*}N.$ Then, $T=\overline{[NDM]}^{\|\cdot\|}$ is a $\sigma$-TRO.

Proof.

We have that

[TABLE]

Thus, $TT^{*}T\subseteq T$, and so $T$ is a TRO. We define the TRO

[TABLE]

Then,

[TABLE]

Let

[TABLE]

be such that

[TABLE]

and

[TABLE]

The elements

[TABLE]

belong to

[TABLE]

and satisfy $\lim_{l}a_{l}x=x,\forall x\in\overline{[ZZ^{*}]}^{\|\cdot\|}.$ Thus, $\overline{[ZZ^{*}]}^{\|\cdot\|}$ is a $\sigma$-unital $C^{*}$- algebra. Now, we have that $Z=\overline{[ZD]}^{\|\cdot\|}$ and

[TABLE]

We can easily see that $D=\overline{[MM^{*}D+N^{*}ND]}^{\|\cdot\|}$, and thus

[TABLE]

Now, apply Lemma 2.3 for

[TABLE]

We obtain that

[TABLE]

$\overline{[M^{*}_{2}M_{2}]}^{\|\cdot\|}=\overline{[ZZ^{*}]}^{\|\cdot\|}=\overline{[EE^{*}]}^{\|\cdot\|}$ is $\sigma$-unital. Lemma 2.3 implies that $\overline{[FF^{*}]}^{\|\cdot\|}$ is $\sigma$-unital, and thus $D$ is $\sigma$-unital $C^{*}$-algebra. Now,

[TABLE]

Lemma 2.4 implies that $T$ is a $\sigma$-TRO. ∎

3. $\sigma$-strong $\Delta$-equivalence

Definition 3.1**.**

Let $A$ and $B$ be operator algebras acting on the Hilbert spaces $H$ and $L$, respectively. We call them $\sigma$-strongly $TRO$-equivalent if there exists a $\sigma$-TRO $M\subseteq B(L,H)$ such that

[TABLE]

In this case, we write $A\sim_{\sigma TRO}B.$

Definition 3.2**.**

Let $A$ and $B$ be operator algebras. We call these $\sigma$-strongly $\Delta$-equivalent if there exist completely isometric homomorphisms $\alpha:A\rightarrow\alpha(A),\;\beta:B\rightarrow\beta(B)$ such that $\alpha(A)\sim_{\sigma TRO}\beta(B).$ In this case, we write $A\sim_{\sigma\Delta}B.$

Theorem 3.1**.**

Let $A,B$ be $\sigma$-strongly $\Delta$-equivalent operator algebras. Then, for every completely isometric homomorphism $\alpha:A\rightarrow\alpha(A)$ there exists a completely isometric homomorphism $\beta:B\rightarrow\beta(B)$ such that $\alpha(A)\sim_{\sigma TRO}\beta(B).$

Proof.

We may assume that $H,L,M$ are as in Definition 3.1. By $Y$, we denote the space $Y=\overline{[BMA]}^{\|\cdot\|}.$ Let $K$ be the $A$-balanced Haagerup tensor product $K=Y\otimes^{h}_{A}H.$ This is a Hilbert space [2]. Define

[TABLE]

By Lemma 2.10 in [9], $\beta$ is a completely isometric homomorphism. From the same article, if $m\in M,$ we define

[TABLE]

The map $\mu:M\rightarrow\mu(M)$ is a TRO homomorphism. Thus, $\mu(M)$ is a $\sigma$-TRO. By Theorem 2.12 in [9], we have that

[TABLE]

The proof is complete. ∎

Theorem 3.2**.**

The $\sigma$-strong $\Delta$-equivalence of operator algebras is an equivalence relation in the class of operator algebras.

Proof.

It suffices to prove the transitivity property. Let $A,B,$ and $C$ be operator algebras such that $A\sim_{\sigma\Delta}B$ and $B\sim_{\sigma\Delta}C.$ Therefore, there exists a $\sigma$-TRO $M$ and completely isometric homomorphisms $\alpha:A\rightarrow\alpha(A),\;\beta:B\rightarrow\beta(B)$ such that

[TABLE]

By Theorem 3.1, there exists a $\sigma$-TRO $N$ and a completely isometric homomorphism $\gamma:C\rightarrow\gamma(C)$ such that

[TABLE]

Let $D$ be the $C^{*}$-algebra generated by the set $\{MM^{*}\}\cup\{N^{*}N\}.$ By Lemma 2.5, the space $T=\overline{[NDM]}^{\|\cdot\|}$ is a $\sigma$-TRO. As in the proof of Theorem 2.1 in [9], we can prove that

[TABLE]

Thus, $A\sim_{\sigma\Delta}C.$ ∎

Theorem 3.3**.**

Let $A,B$ be operator algebras. Then, $A$ and $B$ are $\sigma$-strongly $\Delta$-equivalent if and only if they are stably isomorphic.

Proof.

We assume that $M$ is a $\sigma$-TRO satisfying

[TABLE]

Theorem 4.6 in [11] implies that there exists a completely isometric onto linear map $K_{\infty}(A)\rightarrow K_{\infty}(B).$ By using the Banach-Stone theorem for operator algebras, we may assume that this map is also a homomorphism [1, 4.5.13].

For the converse, suppose that $K_{\infty}(A)$ and $K_{\infty}(B)$ are completely isometrically isomorphic as operator algebras. Let $R_{\infty}$ be the space of infinite rows consisting of compact operators. Then, $R_{\infty}$ is a $\sigma$-TRO, and we have that

[TABLE]

Thus, $A\sim_{\sigma TRO}K_{\infty}(A).$ Therefore, $A\sim_{\sigma\Delta}K_{\infty}(B).$ By the same arguments, $B\sim_{\sigma TRO}K_{\infty}(B).$ Therefore, Theorem 3.2 implies that $A\sim_{\sigma\Delta}B.$ ∎

Corollary 3.4**.**

Rieffel’s strong Morita equivalence of $C^{*}$-algebras is weaker than $\sigma$-strong $\Delta$-equivalence.

Proof.

It is well-known [4] that there exist $C^{*}$-algebras that are strongly Morita equivalent in the sense of Rieffel but are not stably isomorphic. Thus, by Theorem 3.3 these $C^{*}$-algebras cannot be $\sigma$-strongly $\Delta$-equivalent. ∎

Corollary 3.5**.**

Two $\sigma$-unital $C^{*}$-algebras are strongly Morita equivalent in the sense of Rieffel if and only if they are $\sigma$-strongly $\Delta$-equivalent.

Proof.

By [4], two $\sigma$-unital $C^{*}$-algebras are strongly Morita equivalent in the sense of Rieffel if and only if they are stably isomorphic. The conclusion is implied by Theorem 3.3. ∎

4. Strong Morita embeddings

In [10], we defined a new relation $\subset_{\Delta}$ between dual operator algebras: Given two unital dual operator algebras $A$ and $B$, we say that $A\subset_{\Delta}B$ if there exists an orthogonal projection $p\in B$ such that $A$ and $pBp$ are weakly stably isomorphic. In this case, there exists a projection $q\in Z(\Delta(B))$ such that $pBp$ and $qBq$ are weakly stably isomorphic [10, Lemma 2.11]. In the present section, we aim to investigate the strong version of the previously stated relation for operator algebras.

Definition 4.1**.**

Let $A$ and $B$ be operator algebras. We say that $A,$ $\sigma\Delta$-embeds into $B,$ if there exists a projection $p\in Z(\Delta(B^{**}))$ such that $pBp$ is an operator algebra and $A\sim_{\sigma\Delta}pBp.$ In this case, we write $A\subset_{\sigma\Delta}B.$

Remark 4.1**.**

Let $A$ be a $C^{*}$-algebra, and $p$ be a central projection of $A^{**}.$ Because the map $A\rightarrow A^{**},\;a\rightarrow ap$ is a $*$-homomorphism, it has norm-closed range. Thus, $Ap$ is a $C^{*}$-algebra.

In the following, we prove that $\subset_{\sigma\Delta}$ is transitive.

Theorem 4.2**.**

Let $A,B,C$ be operator algebras. If $A\subset_{\sigma\Delta}B$ and $B\subset_{\sigma\Delta}C$, then $A\subset_{\sigma\Delta}C.$

Proof.

Let $p\in Z(\Delta(B^{**})),q\in Z(\Delta(C^{**}))$ be such that $pBp,qCq$ are operator algebras and $A\sim_{\sigma\Delta}pBp,B\sim_{\sigma\Delta}qCq.$ We write

[TABLE]

Then,

[TABLE]

There exist completely isometric homomorphisms

[TABLE]

such that

[TABLE]

There exists a completely isometric homomorphism $\rho_{0}:\hat{B}^{**}\rightarrow\hat{C}^{**}$ such that

[TABLE]

Because $p^{\infty}\in Z(\Delta(\hat{B}^{**}))$ and $\rho_{0}(p^{\infty})\leq q^{\infty}$, there exists $q_{0}\in Z(\Delta(\hat{C}^{**}))$ such that $\rho_{0}(p^{\infty})=q_{0}^{\infty}.$ Now,

[TABLE]

[TABLE]

Thus,

[TABLE]

Because $\rho\circ\theta$ is a completely isometric homomorphism, we have that

[TABLE]

∎

Remark 4.3**.**

Following this theorem, one should expect that $\subset_{\sigma\Delta}$ is a partial order relation in the class of operator algebras if we identify those operator algebras that are $\sigma$-strongly $\Delta$-equivalent. This means that the additional property holds that

[TABLE]

However, this is not true, as we will prove in Section 5.

4.1. The case of $C^{*}$-algebras

In this subsection, we investigate the relation $\subset_{\sigma\Delta}$ in the case of $C^{*}$-algebras.

Theorem 4.4**.**

Let $A,B$ be $C^{*}$-algebras. The following are equivalent:

(i)

[TABLE]

.

(ii) There exists an onto $*$-homomorphism $\theta:K_{\infty}(B)\rightarrow K_{\infty}(A).$

(iii) There exists an ideal $I$ of $B$ such that

[TABLE]

(iv) For every $*$-isomorphism $\alpha:A\rightarrow\alpha(A)$, there exists a $*$-homomorphism (not necessarily faithful) $\beta:B\rightarrow\beta(B)$ such that $\alpha(A)\sim_{\sigma TRO}\beta(B).$

Proof.

(i) $\Rightarrow(ii)$

By Definition 4.1 and Theorem 3.3, there exist a projection $p\in Z(B^{**})$ and a $*$-isomorphism $\rho:K_{\infty}(pB)\rightarrow K_{\infty}(A).$ Define the onto $*$-homomorphism $\tau:K_{\infty}(B)\rightarrow K_{\infty}(pB)$ given by $\tau((b_{i,j})_{i,j})=(pb_{i,j})_{i,j}.$ We denote $\theta=\rho\circ\tau.$

(ii) $\Rightarrow(i)$

Suppose that $\theta^{**}:M_{\infty}(B^{**})\rightarrow M_{\infty}(A^{**})$ is the second dual of $\theta,$ then there exists a projection $q\in Z(B^{**})$ such that

[TABLE]

and $\theta|_{M_{\infty}(B^{**}q)}$ is a $*$-homomorphism. Thus, if $x\in K_{\infty}(B)$, we have that $\theta(xq^{\infty})=\theta(x).$ Therefore,

[TABLE]

which implies that

[TABLE]

(iii) $\Rightarrow(ii)$

If $A\sim_{\sigma\Delta}B/I,$ then

[TABLE]

Because $K_{\infty}(I)$ is an ideal of $K_{\infty}(B),$ there exists an onto $*$-homomorphism $\theta:K_{\infty}(B)\rightarrow K_{\infty}(A).$

(ii) $\Rightarrow(iii)$

Suppose that $\theta:K_{\infty}(B)\rightarrow K_{\infty}(A)$ is an onto $*$-homomorphism. Then, there exists an ideal $J\subseteq K_{\infty}(B)$ such that

[TABLE]

The ideal $J$ is of the form $K_{\infty}(I)$ for an ideal $I$ of $B.$ Thus,

[TABLE]

Therefore, $A\sim_{\sigma\Delta}B/I.$

(iv) $\Rightarrow(iii)$

Suppose that $\alpha:A\rightarrow\alpha(A),\beta:B\rightarrow\beta(B)$ are $*$-homomorphisms such that $Ker\alpha=\{0\}$ and $\alpha(A)\sim_{\sigma TRO}\beta(B).$ Let $I$ be the ideal $Ker\beta.$ Then, $\beta(B)\cong B/I,$ and thus $A\sim_{\sigma\Delta}B/I.$

(iii) $\Rightarrow(iv)$

We assume that $\alpha:A\rightarrow\alpha(A)$ is a faithful $*$-homomorphism, and that $A\sim_{\sigma\Delta}B/I.$ By Theorem 3.1, there exists a faithful $*$-homomorphism $\gamma:B/I\rightarrow\gamma(B/I)$ such that

[TABLE]

If $\pi:B\rightarrow B/I$ is the natural mapping and $\beta=\gamma\circ\pi$, then

[TABLE]

∎

Remark 4.5**.**

If $A$ and $B$ are $W^{*}$-algebras and $\alpha:A\rightarrow B,\;\;\;\beta:B\rightarrow A$ are $w^{*}$-continuous onto $*$-homomorphisms, then $A$ and $B$ are $*$-isomorphic. Indeed, there exist projections $e_{1}\in Z(A),f_{1}\in Z(B)$ such that

[TABLE]

Thus, there exists a projection $e_{2}\in Z(A),e_{2}\leq e_{1}$ such that

[TABLE]

From the proof of Lemma 2.17 in [10], we have that $A\cong Ae_{1},$ and thus $A\cong B.$ In Example 4.9, we will present non-isomorphic $C^{*}$-algebras $A$ and $B$ for which there exist onto $*$-homomorphisms $\alpha:A\rightarrow B,\;\;\;\beta:B\rightarrow A.$ These algebras are not $W^{*}$-algebras.

Remark 4.6**.**

As we have previously mentioned, in [10] we defined an analogous relation $\subset_{\Delta}$ between unital dual operator algebras. We have proven that if $A\subset_{\Delta}B,$ where $A,B$ are unital dual operator algebras, then there exists a central projection $p$ in $\Delta(B)$ and a Hilbert space $H$ such that $A\bar{\otimes}B(H)$ and $(pBp)\bar{\otimes}B(H)$ are isomorphic as dual operator algebras. Here, $\bar{\otimes}$ is the normal spatial tensor product. In the case of $W^{*}$-algebras, we have proven that $A\subset_{\Delta}B$ if and only if there exists a a Hilbert space $H$ and a $w^{*}$-continuous $*$-homomorphism from $B\bar{\otimes}B(H)$ onto $A\bar{\otimes}B(H)).$ We have also proven that if $A$ and $B$ are $W^{*}$-algebras such that $A\subset_{\Delta}B$ and $B\subset_{\Delta}A$, then $A$ and $B$ are stably isomorphic in the weak sense. We present a new proof of this fact here.

Suppose that $A\subset_{\Delta}B$ and $B\subset_{\Delta}A.$ Then, there exist Hilbert spaces $H$ and $K$ and $w^{*}$-continuous $*$-homomorphisms from $B\bar{\otimes}B(H)$ onto $A\bar{\otimes}B(H)$ and from $A\bar{\otimes}B(K)$ onto $B\bar{\otimes}B(K).$ We conclude that there exist $w^{*}$-continuous $*$-homomorphisms from $B\bar{\otimes}B(H)\bar{\otimes}B(K)$ onto $A\bar{\otimes}B(H)\bar{\otimes}B(K)$ and from $A\bar{\otimes}B(K)\bar{\otimes}B(H)$ onto $A\bar{\otimes}B(K)\bar{\otimes}B(H).$ Therefore, by Remark 4.5,

[TABLE]

Because

[TABLE]

we have that

[TABLE]

Thus, $A$ and $B$ are stably isomorphic.

Remark 4.7**.**

The relation $\subset_{\Delta}$ between $W^{*}$-algebras is a partial order relation up to weak stable isomorphism [10]. This means that it has the following properties:

(i) $A\subset_{\Delta}A$.

(ii) $A\subset_{\Delta}B,\;\;\;B\subset_{\Delta}C\Rightarrow A\subset_{\Delta}C.$

(iii)If $A\subset_{\Delta}B$ and $B\subset_{\Delta}A$, then $A$ and $B$ are weakly stably isomorphic. Therefore, it is natural to ask whether $\subset_{\sigma\Delta}$ is a partial order relation up to strong stable isomorphism for $C^{*}$-algebras. Although $\subset_{\sigma\Delta}$ satisfies the properties (i) and (ii), it does not satisfy property (iii), as we show in Example 4.9. Nevertheless, $\subset_{\sigma\Delta}$ satisfies the property described in Theorem 4.18.

Example 4.8**.**

Let $X,Y$ be compact metric spaces, $\theta:X\rightarrow Y$ be a continuous one-to-one function, and $C(X)$ and $C(Y)$ be the algebras of continuous functions from $X$ and $Y$, respectively, into the complex plane $\mathbb{C}$, equipped with the supremum norm. Then, the map

[TABLE]

is an onto $*$-homomorphism, and thus $C(X)\subset_{\sigma\Delta}C(Y).$ Indeed, if $g\in C(X)$ we define

[TABLE]

Because $\theta:X\rightarrow\theta(X)$ is a homeomorphism, $f_{0}$ is continuous. By Tietze’s theorem, there exists $f\in C(Y)$ such that $f|_{\theta(X)}=f_{0}.$ We have that $f\circ\theta(x)=g(x)$ for all $x\in X,$ and thus $\rho(f)=g.$

Example 4.9**.**

There exist commutative $C^{*}$-algebras $A$ and $B$ such that $A\subset_{\sigma\Delta}B,\;\;\;B\subset_{\sigma\Delta}A,$ but $A$ and $B$ are not strongly Morita equivalent. Thus, $A$ and $B$ are not $\sigma\Delta$- equivalent. We denote the following subsets of $\mathbb{C}:$

[TABLE]

We write $A=C(X),\;\;B=C(Y).$ Because $X\subseteq Y$, by Example 4.8 we have that $A\subset_{\sigma\Delta}B.$ All of the closed discs of $\mathbb{C}$ are homeomorphic, and thus there exists a homeomorphism $\theta:Y\rightarrow X_{0},$ where $X_{0}=\{z\in\mathbb{C}:|z-3|\leq 1\}.$ Because $X_{0}\subseteq X,$ Example 4.8 implies that $B\subset_{\sigma\Delta}A.$ If $A$ and $B$ were strongly Morita equivalent, then they would also be $*$-isomorphic. The Stone-Banach theorem implies that $X$ and $Y$ would then be homeomorphic. However, this contradicts the fact that $Y$ is a simply connected set and $X$ is not.

Next, we will prove Theorem 4.18, which states the following:

[TABLE]

for central projections $r\in A^{**},\hat{r}\in B^{**}.$

Lemma 4.10**.**

Let $A,B$ be operator algebras and $\hat{A},\hat{B}$ be unital dual operator algebras such that $\hat{A}=\overline{A}^{w^{*}},\hat{B}=\overline{B}^{w^{*}}.$ Furthermore, let $M$ be a TRO such that

[TABLE]

and let $\alpha:\hat{A}\rightarrow\alpha(\hat{A})$ be a $w^{*}$-continuous completely isometric homomorphism such that $H=\overline{\alpha(A)(H)}.$ Then, there exist a Hilbert space $K,$ a $w^{*}$-continuous completely isometric honomorphism $\beta:\hat{B}\rightarrow B(K)$ such that $K=\overline{\beta(B)(K)}$, and a TRO homomorphism $\mu:M\rightarrow B(H,K)$ such that the following hold:

A) If $a\in A,b\in B,m,n\in M$ such that $a=m^{*}bn$, then $\alpha(a)=\mu(m)^{*}\beta(b)\mu(n).$

B) If $a\in A,b\in B,m,n\in M$ such that $b=man^{*}$, then $\beta(b)=\mu(m)\alpha(a)\mu(n)^{*}.$ Therefore,

[TABLE]

and

[TABLE]

The proof of this lemma can be inferred from the proof of Theorem 2.12 in [9], with the addition of some simple modifications.

Definition 4.2**.**

Let $\hat{A},\hat{B}$ be von Neumann algebras, and $A$ (resp. $B$) be a $C^{*}$-subalgebra of $\hat{A}$ (resp. $\hat{B}$) such that $\hat{A}=\overline{A}^{w^{*}}$ (resp. $\hat{B}=\overline{B}^{w^{*}},$ ). We write $(A,\hat{A})\sim_{\Delta}(B,\hat{B})$ if there exist $w^{*}$-continuous and injective $*$-homomorpisms $\alpha:\hat{A}\rightarrow\alpha(\hat{A}),\;\;\beta:\hat{B}\rightarrow\beta(\hat{B})$ and a $\sigma$-TRO $M$ such that

[TABLE]

Remarks 4.11**.**

(i) If (4.1) holds then

[TABLE]

(ii) Lemma 4.10 implies that if $(A,\hat{A})\sim_{\Delta}(B,\hat{B})$ and $\gamma:\hat{A}\rightarrow\gamma(\hat{A})$ is a $w^{*}$-continuous $*$-isomorphism, then there exists a $w^{*}$-continuous $*$-isomorphism $\delta:\hat{B}\rightarrow\delta(\hat{B})$ and a $\sigma$-TRO $N$ such that

[TABLE]

(iii) The above remark and Theorem 3.2 both imply that if $\hat{A},\hat{B},\hat{C}$ are von Neumann algebras $A,B,C$ are, respectively, $w^{*}$-dense $C^{*}$-subalgebras of these, and $(A,\hat{A})\sim_{\Delta}(B,\hat{B})$ and $(B,\hat{B})\sim_{\Delta}(C,\hat{C}),$ then $(A,\hat{A})\sim_{\Delta}(C,\hat{C}).$

In the following, we assume that $A$ is a $C^{*}$-algebra such that $A\subseteq A^{**}\subseteq B(H)$ for some Hilbert space $H$, and $e_{2}$ is a central projection of $A^{**}.$ We also assume that $A\sim_{\sigma\Delta}Ae_{2}.$

Lemma 4.12**.**

There exist a $w^{*}$-continuous $*$-isomorphism $\theta_{1}:A^{**}\rightarrow\theta_{1}(A^{**})$ and a $\sigma$-TRO $M$ such that

[TABLE]

Proof.

Let $B$ be a $C^{*}$ algebra. We assume that $B\subseteq B^{**}\subseteq B(H).$ Let $\mathcal{K}$ be the algebra of compact operators acting on $l^{2}(\mathbb{N})$, and $p\in\mathcal{K}$ be a rank one projection. We define the $\sigma$-TRO $M=I_{H}\otimes p\mathcal{K}.$ Then, we have that

[TABLE]

where $\otimes$ is the minimal tensor product. Because $\overline{B\otimes p}^{w^{*}}=B^{**}\bar{\otimes}p,\;\;\;\overline{B\otimes\mathcal{K}}^{w^{*}}=B^{**}\bar{\otimes}B(l^{2}(\mathbb{N}))$, here $\bar{\otimes}$ is the spatial tensor product, we have

[TABLE]

Because there exists a $*$-isomorphism from $B^{**}$ onto $B^{**}\bar{\otimes}p$ mapping $B$ onto $B\otimes p$, we can conclude that $(B,B^{**})\sim_{\Delta}(B\otimes\mathcal{K},B^{**}\bar{\otimes}B(l^{2}(\mathbb{N}))).$

Therefore,

[TABLE]

and

[TABLE]

Because $A\sim_{\sigma\Delta}Ae_{2}$, there exists a $*$-isomorphism from $A^{**}\bar{\otimes}B(l^{2}(\mathbb{N}))$ onto $(A^{**}e_{2})\bar{\otimes}B(l^{2}(\mathbb{N}))$ mapping $A\otimes\mathcal{K}$ onto $(Ae_{2})\otimes\mathcal{K}$ and, therefore,

[TABLE]

Now Remark 4.11, (iii), implies that $(A,A^{**})\sim_{\Delta}(Ae_{2},A^{**}e_{2}).$ By Remark 4.11, (ii), for the identity map $id:A^{**}e_{2}\rightarrow A^{**}e_{2}$ there exist a $w^{*}$-continuous $*$-isomorphism $\theta_{1}:A^{**}\rightarrow\theta_{1}(A^{**})$ and a $\sigma$-TRO $M$ such that

[TABLE]

∎

Lemma 4.13**.**

Let $M,\theta_{1}$ be as in Lemma 4.12. Then, there exist $w^{*}$-continuous $*$- isomorphisms $\rho_{k}:A^{**}\rightarrow\rho_{k}(A^{**})$ and TRO homomorphims $\phi_{k}:M\rightarrow\phi_{k}(M),k=0,1,2,...$ where $\rho_{0}=id_{A^{**}},\phi_{0}=id_{M},$ such that if $a\in A^{**},x\in A^{**}e_{2},m,n\in M,$ the equality $\rho_{k}(a)=\phi_{k-1}(m)^{*}\rho_{k-1}(x)\phi_{k-1}(n)$ implies that $\rho_{k+1}(a)=\phi_{k}(m)^{*}\rho_{k}(x)\phi_{k}(n)$ and the equality $\rho_{k-1}(x)=\phi_{k-1}(m)\rho_{k}(a)\phi_{k-1}(n)^{*}$ implies that $\rho_{k}(x)=\phi_{k}(m)\rho_{k+1}(a)\phi_{k}(n)^{*}$ for all $k=1,2,...$ Therefore,

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

for all $k=1,2,...$

Proof.

By Lemma 4.10, given the representation $\theta_{1}|_{A^{**}e_{2}}$, there exists a $*$-isomorphism

[TABLE]

and a TRO homomorphism $\phi_{1}:M\rightarrow\phi_{1}(M)$ such that

[TABLE]

and

[TABLE]

and such that if $a\in A^{**},x\in A^{**}e_{2},m,n\in M$, the equality $\theta_{1}(a)=m^{*}xn$ implies that $\theta_{2}(\theta_{1}(a))=\phi_{1}(m)^{*}\theta_{1}(x)\phi_{1}(n)$ and the equality $x=m\theta_{1}(a)n^{*}$ implies that $\theta_{1}(x)=\phi_{1}(m)\theta_{2}(\theta_{1}(a))\phi_{1}(n)^{*}.$

We write $\rho_{0}=id_{A^{**}},\rho_{1}=\theta_{1},\rho_{2}=\theta_{2}\circ\theta_{1}$ and continue inductively. ∎

Let $M,\theta_{1}$ be as in Lemma 4.12. Given the $*$-isomorphism $\theta_{1}^{-1}:\theta_{1}(A^{**})\rightarrow A^{**},$ Lemma 4.10 implies that there exist a $*$-isomorphism $\sigma_{1}:A^{**}e_{2}\rightarrow\sigma_{1}(A^{**}e_{2})$ and a TRO homomorphism $\chi_{0}:M\rightarrow\chi_{0}(M)$ such that if $\chi(m)=\chi_{0}(m)^{*},\;\;\forall\;m\;\in\;M,$ then

[TABLE]

and

[TABLE]

Furthermore, if $a\in A^{**},m,n\in M,x\in A^{**}e_{2}$ then the equality $\theta_{1}(a)=m^{*}xn$ implies that $a=\chi(m)\sigma_{1}(x)\chi(n)^{*}.$

Lemma 4.14**.**

Let $M,\chi,\theta_{1}$ be as in the previous discussion, then there exists a $w^{*}$-continuous $*$-isomorphism $\tau_{1}:A^{**}\rightarrow\tau_{1}(A^{**})$ and a TRO homomorphism $\psi_{1}:\chi(M)\rightarrow\psi_{1}(\chi(M))$ such that if $a\in A^{**},m,n\in M,x\in A^{**}e_{2}$, then the equality $a=\chi(m)\sigma_{1}(x)\chi(n)^{*}$ implies that $a=\psi_{1}(\chi(m))\tau_{1}(x)\psi_{1}(\chi(n))^{*}$ and $\sigma_{1}(x)=\chi(m)^{*}a\chi(n)$ implies that $\tau_{1}(x)=\psi_{1}(\chi(m))^{*}a\psi_{1}(\chi(n)).$ Thus,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Proof.

Define the $*$-isomorphism $\tau_{1}:A^{**}\rightarrow\sigma_{1}(A^{**}e_{2})\oplus A^{**}e_{2}^{\bot},$ given by $\tau_{1}(a)=\sigma_{1}(ae_{2})\oplus ae_{2}^{\bot},$ and the TRO homomorphism $\psi_{1}:\chi(M)\rightarrow\psi_{1}(\chi(M))$ given by $\psi_{1}(\chi(m))=(\chi(m)\;\;0).$ If $a\in A^{**},m,n\in M,x\in A^{**}e_{2}$ satisfies $a=\chi(m)\sigma_{1}(x)\chi(n)^{*}$, then

[TABLE]

Furthermore, if $\sigma_{1}(x)=\chi(m)^{*}a\chi(n)$, then

[TABLE]

[TABLE]

∎

Lemma 4.15**.**

Let $\tau_{1},M,\chi,\psi_{1}$ be as in Lemma 4.14. Then, there exist $w^{*}$- continuous $*$-isomorphisms $\tau_{k}:A^{**}\rightarrow\tau_{k}(A^{**})$ and TRO homomorphisms $\psi_{k}:\chi(M)\rightarrow\psi_{k}(\chi(M))$ such that if $a\in A^{**},m,n\in M,x\in A^{**}e_{2}$ the equality $a=\psi_{1}(\chi(m))\tau_{1}(x)\psi_{1}(\chi(n))^{*}$ implies that $\tau_{k}(a)=\psi_{k+1}(\chi(m))\tau_{k+1}(x)\psi_{k+1}(\chi(n))^{*}$ and $\tau_{1}(x)=\psi_{1}(\chi(m))^{*}a\psi_{1}(\chi(n))$ implies that $\tau_{k+1}(x)=\psi_{k+1}(\chi(m))^{*}\tau_{k}(a)\psi_{k+1}(\chi(n))$ for all $k=1,2,\ldots$. Thus,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Proof.

Lemma 4.10 implies that given the $*$-isomorphism $\tau_{1}:A^{**}\rightarrow\tau_{1}(A^{**})$, there exist a $w^{*}$-continuous $*$-isomorphism $\tau_{2,0}:A^{**}e_{2}\rightarrow\tau_{2,0}(A^{**}e_{2})$ and a TRO homomorphism $\zeta:\chi(M)\rightarrow\zeta(\chi(M))$ such that if $a\in A^{**},m,n\in M,x\in A^{**}e_{2}$, then the equality $a=\psi_{1}(\chi(m))\tau_{1}(x)\psi_{1}(\chi(n))^{*}$ implies that $\tau_{1}(a)=\zeta(\chi(m))\tau_{2,0}(x)\zeta(\chi(n))^{*}$ and $\tau_{1}(x)=\psi_{1}(\chi(m))^{*}a\psi_{1}(\chi(n))$ implies that $\tau_{2,0}(x)=\zeta(\chi(m))^{*}\tau_{1}(a)\zeta(\chi(n)).$ For every $a\in A^{**},m\in M,$ we define

[TABLE]

If $a\in A^{**},m,n\in M,x\in A^{**}e_{2}$, then the equality $a=\psi_{1}(\chi(m))\tau_{1}(x)\psi_{1}(\chi(n))^{*}$ implies that

[TABLE]

[TABLE]

and the equality $\tau_{1}(x)=\psi_{1}(\chi(m))^{*}a\psi_{1}(\chi(n))$ implies that

[TABLE]

[TABLE]

We continue inductively. ∎

Lemma 4.16**.**

There exist a faithful $*$-homomorphism $\alpha:A^{**}\rightarrow B(L),$ where $L$ is a Hilbert space such that $\overline{\alpha(A)(L)}=L,$ and a $\sigma$-TRO $N\subseteq B(\alpha(e_{2})(L),L)$ such that

[TABLE]

and

[TABLE]

Proof.

We recall the maps $\theta_{1},\tau_{k},\rho_{k}$ from Lemmas 4.13, 4.14, and 4.15. We denote

[TABLE]

for all $a\in A^{**}.$ We also recall the maps $\psi_{k},\phi_{k},\chi$, and for each $m\in M,$ we let $\zeta(m)$ be the $\infty\times\infty$ matrix whose first diagonal under the main diagonal is

[TABLE]

where the other diagonals have zero entries. Clearly, $\zeta(M)$ is a $\sigma$-TRO.

Let $a\in A^{**},x\in A^{**}e_{2},m,n\in M$ be such that $\rho_{1}(a)=\theta_{1}(a)=m^{*}xn.$ Then, by Lemma 4.13 we have that

[TABLE]

Furthermore, following the discussion for the previous Lemma 4.14, we have that $a=\chi(m)\sigma_{1}(x)\chi(n)^{*},$ which by Lemma 4.14 implies that $a=\psi_{1}(\chi(m))\tau_{1}(x)\psi_{1}(\chi(n))^{*}.$ By Lemma 4.15, we have that

[TABLE]

Therefore,

[TABLE]

We conclude that

[TABLE]

Similarly, we can see that

[TABLE]

∎

Lemma 4.17**.**

Let $A$ be a $C^{*}$ algebra and $e_{1},e_{2}\in Z(A^{**})$ be projections such that $Ae_{2}$ is a $C^{*}$- algebra, $A\sim_{\sigma\Delta}Ae_{2}$, and $e_{2}\leq e_{1}\leq e_{0}=id_{A^{**}},e_{2}\neq e_{1}\neq e_{0}.$ Then, there exist central projections $q,p,r\in A^{**}$ such that

[TABLE]

and

[TABLE]

Proof.

From Lemma 4.16, we may assume that

[TABLE]

and there exists a $\sigma$-TRO $M\subseteq B(e_{2}(H),H)$ such that

[TABLE]

and

[TABLE]

By Proposition 2.8 and Theorem 3.3 in [8], there exists a $*$-isomorphism

[TABLE]

such that

[TABLE]

By induction, there exist central projections $\{e_{n}:n\in\mathbb{N}\}\subseteq A^{**}$ such that

[TABLE]

Define

[TABLE]

Then, $e_{0}=p\oplus q.$ If

[TABLE]

then $e_{1}=r\oplus q.$ We define $N=pMr.$ Because

[TABLE]

$N$ is a TRO. Furthermore, the fact that $M$ is a $\sigma$-TRO implies that $N$ is a $\sigma$-TRO. We have that

[TABLE]

Thus, because $pM=M\phi(p),$ we have that

[TABLE]

Similarly, we can prove that $Ap=\overline{[NArN^{*}]}^{\|\cdot\|}.$ Therefore, $Ap\sim_{\sigma\Delta}Ar.$ ∎

Theorem 4.18**.**

Let $A,B$ be $C^{*}$-algebras such that $A\subset_{\sigma\Delta}B,\;\;\;B\subset_{\sigma\Delta}A.$ Assume that $e_{0}=id_{A^{**}},\;\hat{e}_{0}=id_{B^{**}}.$ Then, there exist projections $r\in Z(A^{**}),\hat{r}\in Z(B^{**})$ such that

[TABLE]

Proof.

There exist projections $e_{1}\in Z(A^{**}),f_{1}\in Z(B^{**})$ such that

[TABLE]

Thus, there exists a projection $e_{2}\in Z(A^{**})$ such that $e_{2}\leq e_{1}$ and $Bf_{1}\sim_{\sigma\Delta}Ae_{2}.$ Therefore, $A\sim_{\sigma\Delta}Ae_{2}.$ By Lemma 4.17, there exist projections $p,q,r\in Z(A^{**})$ such that

[TABLE]

and

[TABLE]

Assume that $\psi:K_{\infty}(Ae_{1})\rightarrow K_{\infty}(B)$ is a $*$-isomorphism. Again, by $\psi$ we denote the second dual of $\psi.$ Because $p\leq e_{1},$ there exists $\hat{p}\in Z(B^{**})$ such that $\psi(K_{\infty}(Ae_{1})^{**}p^{\infty})=K_{\infty}(B)^{**}\hat{p}^{\infty}.$ We have that

[TABLE]

Similarly, there exists a projection $\hat{q}\in Z(B^{**})$ such that

[TABLE]

Because $p\perp q$, we have that $\hat{p}\bot\hat{q}.$ Furthermore, because $e_{1}=p\oplus q\Rightarrow\hat{e}_{0}=\hat{p}\oplus\hat{q},$ we conclude that

[TABLE]

and

[TABLE]

We write $\hat{r}$ for $\hat{p}.$ The proof is now complete. ∎

5. Examples in the non-self-adjoint case

In this section, we will present a counterexample of two non-self-adjoint operator algebras $\hat{A},\hat{B}$ such that $\hat{A}\subset_{\sigma\Delta}\hat{B},\;\;\hat{B}\subset_{\sigma\Delta}\hat{A}$ but $\hat{A}$ and $\hat{B}$ are not $\sigma\Delta$-strongly equivalent.

Let $\mathcal{N},\mathcal{M}$ be nests acting on the separable Hilbert spaces $H_{1}$ and $K_{1}$, respectively. These nests are called similar if there exists an invertible operator $s:H_{1}\rightarrow K_{1}$ such that

[TABLE]

In this case, the map

[TABLE]

is a nest isomorphism. This means that $\theta_{s}$ is one-to-one, onto, and order-preserving. We can easily check that $\mathrm{Alg}(\mathcal{M})=s\mathrm{Alg}(\mathcal{N})s^{-1}.$ If $n\in\mathcal{N},$ we write

[TABLE]

In the case where $n_{-}$ is strictly contained in $n,$ the projection $a=n-n_{-}$ is called an atom of $\mathcal{N}.$

Theorem 5.1**.**

[5, 13.20]** The nests $\mathcal{N},\mathcal{M}$ are similar if and only if there exists a nest isomorphism $\theta:\mathcal{N}\rightarrow\mathcal{M}$ such that

[TABLE]

for all $n\in\mathcal{N}.$

The lemmas 5.2 and 5.3 can be inferred from Section 13 in [5]. We present the proofs here for completeness.

Lemma 5.2**.**

Let $\mathcal{N},\mathcal{M}$ be separably-acting nests, and $\theta:\mathcal{N}\rightarrow\mathcal{M}$ be a nest isomorphism preserving the dimensions of the atoms. For every $0<\epsilon<1$, there exists an invertible operator $s,$ a unitary $u$, and a compact operator $k$ such that

[TABLE]

and $\theta=\theta_{s}.$

Proof.

By Theorem [5, 13.20], there exists a compact operator $k,$ a unitary $u$, an invertible operator $s=u+k,$ such that $\theta=\theta_{s}$ and $\|k\|<\frac{\epsilon}{1+\epsilon}.$ Observe that $\|k\|<\epsilon.$ We have that $u^{*}s=I+u^{*}k\Rightarrow\|I-u^{*}s\|<\epsilon.$ Therefore,

[TABLE]

We conclude that

[TABLE]

We have that

[TABLE]

∎

Lemma 5.3**.**

Let $\mathcal{N},\mathcal{M}$ be separably-acting nests and $\theta:\mathcal{N}\rightarrow\mathcal{M}$ be a nest isomorphism preserving the dimensions of the atoms. For every $0<\epsilon<1$, there exists an invertible operator $s,$ a unitary $u,$ and compact operators $k,l$ such that

[TABLE]

and $\theta=\theta_{s}.$

Proof.

Choose $0<\delta<1$ such that $(1+\delta)\delta<\epsilon,\;\;\delta<\epsilon.$ By Lemma 5.2, there exist a unitary $u$ and compact $k$ such that $s=u+k$ is an invertible operator and $\|k\|<\delta,\;\|s^{-1}\|<1+\delta,\;\theta=\theta_{s}.$ Define $l_{0}=-u^{*}ks^{-1}u.$ We have that

[TABLE]

[TABLE]

Because $\|u^{*}k\|<\delta<1,$ the operator $I+u^{*}k$ is invertible, and thus

[TABLE]

By (5.1), we have that

[TABLE]

If $l=l_{0}u^{*},$ then $l$ is a compact operator, and

[TABLE]

Thus, $s^{-1}=u^{*}+l$ and $\|l\|<\epsilon.$ ∎

In the following, we fix similar nests $\mathcal{N}$ and $\mathcal{M}$ acting on the Hilbert spaces $H_{1}$ and $H_{2}$, respectively, and a nest isomorphism $\theta:\mathcal{N}\rightarrow\mathcal{M}$ preserving the dimensions of atoms. Suppose that $a_{i}=n_{i}-(n_{i})_{-},\;\;b_{i}=\theta(n_{i})-\theta(n_{i})_{-},i=1,2,3,...$ are the atoms of $\mathcal{N}$ and $\mathcal{M}$, respectively. We also assume that $p=\vee_{i}a_{i}$, $p$ is strictly contained in $I_{H_{1}},I_{H_{2}}=\vee_{i}b_{i}$, and $dim(a_{i})=dim(b_{i})<+\infty$ for all $i.$ By Lemma 5.3, there exists a sequence of invertible operators $(s_{n})_{n}$ such that $\theta=\theta_{s_{n}},$ a sequence of unitary $(u_{n})_{n}$, and sequences of compact operators $(k_{n})_{n},(l_{n})_{n}$ such that

[TABLE]

for all $n\in\mathbb{N}$ and $\|k_{n}\|\rightarrow 0,\|l_{n}\|\rightarrow 0.$ We can also assume that $\|s_{n}\|<2,\;\;\|s^{-1}_{n}\|<2$ for all $n\in\mathbb{N},$ and

[TABLE]

Lemma 5.4**.**

(i)

[TABLE]

(ii)

[TABLE]

Proof.

We shall prove (i), while statement (ii) follows by symmetry. Fix $i\in\mathbb{N}$, and assume that

[TABLE]

for $x_{j},y_{j}\in H_{1}.$ For all $\xi\in H_{1}$, we have that

[TABLE]

Thus,

[TABLE]

If $\xi\in H_{1}$ for all $k\in\mathbb{N},$ we have that

[TABLE]

Fix $\epsilon>0.$ Then, there exists $k_{0}\in\mathbb{N}$ such that $\sum_{i>k_{0}}^{\infty}\left\|a_{i}(\xi)\right\|^{2}<\epsilon.$ Thus,

[TABLE]

We let $n\rightarrow\infty,$ and we have that

[TABLE]

Thus,

[TABLE]

∎

Lemma 5.5**.**

For every $j,i\in\mathbb{N}$, we have that

[TABLE]

Proof.

Because

[TABLE]

if $\xi\in a_{i}(H_{1})$, then $\xi=n_{i}(\xi)-(n_{i})_{-}(\xi).$ Thus, there exist $\xi_{j},\;\omega_{j}\;\in H_{2},$ such that

[TABLE]

Because $b_{i}=\theta(n_{i})-\theta(n_{i})_{-},$ we have that $b_{i}s_{j}(\xi)=b_{i}(\xi_{j}).$ Therefore,

[TABLE]

However, $s_{j}^{-1}(\theta(n_{i})_{-}(H_{2}))=(n_{i})_{-}(H_{1}).$ Thus, there exists $\phi_{j}\;\in\;H_{1},$ such that

[TABLE]

We have proved that

[TABLE]

which implies that

[TABLE]

Because $a_{i}(n_{i})_{-}=0$, we have that

[TABLE]

Because

[TABLE]

we obtain

[TABLE]

∎

Lemma 5.6**.**

Let $s_{0}$ be as in Lemma 5.4. Then,

[TABLE]

Proof.

We shall prove that $s_{0}^{*}s_{0}=p.$ Because the span of the atoms of $\mathcal{M}$ is $I_{H_{2}},$ the other equality follows from symmetry. By Lemma 5.4, we have that

[TABLE]

Thus,

[TABLE]

Because $p=\vee_{i}a_{i},$ Lemma 5.5 implies that $s_{0}^{*}s_{0}=p.$ ∎

Suppose that $A$ (resp. $B$) is the subalgebra of compact operators of the algebra $\mathrm{Alg}(\mathcal{N})$ (resp. $\mathrm{Alg}(\mathcal{M})$ ). It is well-known that $\mathrm{Alg}(\mathcal{N})=A^{**},\mathrm{Alg}(\mathcal{M})=B^{**}.$ We define a map $\rho:B\rightarrow A$ such that

[TABLE]

Because $s_{0}s_{0}^{*}=I_{H_{2}}$, this map is a homomorphism. If $k\in A$, then

[TABLE]

Thus $\rho(B)=pAp.$ Because $p\in\Delta(A^{**})^{\prime}=Z(\Delta(A^{**}))$, we have that $B\subset_{\sigma\Delta}A.$

In the following, we additionally assume that the dimensions of the atoms of $\mathcal{N}$ and $\mathcal{M}$ are one and that $\Delta(A^{**}),\;\Delta(B^{**})$ are maximal abelian self-adjoint algebras (MASAs). Such nests exist, see, for instance, Example 13.15 in [5]. We denote the algebras

[TABLE]

Because $B\subset_{\sigma\Delta}A$, we have that $\hat{B}\subset_{\sigma\Delta}\hat{A}.$ Furthermore,

[TABLE]

Thus, $\hat{A}\subset_{\sigma\Delta}\hat{B}.$ If $\subset_{\sigma\Delta}$ was a partial-order relation for non-self-adjoint algebras, then up to stable isomorphism we should have that $\hat{A}\sim_{\sigma\Delta}\hat{B}.$ Thus, the algebras

[TABLE]

would be weakly stably isomorphic. Because $\Omega$ and $\Xi$ are CSL algebras (see the definition of a CSL algebra in [5]), it follows from Theorem 3.2 in [7] and Theorem 3.3 in [8] that there would exists a $*$-isomorphism

[TABLE]

such that $\theta(\mathrm{Lat}(\Omega))=\mathrm{Lat}(\Xi).$ However, $\Delta(\Omega)$ and $\Delta(\Xi)$ are MASAs, and thus there exists a unitary $u$ such that

[TABLE]

Therefore,

[TABLE]

There exist completely contractive homomorphisms $\rho_{k}:B^{**}\rightarrow A^{**},\;\;k=1,2,...$ such that

[TABLE]

Suppose that

[TABLE]

Because

[TABLE]

for all $i,$ we have that

[TABLE]

Thus,

[TABLE]

for orthogonal projections $\hat{p}_{i}\in\Delta(B^{**}),i\in\mathbb{N}.$ Observe that $\hat{p}_{i}\hat{p}_{j}=0$ for $i\neq j.$ If $x\in B^{**},$ then

[TABLE]

Thus,

[TABLE]

We conclude that

[TABLE]

Thus, for all $x\in B^{**},$ we have that

[TABLE]

Therefore, $\hat{p}_{i}$ is in the center of $B^{**}.$ However, as a nest algebra, $B^{**}$ has a trivial center. We can, therefore, conclude that there exists $i$ such that

[TABLE]

and $\hat{p}_{j}=0$ for all $j\neq i.$ We obtain that

[TABLE]

By the same arguments, for the nest algebra $A^{**}$ there exists exactly one of the algebras $q_{1}B^{**}q_{1},q_{2}A^{**}q_{2},q_{3}A^{**}q_{3},...$ with $q_{1}\in B^{**},q_{k}\in A^{**},k\geq 2$ such that

[TABLE]

or

[TABLE]

Here, in the left-hand side, $A^{**}$ is in the $i$-th position. The equality (5.2) implies that

[TABLE]

We have proven that

[TABLE]

and

[TABLE]

We conclude that

[TABLE]

Thus, the nest algebras $A^{**}$ and $B^{**}$ are completely isometrically isomorphic. It follows that their diagonals $\Delta(A^{**})$ and $\Delta(B^{**})$ are $*$-isomorphic. However, $\Delta(B^{**})$ is an atomic MASA, and $\Delta(A^{**})$ is a MASA with a nontrivial continuous part. This contradiction shows that $\hat{A}$ and $\hat{B}$ are not $\sigma$-strongly $\Delta$-equivalent.