Equivalence bundles over a finite group and strong Morita equivalence for unital inclusions of unital $C^*$-algebras

TL;DR

This paper investigates the relationship between $C^*$-algebraic bundles over finite groups and strong Morita equivalence of their associated unital inclusions, establishing conditions for equivalence and automorphisms.

Contribution

It introduces conditions under which equivalence bundles imply strong Morita equivalence of algebra inclusions and characterizes automorphisms related to these equivalences.

Findings

Equivalence bundles over finite groups induce strong Morita equivalence of algebra inclusions.

Strong Morita equivalence of inclusions implies existence of automorphisms and equivalence bundles.

Conditions for saturated bundles and trivial commutant are established for equivalence.

Abstract

Let $\mathcal{A}= \{A_t \}_{t \in G}$ and $\mathcal{B}= \{B_t \}_{t\in G}$ be $C^*$-algebraic bundles over a finite group $G$. Let $C=\oplus_{t \in G}A_t$ and $D=\oplus_{t\in G}B_t$. Also, let $A=A_e$ and $B=B_e$, where $e$ is the unit element in $G$. We suppose that $C$ and $D$ are unital and $A$ and $B$ have the unit elements in $C$ and $D$, respectively. In this paper, we shall show that if there is an equivalence $\mathcal{A}-\mathcal{B}$-bundle over $G$ with some properties, then the unital inclusions of unital $C^*$-algebras $A \subset C$ and $B \subset D$ induced by $\mathcal{A}$ and $\mathcal{B}$ are strongly Morita equivalent. Also, we suppose that $\mathcal{A}$ and $\mathcal{B}$ are saturated and that $A' \cap C= \mathbf{C} 1$. We shall show that if $A \subset C$ and $B \subset D$ are strongly Morita equivalent, then there are an automorphism $f$ of $G$ and an equivalence bundle $\mathcal{A}-\mathcal{B}^f $-bundle over $G$ with the some properties, where $\mathcal{B}^f$ is the $C^*$-algebraic bundle induced by $\mathcal{B}$ and $f$, which is defined by $\mathcal{B}^f = \{B_{f(t)} \}_{t \in G}$. Furthermore, we shall give an application.

Equations

$$E^A(x)=x_e$$

$${}_C⟨X_t,X_s⟩=A_{t{s}^{-1}},⟨X_t,X_s{⟩}_D=B_{t^{-1}s}$$

$$\{{}_C⟨x,y⟩\in A_{t{s}^{-1}}|x\in X_t,y\in X_s\}$$

$${}_C⟨X_t,X_s⟩=A_{t{s}^{-1}},⟨X_t,X_s{⟩}_D=B_{t^{-1}s}$$

$${}_C⟨Y,X⟩=C,⟨Y,X{⟩}_D=D.$$

$${}_C⟨y,x⟩=\sum _{t\in G}{}_C⟨y_t,x⟩,⟨y,x{⟩}_D=\sum _{t\in G}⟨y_t,x{⟩}_D.$$

$${}_C⟨Y,X⟩=C,⟨Y,X{⟩}_D=D.$$

$$(y,d)\in Y\times D\mapsto y\cdot d\in Y,(y,z)\in Y\times Y\mapsto ⟨y,z{⟩}_D\in D$$

$$(c,y)\in C\times Y\mapsto c\cdot y\in Y,(y,z)\in Y\times Y\mapsto {}_C⟨y,z⟩\in C,$$

$${}_C⟨X_t,X_s⟩=A_{t{s}^{-1}},⟨X_t,X_s{⟩}_D=B_{t^{-1}s}$$

$$A{⋊}_{\alpha }G=\{\sum _{t\in G}{a}_t{u}_t|a_t\in A\text{for any t∈G}\}.$$

$$X{⋊}_{\lambda }G=\{\sum _{t\in G}{x}_t{w}_t|x_t\in X\text{for any t∈G}\},$$

$$(a{u}_t)\cdot (x{w}_s)=(a\cdot {\lambda }_t(x))w_{ts}(x{w}_s)\cdot (b{v}_t)=(x\cdot {\beta }_s(b))v_{st}$$

$${}_C⟨x{w}_t,y{w}_s⟩={}_A⟨x,{\lambda }_{t{s}^{-1}}(y)⟩u_{t{s}^{-1}},⟨x{w}_t,y{w}_s{⟩}_D={\beta }_{t^{-1}}(⟨x,y{⟩}_B)v_{t^{-1}s}$$

$${}_C⟨X_t,X_s⟩=A_{t{s}^{-1}},⟨X_t,X_s{⟩}_D=B_{t^{-1}s}$$

$$(x,y)\in Y_{{\alpha }_t^{\mathcal{A}}}\times Y_{{\alpha }_s^{\mathcal{A}}}$$

$$x\in Y_{{\alpha }_t^{\mathcal{A}}}$$

$$Y_{{\alpha }_t^{\mathcal{A}}}=e_{A}C{e}_{A}C{e}_{t^{-1}}=e_{A}AC{e}_{t^{-1}}=e_{A}C{e}_{t^{-1}}.$$

$$e_{A}x{e}_{t^{-1}}$$

$$=\sum _i{x}_t{x}_i^{t^{-1}}{e}_A{x}_i^{t^{-1}\ast }=e_A{x}_t\sum _i{x}_i^{t^{-1}}{x}_i^{t^{-1}\ast }=e_A{x}_t.$$

$${\pi }_t(x)=e_{A}x$$

$$||{\pi }_t(x)|{|}^2=||e_{A}x{x}^{\ast }{e}_A||=||E^A(x{x}^{\ast })e_A||=||E^A(x{x}^{\ast })||=||x{x}^{\ast }||=||x|{|}^2.$$

$${\pi }_t(x)\bullet {\pi }_s(y)$$

$$=e_A\sum _{i}x{x}_i^{t^{-1}}{x}_i^{t^{-1}\ast }y=e_{A}xy={\pi }_{ts}(xy),$$

$${\pi }_t(x{)}^{♯}$$

$$=\sum _i{x}^{\ast }{x}_i^t{e}_A{x}_i^{t\ast }=e_A\sum _i{x}^{\ast }{x}_i^t{x}_i^{t\ast }=e_A{x}^{\ast }={\pi }_{t^{-1}}(x^{\ast }).$$

$$e_{A}x{\alpha }_t^{\mathcal{A}}(e_A)\diamond [e_A\cdot z\cdot {\beta }_s(e_B)]$$

$$=e_{A}x{\alpha }_t^{\mathcal{A}}(e_A)\cdot {\lambda }_t(z)\cdot {\beta }_{ts}(e_B)$$

$$=e_A\cdot [x{\alpha }_t^{\mathcal{A}}(e_A)\cdot {\lambda }_t(z)]\cdot {\beta }_{ts}(e_B),$$

$${}_C⟨e_A\cdot z\cdot {\beta }_t(e_B),e_A\cdot w\cdot {\beta }_s(e_B)⟩$$

$$=e_A{}_{C_1}⟨z\cdot {\beta }_t(e_B),{\lambda }_{t{s}^{-1}}(w)\cdot {\beta }_t(e_B)⟩{\alpha }_{t{s}^{-1}}^{\mathcal{A}}(e_A),$$

Full text

1. Intrtoduction

Let $\mathcal{A}=\{A_{t}\}_{t\in G}$ be a $C^{*}$-algebraic bundle over a finite group $G$. Let $C=\oplus_{t\in G}A_{t}$ and $A_{e}=A$, where $e$ is the unit element in $G$. We suppose that $C$ is unital and that $A$ has the unit element in $C$. Then we obtain a unital inclusion of unital $C^{*}$-algebras, $A\subset C$. We call it the unital inclusion of unital $C^{*}$-algebras induced by a $C^{*}$-algebraic bundle $\mathcal{A}=\{A_{t}\}_{t\in G}$. Let $E^{A}$ be the canonical conditional expectation from $C$ onto $A$ defined by

[TABLE]

for all $x=\sum_{t\in G}x_{i}\in C$.

Definition 1.1*.*

Let $\mathcal{A}=\{A_{t}\}_{t\in G}$ be a $C^{*}$-algebraic bundle over a finite group $G$. We say that $\mathcal{A}$ is saturated if $\overline{A_{t}A_{t}^{*}}=A$ for all $t\in G$.

Since $A$ is unital, in our case we do not need to take the closure in Definition 1.1. If $\mathcal{A}$ is saturated, by [10, Corollary 3.2], $E^{A}$ is of index-finite type and its Watatani index, ${\rm{Ind}}_{W}(E^{A})=|G|$, where $|G|$ is the order of $G$.

Let $\mathcal{B}=\{B_{t}\}_{t\in G}$ be another $C^{*}$-algebraic bundle over $G$. Let $D=\oplus_{t\in G}B_{t}$ and $B=B_{e}$. Also, we suppose that $\mathcal{B}$ has the same conditions as $\mathcal{A}$. Let $B\subset D$ be the unital inclusion of unital $C^{*}$-algebras induced by $\mathcal{B}$.

Let $\mathcal{X}=\{X_{t}\}_{t\in G}$ be an $\mathcal{A}-\mathcal{B}$-equivalence bundle defined by Abadie and Ferraro [1, Definition 2.2]. Moreover, we suppose that

[TABLE]

for any $t,s\in G$, where ${}_{C}\langle X_{t}\,,\,X_{s}\rangle$ means the linear span of the set

[TABLE]

and $\langle X_{t}\,,\,X_{s}\rangle_{D}$ means the linear span of the similar set to the above. The above two properties are stronger than the properties (7R) and (7L) in [1, Definition 2.1].

In the present paper, we shall show that if there is an $\mathcal{A}-\mathcal{B}$-equivalence bundle $\mathcal{X}=\{X_{t}\}_{t\in G}$ such that ${}_{C}\langle X_{t}\,,\,X_{s}\rangle=A_{ts^{-1}}$ and $\langle X_{t}\,,\,X_{s}\rangle_{D}=B_{t^{-1}s}$ for any $t,s\in G$, then the unital inclusions of unital $C^{*}$-algebras $A\subset C$ and $B\subset D$ induced by $\mathcal{A}$ and $\mathcal{B}$ are strongly Morita equivalent. Also, we suppose that $\mathcal{A}$ and $\mathcal{B}$ are saturated and that $A^{\prime}\cap C=\mathbf{C}1$. We shall show that if $A\subset C$ and $B\subset D$ are strongly Morita equivalent, then there are an automorphism $f$ of $G$ and an $\mathcal{A}-\mathcal{B}^{f}$-equivalence bundle $\mathcal{X}=\{X_{t}\}_{t\in G}$ such that ${}_{C}\langle X_{t}\,,\,X_{s}\rangle=A_{ts^{-1}}$ and $\langle X_{t}\,,\,X_{s}\rangle_{D}=B_{f(t^{-1}s)}$ for any $t,s\in G$, where $\mathcal{B}^{f}$ is the $C^{*}$-algebraic bundle induced by $\mathcal{B}=\{B_{t}\}_{t\in G}$ and $f$, which is defined by $\mathcal{B}^{f}=\{B_{f(t)}\}_{t\in G}$.

Let $A$ and $B$ be unital $C^{*}$-algebras and $X$ an $A-B$-equivalence bimodule. Then we denote its left $A$-action and right $B$-action on $X$ by $a\cdot x$ and $x\cdot b$ for any $a\in A$, $b\in B$ and $x\in X$, respectively. Also, we mean by the words “Hilbert $C^{*}$-bimodules” Hilbert $C^{*}$-bimodules in the sense of Brown, Mingo and Shen [3].

2. Equivalence bundles over a finite group

Let $\mathcal{A}=\{A_{t}\}_{t\in G}$ and $\mathcal{B}=\{B_{t}\}_{t\in G}$ be $C^{*}$-algebraic bundles over a finite group $G$. Let $e$ be the unit element in $G$. Let $C=\oplus_{t\in G}A_{t}$, $D=\oplus_{t\in G}B_{t}$ and $A=A_{e}$, $B=B_{e}$. We suppose that $C$ and $D$ are unital and that $A$ and $B$ have the unit elements in $C$ and $D$, respectively. Let $\mathcal{X}=\{X_{t}\}_{t\in G}$ be an $\mathcal{A}-\mathcal{B}$-equivalence bundle over $G$ such that

[TABLE]

for any $t,s\in G$. Let $Y=\oplus_{t\in G}X_{t}$ and $X=X_{e}$. Then $Y$ is a $C-D$-equivalence bimodule by Abadie and Ferraro [1, Definitions 2.1, 2.2 ], Also, $X$ is an $A-B$-equivalence bimodule since ${}_{C}\langle X\,,\,X\rangle=A$ and $\langle X\,,\,X\rangle_{D}=B$.

Proposition 2.1**.**

Let $\mathcal{A}=\{A_{t}\}_{t\in G}$ and $\mathcal{B}=\{B_{t}\}_{t\in G}$ be $C^{*}$-algebraic bundles over a finite group $G$. Let $C=\oplus_{t\in G}A_{t}$ and $D=\oplus_{t\in G}B_{t}$ Also, let $A=A_{e}$ and $B=B_{e}$, where $e$ is the unit element in $G$. We suppose that $C$ and $D$ are unital and that $A$ and $B$ have the unit elements in $C$ and $D$, respectively. Also, we suppose that there is an $\mathcal{A}-\mathcal{B}$- equivalence bundle $\mathcal{X}=\{X_{t}\}_{t\in G}$ over $G$ such that

[TABLE]

for any $t,s\in G$. Then the unital inclusions of unital $C^{*}$-algebras $A\subset C$ and $B\subset D$ are strongly Morita equivalent.

Proof.

Let $Y=\oplus_{t\in G}X_{t}$ and $X=X_{e}$. By the above discussions and [11, Definition 2.1], we have only to show that

[TABLE]

Let $x\in X$ and $y=\sum_{t\in G}y_{t}\in Y$, where $y_{t}\in X_{t}$ for any $t\in G$. Then

[TABLE]

We note that ${}_{C}\langle y_{t}\,,x\rangle\in A_{t}$ and $\langle y_{t}\,,x\rangle_{D}\in B_{t}$ for any $t\in G$. Since ${}_{D}\langle X_{t}\,,\,X_{s}\rangle=A_{ts^{-1}}$ and $\langle X_{t}\,,\,X_{s}\rangle_{D}=B_{t^{-1}s}$ for any $t,s\in G$, by the above computations, we can see that

[TABLE]

Therefore, we obtain the conclusion. ∎

Next, we shall give an example of an equivalence bundle $\mathcal{X}=\{X_{t}\}_{t\in G}$ over $G$ satisfying the above properties. In order to do this, we prepare a lemma. Let $\mathcal{A}=\{A_{t}\}_{t\in G}$ and $\mathcal{B}=\{B_{t}\}_{t\in G}$ be as above. Let $\mathcal{X}=\{X_{t}\}_{t\in G}$ be a complex Banach bundle over $G$ with the maps defined by

[TABLE]

where $Y=\oplus_{t\in G}X_{t}$.

Lemma 2.2**.**

With the above notation, we suppose that by the above maps, $Y$ is a $C-D$-equivalence bimodule satisfying that

[TABLE]

for any $t,s\in G$. If $\mathcal{X}$ satisfies Conditions $(1R)$-$(3R)$ and $(1L)$-$(3L)$ in [1, Definition 2.1], then $\mathcal{X}$ is an $\mathcal{A}-\mathcal{B}$-equivalence bundle.

Proof.

Since $Y$ is a $C-D$-equivalence bimodule, $\mathcal{X}$ has Conditions (4R)-(6R) and (4L)-(6L) in [1, Definiton 2.1] except that $X_{t}$ is complete with the norms $||\langle-,-\rangle_{D}||^{\frac{1}{2}}=||{}_{C}\langle-,-\rangle||^{\frac{1}{2}}$ for any $t\in G$. But we know that if $Y$ is complete with two different norms, then the two norms are equivalent. Hence $X_{t}$ is complete with the norms $||\langle-,-\rangle_{D}||^{\frac{1}{2}}=||{}_{C}\langle-,-\rangle||^{\frac{1}{2}}$ for any $t\in G$. Furthermore, since

[TABLE]

for any $t,s\in G$, $\mathcal{X}$ has Conditions (7R) and (7L) in [1, Definiton 2.1]. Therefore, we obtain the conclusion. ∎

We give an example of an $\mathcal{A}-\mathcal{B}$-equivalence bundle $\mathcal{X}=\{X_{t}\}_{t\in G}$ such that

[TABLE]

for any $t,s\in G$.

Example 2.3**.**

Let $G$ be a finite group. Let $\alpha$ be an action of $G$ on a unital $C^{*}$-algebra $A$. Let $u_{t}$ be implementing unitary elements of $\alpha$, that is, $\alpha_{t}={\rm{Ad}}(u_{t})$ for any $t\in G$. Then the crossed product of $A$ by $\alpha$, $A\rtimes_{\alpha}G$ is:

[TABLE]

Let $A_{t}=Au_{t}$ for any $t\in G$. By routine computations, we see that $\mathcal{A}_{\alpha}=\{A_{t}\}_{t\in G}$ is a $C^{*}$-algebraic bundle over $G$. We call $\mathcal{A}_{\alpha}$ the $C^{*}$-algebraic bundle over $G$ induced by an action $\alpha$. Let $\beta$ be an action of $G$ on a unital $C^{*}$-algebra $B$ and let $\mathcal{A}_{\beta}=\{B_{t}\}_{t\in G}$ induced by $\beta$, where $B_{t}=Bv_{t}$ for any $t\in G$ and $v_{t}$ are implementing unitary elements of $\beta$. We suppose that $\alpha$ and $\beta$ are strongly Morita equivalent with respect to an action $\lambda$ of $G$ on an $A-B$-equivalence bimodule $X$. Let $X\rtimes_{\lambda}G$ be the crossed product of $X$ by $\lambda$ defined in Kajiwara and Watatani [6, Definition 1.4], that is, the direct sum of $n$-copies of $X$ as a vector space, where $n$ is the order of $G$. And its elements are written as formal sums so that

[TABLE]

where $w_{t}$ are indeterminates for all $t\in G$. Let $C=A\rtimes_{\alpha}G$, $D=B\rtimes_{\beta}G$ and $Y=X\rtimes_{\lambda}G$. Then by [6, Proposition 1.7], $Y$ is a $C-D$-equivalence bimodule, where we define the left $C$-action and the right $D$-action on $Y$ by

[TABLE]

for any $a\in A$, $b\in B$, $x\in X$ and $t,s\in G$ and we define the left $C$-valued inner product and the right $D$-valued inner product on $Y$ by extending linearly the following:

[TABLE]

for any $x,y\in X$, $t,s\in G$. Let $X_{t}=Xw_{t}$ for any $t\in G$ and $\mathcal{X}_{\lambda}=\{X_{t}\}_{t\in G}$. Then $Y=\oplus_{t\in G}X_{t}$. Also, $\mathcal{X}_{\lambda}$ has Conditions (1R)-(3R) and (1L)-(3L) in [1, Definition 2.1]. Furthermore, $X$ is an $A-B$-equivalence bimodule, $\mathcal{X}_{\lambda}$ and satisfies

[TABLE]

for any $t,s\in G$. Therefore, $\mathcal{X}_{\lambda}$ is an $\mathcal{A}_{\alpha}-\mathcal{A}_{\beta}$- equivalence bundle by Lemma 2.2.

3. Saturated $C^{*}$-algebraic bundles over a finite group

Let $\mathcal{A}=\{A_{t}\}_{t\in G}$ be a saturated $C^{*}$-algebraic bundle over a finite group $G$. Let $e$ be the unit element in $G$. Let $C=\oplus_{t\in G}A_{t}$ and $A=A_{e}$. We suppose hat $C$ is unital and that $A$ has the unit element in $C$. Let $E^{A}$ be the canonical conditional expectation from $C$ onto $A$ defined in Section 1, which is of Watatani index-finite type. Let $C_{1}$ be the $C^{*}$-basic construction of $C$ and $e_{A}$ the Jones’ projection for $E^{A}$. By [10, Lemma 3.7], there is an action $\alpha^{\mathcal{A}}$ of $G$ on $C_{1}$ induced by $\mathcal{A}$ defined as follows: Since $\mathcal{A}$ is saturated and $A$ is unital, there is a finite set $\{x_{i}^{t}\}_{i=1}^{n_{t}}\subset A_{t}$ such that $\sum_{i=1}^{n_{t}}x_{i}^{t}x_{i}^{t*}=1$ for any $t\in G$. Let $e_{t}=\sum_{i=1}^{n_{t}}x_{i}^{t}e_{A}x_{i}^{t*}$ for all $t\in G$. Then by [10, Lemmas 3.3, 3.5 and Remark 3.4], $\{e_{t}\}_{t\in G}$ are mutually orthogonal projections in $A^{\prime}\cap C_{1}$, which are independent of the choice of $\{x_{i}^{t}\}_{i=1}^{n_{t}}$, with $\sum_{t\in G}e_{t}=1$ such that $C$ and $e_{t}$ generate the $C^{*}$-algebra $C_{1}$ for all $t\in G$. We define $\alpha^{\mathcal{A}}$ by $\alpha_{t}^{\mathcal{A}}(c)=c$ and $\alpha_{t}^{\mathcal{A}}(e_{A})=e_{t^{-1}}$ for any $t\in G$, $c\in C$. Let $\mathcal{A}_{1}=\{Y_{\alpha_{t}^{\mathcal{A}}}\}_{t\in G}$ be the $C^{*}$-algebraic bundle over $G$ induced by the action $\alpha^{\mathcal{A}}$ of $G$ which is defined in [10, Sections 5, 6], that is, let $Y_{\alpha_{t}^{\mathcal{A}}}=e_{A}C_{1}\alpha_{t}^{\mathcal{A}}(e_{A})=e_{A}C_{1}e_{t^{-1}}$ for any $t\in G$. The product $\bullet$ and the involution $\sharp$ in $\mathcal{A}_{1}$ are defined as follows:

[TABLE]

Lemma 3.1**.**

With the above notation, $\mathcal{A}$ and $\mathcal{A}_{1}$ are isomorphic as $C^{*}$-algebraic bundles over $G$.

Proof.

Since $C_{1}=Ce_{A}C$, for any $t\in G$

[TABLE]

Let $x$ be any element in $C$. Then we can write that $x=\sum_{s\in G}x_{s}$, where $x_{s}\in A_{s}$. Hence

[TABLE]

Thus $Y_{\alpha_{t}^{\mathcal{A}}}=e_{A}Ce_{t^{-1}}=e_{A}A_{t}$ for any $t\in G$. Let $\pi_{t}$ be the map from $A_{t}$ to $Y_{\alpha_{t}^{\mathcal{A}}}$ defined by

[TABLE]

for any $x\in A_{t}$ and $t\in G$. By the above discussions $\pi_{t}$ is a linear map from $A_{t}$ onto $Y_{\alpha_{t}^{\mathcal{A}}}$. Then

[TABLE]

Hence $\pi_{t}$ is injective for any $t\in G$. Thus $A_{t}\cong e_{A}C_{1}\alpha_{t}^{\mathcal{A}}(e_{A})$ as Banach spaces for any $t\in G$. Also, for any $x\in A_{t}$, $y\in A_{s}$, $t,s\in G$,

[TABLE]

[TABLE]

Therefore, $\mathcal{A}=\{A_{t}\}_{t\in G}$ and $\mathcal{A}_{1}=\{Y_{\alpha_{t}^{\mathcal{A}}}\}_{t\in G}$ are isomorphic as $C^{*}$-algebraic bundles over $G$. ∎

4. Strong Morita equivalence for unital inclusions of unital $C^{*}$-algberas

Let $\mathcal{A}=\{A_{t}\}_{t\in G}$ and $\mathcal{B}=\{B_{t}\}_{t\in G}$ be saturated $C^{*}$-algebraic bundles over a finite group $G$. Let $e$ be the unit element in $G$. Let $C=\oplus_{t\in G}A_{t}$, $D=\oplus_{t\in G}B_{t}$ and $A=A_{e}$, $B=B_{e}$. We suppose that $C$ and $D$ are unital and that $A$ and $B$ have the unit elements in $C$ and $D$, respectively. Let $E^{A}$ and $E^{B}$ be the canonical conditional expectations from $C$ and $D$ onto $A$ and $B$ defined in Section 1, respectively. They are of Watatani index-finite type. Let $A\subset C$ and $B\subset D$ be the unital inclusions of unital $C^{*}$-algebras induced by $\mathcal{A}$ and $\mathcal{B}$, respectively. We suppose that $A\subset C$ and $B\subset D$ are strongly Morita equivalent with respect to a $C-D$-equivalence bimodule $Y$ and its closed subspace $X$. Also, we suppose that $A^{\prime}\cap C=\mathbf{C}1$. Then by [11, Lemma 10.3], $B^{\prime}\cap D=\mathbf{C}1$ and by [8, Lemma 4.1] and its proof, there is the unique conditional expectation $E^{X}$ from $Y$ onto $X$ with respect to $E^{A}$ and $E^{B}$. Let $C_{1}$ and $D_{1}$ be the $C^{*}$-basic constructions of $C$ and $D$ and $e_{A}$ and $e_{B}$ the Jones’ projections for $E^{A}$ and $E^{B}$, respectively. Then by [10, Lemma 3.7], there are actions $\alpha^{\mathcal{A}}$ and $\alpha^{\mathcal{B}}$ of $G$ on $C_{1}$ and $D_{1}$ induced by $\mathcal{A}$ and $\mathcal{B}$, respectively. Furthermore, let $C_{2}$ and $D_{2}$ be the $C^{*}$-basic constructions of $C_{1}$ amd $D_{1}$ for the dual conditional expectations $E^{C}$ of $E^{A}$ and $E^{D}$ of $E^{B}$, which are isomorphic to $C_{1}\rtimes_{\alpha^{\mathcal{A}}}G$ and $D_{1}\rtimes_{\alpha^{\mathcal{B}}}G$, respectively. We identify $C_{2}$ and $D_{2}$ with $C_{1}\rtimes_{\alpha^{\mathcal{A}}}G$ and $D_{1}\rtimes_{\alpha^{\mathcal{B}}}G$, respectively. By [11, Corollary 6.3], the unital inclusions $C_{1}\subset C_{2}$ and $D_{1}\subset D_{2}$ are strongly Morita equivalent with respect to a $C_{2}-D_{2}$-equivalence bimodule $Y_{2}$ and its closed subspace $Y_{1}$, where $Y_{1}$ and $Y_{2}$ are the $C_{1}-D_{1}$-equivalence bimodule and the $C_{2}-D_{2}$-equivalence bimodule defined in [11, Section 6], respectively and $Y_{1}$ is regarded as a closed subspace of $Y_{2}$ in the same way as in [11, Section 6]. Also, ${C_{1}}^{\prime}\cap C_{2}=\mathbf{C}1$ by the proof of Watatani [15, Proposition 2.7.3] since $A^{\prime}\cap C=\mathbf{C}1$. Hence by [12, Corollary 6.5], there is an automorphism $f$ of $G$ such that $\alpha^{\mathcal{A}}$ is strongly Morita equivalent to $\beta$, where $\beta$ is the action of $G$ on $D_{1}$ induced by $\alpha^{\mathcal{B}}$ and $f$, which is defined by $\beta_{t}(d)=\alpha_{f(t)}^{\mathcal{B}}(d)$ for any $t\in G$ and $d\in D_{1}$. Let $\lambda$ be an action of $G$ on a $C_{1}-D_{1}$-equivalence bimodule $Z$ with respect to $(C_{1},D_{1},\alpha^{\mathcal{A}},\beta)$.

Let $\mathcal{A}_{1}=\{Y_{\alpha_{t}^{\mathcal{A}}}\}_{t\in G}$ and $\mathcal{B}_{1}=\{Y_{\alpha_{t}^{\mathcal{B}}}\}_{t\in G}$ be the $C^{*}$-algebraic bundles over $G$ induced by the actions $\alpha^{\mathcal{A}}$ and $\alpha^{\mathcal{B}}$, which are defined in Section 3. Furthermore, let $\mathcal{B}^{f}=\{B_{f(t)}\}_{t\in G}$ be the $C^{*}$-algebraic bundle over $G$ induced by $\mathcal{B}$ and $f$ and let $\mathcal{B}_{1}^{f}=\{Y_{\beta_{t}}\}_{t\in G}$ be the $C^{*}$-algebraic bundle over $G$ induced by the action $\beta$, which is defined in Section 3. We shall construct an $\mathcal{A}_{1}-\mathcal{B}_{1}^{f}$-equivalence bundle $\mathcal{Z}=\{Z_{t}\}_{t\in G}$ over $G$. Let $Z_{t}=e_{A}\cdot Z\cdot\beta_{t}(e_{B})$ for any $t\in G$ and let $W=\oplus_{t\in G}Z_{t}$. Also, by Lemma 3.1 and its proof $\oplus_{t\in G}Y_{\alpha_{t}^{\mathcal{A}}}\cong C$ and $\oplus_{t\in G}Y_{\beta_{t}}\cong D$ as $C^{*}$-algebras. We identify $\oplus_{t\in G}Y_{\alpha_{t}^{\mathcal{A}}}$ and $\oplus_{t\in G}Y_{\beta_{t}}$ with $C$ and $D$, respectively. We define the left $C$-action $\diamond$ and the left $C$-valued inner product ${}_{C}\langle-,-\rangle$ on $W$ by

[TABLE]

[TABLE]

where $e_{A}x\alpha_{t}^{\mathcal{A}}(e_{A})\in e_{A}C_{1}\alpha_{t}^{\mathcal{A}}(e_{A})$,   $e_{A}\cdot z\cdot\beta_{s}(e_{B}),\,e_{A}\cdot w\cdot\beta_{s}(e_{B})\in Z_{s}$,   $e_{A}\cdot z\cdot\beta_{t}(e_{B})\in Z_{t}$. Also, we define the right $D$-action, which is also denoted by the same symbol $\diamond$ and the $D$-valued inner product $\langle-,-\rangle_{D}$ on $W$ by

[TABLE]

[TABLE]

where $e_{B}x\beta_{s}(e_{B})\in e_{B}D_{1}\beta_{s}(e_{B})$, $e_{A}\cdot z\cdot\beta_{t}(e_{B})\in Z_{t}$, $e_{A}\cdot w\cdot\beta_{s}(e_{B})\in Z_{s}$. By the above definitions, $\mathcal{Z}$ has Conditions (1R)-(3R) and (1L)-(3L) in [1, Definition 2.1]. We show that $\mathcal{Z}$ has Conditions (4R) and (4L) in [1, Definition 2.1] and that $\mathcal{Z}$ is an $\mathcal{A}_{1}-\mathcal{B}_{1}^{f}$-bundle in the same way as in Example 2.3.

Lemma 4.1**.**

With the above notation, $\mathcal{Z}$ has Conditions $(4R)$ and $(4L)$ in [1, Definition 2.1].

Proof.

Let $e_{A}\cdot z\cdot\beta_{t}(e_{B})\in Z_{t}$, $e_{A}\cdot w\cdot\beta_{s}(e_{B})\in Z_{s}$ and $e_{B}x\beta_{r}(e_{B})\in e_{B}D_{1}\beta_{r}(e_{B})$, where $t,s,r\in G$. Then

[TABLE]

Also,

[TABLE]

Hence $\mathcal{Z}$ has Condition (4R) in [1, Definition 2.1]. Next, let $e_{A}\cdot z\cdot\beta_{t}(e_{B})\in Z_{t}$, $e_{A}\cdot w\cdot\beta_{s}(e_{B})\in Z_{s}$ and $e_{A}x\alpha_{r}^{\mathcal{A}}(e_{A})\in e_{A}C_{1}\alpha_{r}^{\mathcal{A}}(e_{A})$, where $t,s,r\in G$. Then

[TABLE]

Also,

[TABLE]

Hence $\mathcal{Z}$ has Conditoin (4L) in [1, Definition 2.1]. ∎

By Lemma 4.1, $W$ is a $C-D$-bimodule having Properties (1)-(6) in [6, Lemma 1.3]. In order to prove that $\mathcal{Z}$ has Conditions (5R), (6R) and (5L), (6L) in [1, Definition 2.1] using [6, Lemma 1.3], we show that $W$ has Properties (7)-(10) in [6, Lemma 1.3].

Lemma 4.2**.**

*With the above notation, $W$ has the following: $(1)$ $(e_{A}x\alpha_{t}^{\mathcal{A}}(e_{A})\diamond[e_{A}\cdot z\cdot\beta_{s}(e_{B})])\diamond e_{B}y\beta_{r}(e_{B})$

      $=e_{A}x\alpha_{t}^{\mathcal{A}}(e_{A})\diamond([e_{A}\cdot z\cdot\beta_{s}(e_{B})]\diamond e_{B}y\beta_{r}(e_{B}))$, $(2)$ $\langle e_{A}x\alpha_{t}^{\mathcal{A}}(e_{A})\diamond[e_{A}\cdot z\cdot\beta_{s}(e_{B})]\,,\,e_{A}\cdot w\cdot\beta_{r}(e_{B})\rangle_{D}$

      $=\langle e_{A}\cdot z\cdot\beta_{s}(e_{B})\,,\,(e_{A}x\alpha_{t}^{\mathcal{A}}(e_{A}))^{\sharp}\diamond[e_{A}\cdot w\cdot\beta_{r}(e_{B})]\rangle_{D}$, $(3)$ ${}_{C}\langle e_{A}\cdot z\cdot\beta_{s}(e_{B})\,,[e_{A}\cdot w\cdot\beta_{r}(e_{B})]\diamond e_{B}y\beta_{t}(e_{B})\rangle$

      $={}_{C}\langle[e_{A}\cdot z\cdot\beta_{s}(e_{B})]\diamond(e_{B}y\beta_{t}(e_{B}))^{\sharp}\,,e_{A}\cdot w\cdot\beta_{r}(e_{B})\rangle$,*

where $x\in C_{1}$, $y\in D_{1}$, $z,w\in Z$, $t,s,r\in G$.

Proof.

We show the lemma by routine computations. Let $x\in C_{1}$, $y\in D_{1}$, $z,w\in Z$, $t,s,r\in G$.

We prove (1):

[TABLE]

We prove (2):

[TABLE]

We prove (3):

[TABLE]

Therefore, we obtain the conclusion. ∎

By Lemma 4.2, $W$ has Properties (7), (8) in [6, Lemma 1.3].

Lemma 4.3**.**

With the above notation, there are finite subsets $\{u_{i}\}_{i}$ and $\{v_{j}\}_{j}$ of $W$ such that

[TABLE]

for any $x\in W$.

Proof.

Since $Z$ is a $C_{1}-D_{1}$-equivalence bimodule, there are finite subsets $\{z_{i}\}_{i}$ and $\{w_{j}\}_{j}$ of $Z$ such that

[TABLE]

for any $z\in Z$. Then for any $z\in Z$, $s\in G$,

[TABLE]

since $\sum_{t\in G}\beta_{t}(e_{B})=1$ by [10, Remark 3.4]. Also, for any $z\in Z$, $s\in G$,

[TABLE]

since $\sum_{t\in G}\alpha_{st^{-1}}^{\mathcal{A}}(e_{A})=1$ for any $s\in G$ by [10, Remark 3.4]. Therefore, we obtain the conclusion. ∎

Remark 4.4*.*

By Lemma 4.2, $\{e_{A}\cdot z_{i}\cdot\beta_{t}(e_{B})\}_{i,t}$ is a right $D$-basis and $\{e_{A}\cdot\lambda_{t}(w_{j})\cdot\beta_{t}(e_{B})\}_{j,t}$ is a left $C$-basis of $W$ in the sense of Kajiwara and Watatani [7].

By Lemma 4.2, $W$ has Properties (9), (10) in [6, Lemma 1.3]. Hence by [6, Lemma 1.3], $W$ is a Hilbert $C-D$- bimodule in the sense of [6, Definition 1.1]. Thus, $\mathcal{Z}$ has Conditions (5R), (6R) and (5L), (6L) in [1, Definition 2.1].

Lemma 4.5**.**

With the above notation, $\mathcal{Z}$ is an $\mathcal{A}_{1}-\mathcal{B}_{1}^{f}$-equivalence bundle such that

[TABLE]

for any $t,s\in G$.

Proof.

First, we show that the left $C$-valued inner product and the right $D$-valued inner product on $W$ are compatible. Let $y,z,w\in Z$ and $t,s,r\in G$. Since $Z$ is a $C_{1}-D_{1}$-equivalence bimodule,

[TABLE]

Hence the left $C$-valued inner product and the right $D$-valued inner product are compatible. Next, we show that

[TABLE]

for any $t,s\in G$. Let $t,s\in G$. Since $E^{B}$ is of Watatani index-finite type, there is a quasi-basis $\{(d_{j},d_{j}^{*})\}\subset D\times D$ for $E^{B}$. Thus $\sum_{j}d_{j}e_{B}d_{j}^{*}=1$. Since $Z$ is a $C_{1}-D_{1}$-equivalence bimodule, there is a finite subset $\{z_{i}\}$ of $Z$ such that $\sum_{i}{}_{C_{1}}\langle z_{i},z_{i}\rangle=1$. Let $c\in C$. Then

[TABLE]

Hence we obtain that ${}_{C}\langle Z_{t},\,Z_{s}\rangle=Y_{\alpha_{ts^{-1}}^{\mathcal{A}}}$ for any $t,s\in G$. Also, since $E^{A}$ is of Watatani index-finite type, there is a quasi-basis $\{(c_{j},c_{j}^{*})\}\subset C\times C$ for $E^{A}$. Thus $\sum_{j}c_{j}e_{A}c_{j}^{*}=1$. Since $Z$ is a $C_{1}-D_{1}$-equivalence bimodule, there is a finite subset $\{w_{i}\}$ of $Z$ such that $\sum_{i}\langle w_{i},w_{i}\rangle_{D_{1}}=1$. Let $d\in D_{1}$. Then

[TABLE]

Hence we obtain that $\langle Z_{t},\,Z_{s}\rangle_{D}=Y_{\beta_{t^{-1}s}}$ for any $t,s\in G$. Therefore, we obtain the conclusion. ∎

Combining the above lemmas, we obtain the following:

Proposition 4.6**.**

With the above notation, $\mathcal{Z}$ is an $\mathcal{A}_{1}-\mathcal{B}_{1}^{f}$-equivalence bundle over $G$.

Proof.

This is immediate by Lemmas 4.1, 4.2, 4.3, 4.5. ∎

Theorem 4.7**.**

Let $\mathcal{A}=\{A_{t}\}_{t\in G}$ and $\mathcal{B}=\{B_{t}\}_{t\in G}$ be saturated $C^{*}$-algebraic bundles over a finite group $G$. Let $e$ be the unit element in $G$. Let $C=\oplus_{t\in G}A_{t}$, $D=\oplus_{t\in G}B_{t}$ and $A=A_{e}$, $B=B_{e}$. We suppose that $C$ and $D$ are unital and that $A$ and $B$ have the unit elements in $C$ and $D$, respectively. Let $A\subset C$ and $B\subset D$ be the unital inclusions of unital $C^{*}$-algebras induced by $\mathcal{A}$ and $\mathcal{B}$, respectively. Also, we suppose that $A^{\prime}\cap C=\mathbf{C}1$. If $A\subset C$ and $B\subset D$ are strongly Morita equivalent, then there are an automorphism $f$ of $G$ and an $\mathcal{A}-\mathcal{B}^{f}$-equivalence bundle $\mathcal{Z}=\{Z_{t}\}_{t\in G}$ satisfying that

[TABLE]

for any $t,s\in G$, where $\mathcal{B}^{f}$ is the $C^{*}$-algebraic bundle over $G$ induced by $\mathcal{B}$ and $f$ defined by $\mathcal{B}^{f}=\{B_{f(t)}\}_{t\in G}$.

Proof.

This is immediate by Lemma 3.1 and Proposition 4.6. ∎

5. Application

Let $A$ and $B$ be unital $C^{*}$-algebras and $X$ a Hilbert $A-B$-bimodule. Let $\widetilde{X}$ be its dual Hilbert $B-A$-bimodule. For any $x\in X$, $\widetilde{x}$ denotes the element in $\widetilde{X}$ induced by $x\in X$.

Lemma 5.1**.**

Let $A$, $B$ and $C$ be unital $C^{*}$-algebras. Let $X$ be a Hilbert $A-B$-bimodule and $Y$ a Hilbert $B-C$-bimodule. Then $\widetilde{X\otimes_{B}Y}\cong\widetilde{Y}\otimes_{B}\widetilde{X}$ as Hilbert $C-A$-bimodules.

Proof.

Let $\pi$ be the map from $\widetilde{X\otimes_{B}Y}$ to $\widetilde{Y}\otimes_{B}\widetilde{X}$ defined by

[TABLE]

for any $x\in X$, $y\in Y$. Then by routine computaions, we can see that $\pi$ is a Hilbert $C-A$-bimodule isomorphism of $\widetilde{X\otimes_{B}Y}$ onto $\widetilde{Y}\otimes_{B}\widetilde{X}$. ∎

We identify $\widetilde{X\otimes_{B}Y}$ with $\widetilde{Y}\otimes_{B}\widetilde{X}$ by the isomorphism $\pi$ defined in the proof of Lemma 5.1. Next, we give the definition of an involutive Hilbert $A-A$-bimodule modifying [9].

Definition 5.1*.*

We say that a Hilbert $A-A$-bimodule $X$ is involutive if there exists a conjugate linear map $x\in X\mapsto x^{\natural}\in X$ such that

(1) $(x^{\natural})^{\natural}=x$,   $x\in X$,

(2) $(a\cdot x\cdot b)^{\natural}=b^{*}\cdot x^{\natural}\cdot a^{*}$,   $x\in X$, $a,b\in A$,

(3) ${}_{A}\langle x,\,y^{\natural}\rangle=\langle x^{\natural},\,y\rangle_{A}$,   $x,y\in X$.

We call the above conjugate linear map $\natural$ an involution on $X$. If $X$ is full with the both inner products, $X$ is an involutive $A-A$-equivalence bimodule. For each involutive Hilbert $A-A$-bimodule, let $L_{X}$ be the linking $C^{*}$-algebra induced by $X$ and $C_{X}$ the $C^{*}$-subalgebra of $L_{X}$, which is defined in [9], that is,

[TABLE]

We note that $C_{X}$ acts on $X\oplus A$ (See Brown, Green and Rieffel [2] and Rieffel [14]). The norm of $C_{X}$ is defined as the operator norm on $X\oplus A$.

Let $A$ be a unital $C^{*}$-algebra and $X$ an involutive Hilbert $A-A$-bimodule. Let $\widetilde{X}$ be its dual Hilbert $A-A$-bimodule. We define the map ♮ on $\widetilde{X}$ by $(\widetilde{x})^{\natural}=\widetilde{(x^{\natural})}$ for any $\widetilde{x}\in\widetilde{X}$.

Lemma 5.2**.**

With the above notation, the above map ♮ is an involution on $\widetilde{X}$.

Proof.

This is immediate by direct computations. ∎

For each involutive Hilbert $A-A$-bimodule $X$, we regard $\widetilde{X}$ as an involutive $A-A$-bimodule in the same manner of Lemma 5.2.

Let $\mathbf{Z}_{2}=\mathbf{Z}/2\mathbf{Z}$ and $\mathbf{Z}_{2}$ consists of the unit element [math] and $1$. Let $X$ be an involutive Hilbert $A-A$-bimodule. We construct a $C^{*}$-algebraic bundle over $\mathbf{Z}_{2}$ induced by $X$. Let $A_{0}=A$ and $A_{1}=X$. Let $\mathcal{A}_{X}=\{A_{t}\}_{t\in\mathbf{Z}_{2}}$. We define a product $\bullet$ and an involution $\sharp$ as follows:

(1) $a\bullet b=ab$  $a,b\in A$,

(2) $a\bullet x=a\cdot x$,  $x\bullet a=x\cdot a$  $a\in A$, $x\in X$,

(3) $x\bullet y={}_{A}\langle x,y^{\natural}\rangle=\langle x^{\natural},y\rangle_{A}$  $x,y\in X$,

(4) $a^{\sharp}=a^{*}$  $a\in A$,

(5) $x^{\sharp}=x^{\natural}$  $x\in X$.

Then $A\oplus X$ is a $*$-algebra and by routine computations, $A\oplus X$ is isomorphic to $C_{X}$ as $*$-algebras. We identify $A\oplus X$ with $C_{X}$ as $*$-algebras. We define a norm of $A\oplus X$ as the operator norm on $X\oplus A$. Hence $\mathcal{A}_{X}$ is a $C^{*}$-algebraic bundle over $\mathbf{Z}_{2}$. Thus, we obtain a correspondence from the involutive Hilbert $A-A$-bimodules to the $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$. Next, let $\mathcal{A}=\{A_{t}\}_{t\in\mathbf{Z}_{2}}$ be a $C^{*}$-algebraic bundle over $\mathbf{Z}_{2}$. Then $A_{1}$ ia an involutive Hilbert $A-A$-bimodule. Hence we obtain a correspondence from the $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$ to the involutive Hilbert $A-A$-bimodules. Clearly the above two correspondences are the inverse correspondences of each other. Furthermore, the inclusion of unital $C^{*}$-algebras $A\subset C_{X}$ induced by $X$ and the inclusion of unital $C^{*}$-algebras $A\subset A\oplus X$ induced by the $C^{*}$-algebraic bundle $\mathcal{A}_{X}$ coincide.

Lemma 5.3**.**

Let $X$ and $Y$ be involutive Hilbert $A-A$-bimodules and $\mathcal{A}_{X}$ and $\mathcal{A}_{Y}$ the $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$ induced by $X$ and $Y$, respectively. Then $\mathcal{A}_{X}\cong\mathcal{A}_{Y}$ as $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$ if and only if $X\cong Y$ as involutive Hilbert $A-A$-bimodules.

Proof.

We suppose that $\mathcal{A}_{X}\cong\mathcal{A}_{Y}$ as $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$. Then there is a $C^{*}$-algebraic bundle isomorphism $\{\pi_{t}\}_{t\in\mathbf{Z}_{2}}$ of $\mathcal{A}_{X}$ onto $\mathcal{A}_{Y}$. We identify $A$ with $\pi_{0}(A)$. Then $\pi_{1}$ is an involutive Hilbert $A-A$-bimodule isomorphism of $X$ onto $Y$. Next, we suppose that there is an involutive Hilbert $A-A$-bimodule isomorphism $\pi$ of $X$ onto $Y$. Let $\pi_{0}={\rm{id}}_{A}$ and $\pi_{1}=\pi$. Then $\{\pi_{t}\}_{t\in\mathbf{Z}_{2}}$ is a $C^{*}$-algebraic bundle isomorphism $\mathcal{A}_{X}$ onto $\mathcal{A}_{Y}$. ∎

Lemma 5.4**.**

Let $X$ be an involutive Hilbert $A-A$-bimodule and $\mathcal{A}_{X}$ the $C^{*}$-algebraic bundle over $\mathbf{Z}_{2}$ induced by $X$. Then $X$ is full with the both inner products if and only if $\mathcal{A}_{X}$ is saturated.

Proof.

We suppose that $X$ is full with the both inner products. Then

[TABLE]

Also,

[TABLE]

by [3, Proposition1.7]. Clearly $A_{0}\bullet A_{0}=AA=A=A_{0}$. Hence $\mathcal{A}_{X}$ is saturated. Next, we suppose that $\mathcal{A}_{X}$ is saturated. Then

[TABLE]

Thus $X$ is full with the both inner products. ∎

Remark 5.5*.*

Let $X$ be an involutive Hilbert $A-A$-bimodule. Then by the above proof, we see that $X$ is full with the left $A$-valued inner product if and only if $X$ is full with the right $A$-valued inner product.

Lemma 5.6**.**

Let $A$ and $B$ be unital $C^{*}$-algebras and $M$ an $A-B$-equivalence bimodule. Let $X$ be an involutive Hilbert $A-A$-bimodule. Then $\widetilde{M}\otimes_{A}X\otimes_{A}M$ is an involutive Hilbert $B-B$-bimodule whose involution $\natural$ is defined by

[TABLE]

for any $m,n\in M$, $x\in X$.

Proof.

This is immediate by routine computations. ∎

Let $A,B,X$ and $M$ be as in Lemma 5.6. Let $Y$ be an involutive Hilbert $B-B$-bimodule. We suppose that there is an involutive Hilbert $B-B$-bimodule isomorphism $\Phi$ of $\widetilde{M}\otimes_{A}X\otimes_{A}M$ onto $Y$. Let $\widetilde{\Phi}$ be the linear map from $\widetilde{M}\otimes_{A}\widetilde{X}\otimes_{A}M$ onto $\widetilde{Y}$ defined by

[TABLE]

for any $m,n\in M$, $x\in X$.

Lemma 5.7**.**

With the above notation, $\widetilde{\Phi}$ is an involutive Hilbert $B-B$-bimodule isomorphism of $\widetilde{M}\otimes_{A}\widetilde{X}\otimes_{A}M$ onto $\widetilde{Y}$.

Proof.

This is immediate by routine computations ∎

Again, let $A,B,X$ and $M$ be as in Lemma 5.6. Let $Y$ be an involutive Hilbert $B-B$-bimodule. We suppose that there is an involutive Hilbert $B-B$-bimodule isomorphism $\Phi$ of $\widetilde{M}\otimes_{A}X\otimes_{A}M$ onto $Y$. Then there is a finite subset $\{u_{i}\}$ of $M$ with $\sum_{i}{}_{A}\langle u_{i}\,,\,u_{i}\rangle=1$. We identify $A$ and $X$ with $M\otimes_{B}\widetilde{M}$ and $A\otimes_{A}X$ by the isomorphisms defined by

[TABLE]

Let $x\in X$, $m\in M$. For any $x\otimes m\in X\otimes_{A}M$,

[TABLE]

Hence there is a linear map $\Psi$ from $X\otimes_{A}M$ to $M\otimes_{B}Y$ defined by

[TABLE]

for any $x\in X$, $m\in M$. By the definition of $\Psi$, we can see that $\Psi$ is a Hilbert $A-B$-bimodule isomorphism of $X\otimes_{A}M$ onto $M\otimes_{B}Y$.

Lemma 5.8**.**

With the above notation, the Hilbert $A-B$-bimodule isomorphism $\Psi$ of $X\otimes_{A}M$ onto $M\otimes_{B}Y$ is independent of the choice of a finite subset $\{u_{i}\}$ of $M$ with $\sum_{i}{}_{A}\langle u_{i}\,,\,u_{i}\rangle=1$.

Proof.

Let $\{v_{j}\}$ be another finite subset of $M$ with $\sum_{j}{}_{A}\langle v_{j}\,,\,v_{j}\rangle=1$. Then for any $x\in X$, $m\in M$,

[TABLE]

Therefore, we obtain the conclusion. ∎

Similarly let $\widetilde{\Psi}$ be the Hilbert $A-B$-bimodule isomorphism of $\widetilde{X}\otimes_{A}M$ onto $M\otimes_{B}\widetilde{Y}$ defined by

[TABLE]

for any $x\in X$, $m\in M$. We compute the inverse map of $\Psi$, which is a Hilbert $A-B$-bimodule isomorphism of $M\otimes_{B}Y$ onto $X\otimes_{A}M$. Let $\Theta$ be the linear map from $M\otimes_{B}Y$ to $X\otimes_{A}M$ defined by

[TABLE]

for any $m\in M$, $y\in Y$, where we identify $M\otimes_{B}\widetilde{M}\otimes_{A}X\otimes_{A}M$ with $X\otimes_{A}M$ as Hilbert $A-B$-bimodules by the map

[TABLE]

Lemma 5.9**.**

With the above notation, $\Theta$ is a Hilbert $A-B$-bimodule isomorphism of $M\otimes_{B}Y$ onot $X\otimes_{A}M$ such that $\Theta\circ\Psi={\rm{id}}_{X\otimes_{A}M}$ and $\Psi\circ\Theta={\rm{id}}_{M\otimes_{B}Y}$.

Proof.

Let $m,m_{1}\in M$, $y,y_{1}\in Y$. Then

[TABLE]

Hence $\Theta$ preserves the left $A$-valued inner products. Also,

[TABLE]

Hence $\Theta$ preserves the right $B$-valued inner products. Furthermore, for any $x\in X$, $m\in M$,

[TABLE]

since we identify $M\otimes\widetilde{M}$ with $A$ as $A-A$-equivalence bimodules by the map $m\otimes\widetilde{n}\in M\otimes_{B}\widetilde{M}\mapsto{}_{A}\langle m,n\rangle\in A$. Hence $\Theta\circ\Psi={\rm{id}}_{X\otimes_{A}M}$. Hence $\Psi\circ\Theta\circ\Psi=\Psi$ on $X\otimes_{A}M$. Since $\Psi$ is surjective, $\Psi\circ\Theta={\rm{id}}_{M\otimes_{B}Y}$. Therefore, by the remark after [5, Definition 1.1.18], $\Theta$ is a Hilbert $A-B$-bimodule isomorphism of $M\otimes_{B}Y$ onto $X\otimes_{A}M$ such that $\Theta\circ\Psi={\rm{id}}_{X\otimes_{A}M}$ and $\Psi\circ\Theta={\rm{id}}_{M\otimes_{B}Y}$. ∎

Similarly, we see that the inverse map of $(\widetilde{\Psi})^{-1}$ is defined by

[TABLE]

for any $m\in M$, $y\in Y$, where we identify $M\otimes_{B}\widetilde{M}\otimes_{A}\widetilde{X}\otimes_{A}M$ with $\widetilde{X}\otimes_{A}M$ as Hilbert $A-B$-bimodules by the map

[TABLE]

We prepare some lemmas in order to show Proposition 5.13.

Lemma 5.10**.**

Let $A$ and $B$ be unital $C^{*}$-algebras. Let $X$ and $Y$ be an involutive Hilbert $A-A$-bimodule and an involutive Hilbert $B-B$-bimodule, respectively. Let $\mathcal{A}_{X}=\{A_{t}\}_{t\in\mathbf{Z}_{2}}$ and $\mathcal{A}_{Y}=\{B_{t}\}_{t\in\mathbf{Z}_{2}}$ be $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$ induced by $X$ and $Y$, respectively. We suppose that there is an $\mathcal{A}_{X}-\mathcal{A}_{Y}$-equivalence bundle $\mathcal{M}=\{M_{t}\}_{t\in\mathbf{Z}_{2}}$ over $\mathbf{Z}_{2}$ such that

[TABLE]

for any $t,s\in\mathbf{Z}_{2}$, where $C=A\oplus X$ and $D=B\oplus Y$. Then there is an $A-B$-equivalence bimodule $M$ such that $Y\cong\widetilde{M}\otimes_{A}X\otimes_{A}M$ as involutive Hilbert $B-B$-bimodules.

Proof.

By the assumptions, $M_{0}$ is an $A-B$-equivalence bimodule. Let $M=M_{0}$. Then by Lemma 5.6, $\widetilde{M}\otimes_{A}X\otimes_{A}M$ is an involutive Hilbert $B-B$-bimodule whose involution is defined by $(\widetilde{m}\otimes x\otimes n)^{\natural}=\widetilde{n}\otimes x^{\natural}\otimes m$ for any $m,n\in M$, $x\in X$. We show that $Y\cong\widetilde{M}\otimes_{A}X\otimes_{A}M$ as involutive Hilbert $B-B$-bimodules. Let $\Phi$ be the map from $\widetilde{M}\otimes_{A}X\otimes_{A}M$ to $Y$ defined by

[TABLE]

for any $m,n\in M$, $x\in X$. Since $A_{1}=X$ and $M=M_{0}$, $X\cdot M_{0}\subset M_{1}$. And $\langle M_{0}\,,\,M_{1}\rangle_{D}\in B_{1}=Y$. Hence $\Phi$ is a map from $\widetilde{M}\otimes_{A}X\otimes_{A}M$ to $Y$. Clearly, $\Phi$ is a linear and $B-B$-bimodule map. We show that $\Phi$ is surjective. Indeed,

[TABLE]

by [3, Proposition 1.7]. Hence $\langle M\,,\,X\cdot M\rangle_{D}=\langle M\,,\,M_{1}\rangle_{D}=Y$. Thus, $\Phi$ is surjective. Let $m,n,m_{1},n_{1}\in M$, $x,x_{1}\in X$. Then

[TABLE]

Hence $\Phi$ preserves the left $B$-valued inner products. Also,

[TABLE]

Hence $\Phi$ preserves the right $B$-valued inner products. Furthermore,

[TABLE]

On the other hand,

[TABLE]

Hence $\Phi$ preserves the involutions $\natural$. Therefore, $Y\cong\widetilde{M}\otimes_{A}X\otimes_{A}M$ as involutive Hilbert $B-B$-bimodules. ∎

Let $A$ and $B$ be unital $C^{*}$-algebras. Let $X$ and $Y$ be an involutive Hilbert $A-A$-bimodule and an involutive Hilbert $B-B$-bimodule, respectively. We suppose that there is an $A-B$-equivalence bimodule $M$ such that

[TABLE]

as involutive Hilbert $B-B$-bimodules. Let $\Phi$ be an involutive Hilbert $B-B$-bimodule isomorphism of $\widetilde{M}\otimes_{A}X\otimes_{A}M$ onto $Y$. Then by the above discussions, there are the Hilbert $A-B$-bimodule isomorphisms $\Psi$ of $X\otimes_{A}M$ onto $M\otimes_{B}Y$ and $\widetilde{\Psi}$ of $\widetilde{X}\otimes_{A}M$ onto $M\otimes_{B}\widetilde{Y}$, respectively. We construct a $C_{X}-C_{Y}$-equivalence bimodule from $M$. Let $C_{M}$ be the linear span of the set

[TABLE]

We define the left $C_{X}$-action on $C_{M}$ by

[TABLE]

for any $a\in A$, $m_{1},m_{2}\in M$, $x,z\in X$, where we regard the tensor product as a left $C_{X}$-action on $C_{M}$ in the formal manner. But we identify $A\otimes_{A}M$ and $X\otimes_{A}\widetilde{X}$, $\widetilde{X}\otimes_{A}X$ with $M$ and closed two-sided ideals of $A$ by the isomorphism and the monomorphisms defined by

[TABLE]

Hence we obtain that

[TABLE]

We define the right $C_{Y}$-action on $C_{M}$ by

[TABLE]

for any $b\in B$, $x\in X$, $y\in Y$, $m_{1},m_{2}\in M$, where we regard the tensor product as a right $C_{Y}$-action on $C_{M}$ in the formal manner. But we identify $X\otimes_{A}M$ and $\widetilde{X}\otimes_{A}M$ with $M\otimes_{B}Y$ and $M\otimes_{B}\widetilde{Y}$ by $\Psi$ and $\widetilde{\Psi}$, respectively. Hence we obtain that

[TABLE]

Furthermore, we identify $M\otimes_{B}B$ and $Y\otimes_{B}\widetilde{Y}$, $\widetilde{Y}\otimes_{B}Y$ with $M$ and closed two-sided ideals of $B$ by the isomorphism and the monomorphisms defined by

[TABLE]

respectively. Then

[TABLE]

Indeed, for any $\epsilon>0$, there finite sets $\{n_{k}\},\{l_{k}\}\subset M$ and $\{z_{k}\}\subset X$ such that

[TABLE]

Also,

[TABLE]

Thus

[TABLE]

and

[TABLE]

Furthermore,

[TABLE]

and

[TABLE]

where we identify $A\otimes_{A}M$ and $X\otimes_{A}\widetilde{X}$, $\widetilde{X}\otimes_{A}X$ with $M$ and closed two-sided ideals of $A$ by the isomorphism and the monomorphisms defined by

[TABLE]

Hence

[TABLE]

It follows that

[TABLE]

Since $\epsilon$ is arbitrary, we obtain that

[TABLE]

Before we define a left $C_{X}$-valued inner product and a right $C_{Y}$-valued inner product on $C_{M}$, we define a conjugate linear map on $C_{M}$,

[TABLE]

by

[TABLE]

for any $m_{1},m_{2}\in M$, $x\in X$. Since we identify $\widetilde{X\otimes_{A}M}$ and $\widetilde{\widetilde{X}\otimes_{A}M}$ with $\widetilde{M}\otimes_{A}\widetilde{X}$ and $\widetilde{M}\otimes_{A}X$ by Lemma 5.1, respectively, we obtain that

[TABLE]

We define the left $C_{X}$-valued inner product on $C_{M}$ by

[TABLE]

for any $m_{1},m_{2},n_{1},n_{2}\in M$, $x,z\in X$, where we regard the tensor product as a product in $C_{M}$ in the formal manner. Identifying in the same way as above,

[TABLE]

We define the right $C_{Y}$-valued inner product on $C_{M}$ by

[TABLE]

for any $m_{1},m_{2},n_{1},n_{2}\in M$, $x,z\in X$, where we regard the tensor product as a product in $C_{M}$ in the formal manner. Identifying in the same way as above and by the isomorphism $\Psi$ and $\widetilde{\Psi}$,

[TABLE]

Here,

[TABLE]

Also,

[TABLE]

Thus

[TABLE]

Hence

[TABLE]

By the above definitions $C_{M}$ has the left $C_{X}$-and the right $C_{Y}$-actions and the left $C_{X}$-valued inner product and the right $C_{Y}$-inner product.

Let $C_{M}^{\prime}$ be the linear span of the set

[TABLE]

In the similar way to the above, we define a left $C_{X}$-and a right $C_{Y}$-actions on $C_{M}^{\prime}$ and a left $C_{X}$-valued inner product and a right $C_{Y}$-valued inner product. But identifying $X\otimes_{A}M$ and $\widetilde{X}\otimes_{A}M$ with $M\otimes_{B}Y$ and $M\otimes_{B}\widetilde{Y}$ by $\Psi$ and $\widetilde{\Psi}$, respectively, we can see that each of them coincides with the other by routine computations. For example, we show that the right $C_{Y}$-actions on $C_{M}$ and $C_{M}^{\prime}$ coincide by $\Psi$ and $\widetilde{\Psi}$. Indeed, for any $m_{1},m_{2}\in M$, $x\in X$, $b\in B$, $y\in Y$,

[TABLE]

Regarding elements in $X\otimes_{A}M$ and $\widetilde{X}\otimes_{A}M$ as elements in $M\otimes_{B}Y$ and $M\otimes_{B}\widetilde{Y}$ by the isomorphisms $\Psi$ and $\widetilde{\Psi}$ defined as above, respectively,

[TABLE]

On the other hand,

[TABLE]

Hence the right $C_{Y}$-actions on $C_{M}$ and $C_{M}^{\prime}$ coincide. Similarly, we can see that the left $C_{X}$-actions on $C_{M}$ and $C_{M}^{\prime}$ coincide. Also, we can see that the left $C_{X}$-valued inner products on $C_{M}$ and $C_{M}^{\prime}$ coincide. Indeed, let $x,z\in X$, $m_{1},m_{2},n_{1},n_{2}\in M$. Then

[TABLE]

On the other hand,

[TABLE]

Here,

[TABLE]

Also,

[TABLE]

Since we identify $\widetilde{M}\otimes_{A}X\otimes_{A}M$ with $Y$ by the involutive Hilbert $B-B$-bimodule isomorphism $\Phi$,

[TABLE]

Similarly,

[TABLE]

Also, since we identify $\widetilde{M}\otimes_{A}\widetilde{X}\otimes_{A}M$ with $\widetilde{Y}$ by the involutive Hilbert $B-B$-bimodule isomorphism $\widetilde{\Phi}$, we see that

[TABLE]

Furthermore,

[TABLE]

Thus, the left $C_{X}$-valued inner products on $C_{M}$ and $C_{M}^{\prime}$ coincide. Similarly we can see that the right $C_{Y}$-valued inner products on $C_{M}$ and $C_{M}^{\prime}$ coincide. Hence we obtain the following lemma:

Lemma 5.11**.**

With the above notation, $C_{M}$ is a $C_{X}-C_{Y}$-equivalence bimodule.

Proof.

By the definitions of the left $C_{X}$-action and the left $C_{X}$-valued inner product on $C_{M}$, we can see that Conditions (a)-(d) in [7, Proposition 1.12] hold. By the definitions of the right $C_{Y}$-action and the right $C_{Y}$-valued inner product on $C_{M}$, we can also see that the similar conditions to Conditions (a)-(d) in [7, Proposition 1.12] hold. Furthermore, we can easily see that the associativity of the left $C_{X}$-valued inner product and the right $C_{Y}$-valued inner product holds. Since $M$ is an $A-B$-equivalence bimodule, there are finite subsets $\{u_{i}\}_{i=1}^{n}$ and $\{v_{j}\}_{j=1}^{m}$ of $M$ such that

[TABLE]

Let $U_{i}=\begin{bmatrix}u_{i}&0\\ 0&u_{i}\end{bmatrix}$ for any $i$ and let $V_{j}=\begin{bmatrix}v_{j}&0\\ 0&v_{j}\end{bmatrix}$ for any $j$. Then $\{U_{i}\}$ and $\{V_{j}\}$ are finite subsets of $C_{M}$ and

[TABLE]

Similarly $\sum_{j=1}^{m}\langle V_{j}\,,\,V_{j}\rangle_{C_{Y}}=1_{C_{Y}}$. Thus, since the associativity of the left $C_{X}$-valued inner product and the right $C_{Y}$-valued inner product on $C_{M}$ holds, we can see that $\{U_{i}\}$ and $\{V_{j}\}$ are a right $C_{Y}$-basis and a left $C_{X}$-basis of $C_{M}$, respectively. Hence by [7, Proposition 1.12], $C_{M}$ is a $C_{X}-C_{Y}$-equivalence bimodule. ∎

Lemma 5.12**.**

Let $A$ and $B$ be unital $C^{*}$-algebras. Let $X$ and $Y$ be an involutive Hilbert $A-A$-bimodule and an involutive Hilbert $B-B$-bimodule, respectively. Let $\mathcal{A}_{X}=\{A_{t}\}_{t\in\mathbf{Z}_{2}}$ and $\mathcal{A}_{Y}=\{B_{t}\}_{t\in\mathbf{Z}_{2}}$ be $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$ induced by $X$ and $Y$, respectively. We suppose that there is an $A-B$-equivalence bimodule $M$ such that

[TABLE]

as involutive Hilbert $B-B$-bimodules. Then there is an $\mathcal{A}_{X}-\mathcal{A}_{Y}$-equivalence bundle $\mathcal{M}=\{M_{t}\}_{t\in\mathbf{Z}_{2}}$ over $\mathbf{Z}_{2}$ such that

[TABLE]

for any $t,s\in\mathbf{Z}_{2}$, where $C=A\oplus X$ and $D=B\oplus Y$.

Proof.

Let $C_{M}$ be the $C_{X}-C_{Y}$-equivalence bimodule induced by $M$, which is defined in the above. We identify $M\oplus(X\otimes_{A}M)$ with $C_{M}$ as vector spaces over $\mathbf{C}$ by the isomorphism defined by

[TABLE]

Since we identify $C=A\oplus X$ and $D=B\oplus Y$ with $C_{X}$ and $C_{Y}$, respectively, $M\oplus(X\otimes_{A}M)$ is a $C-D$-equivalence bimodule by above identifications and Lemma 5.11. Let $M_{0}=M$ and $M_{1}=X\otimes_{A}M$. We note that $X\otimes_{A}M$ is identified with $M\otimes_{B}Y$ by the Hilbert $B-B$-bimodule isomorphism $\Psi$. Let $\mathcal{M}=\{M_{t}\}_{t\in\mathbf{Z}_{2}}$. Then by routine computations, $\mathcal{M}$ is an $\mathcal{A}_{X}-\mathcal{A}_{Y}$-equivalence bundle over $\mathbf{Z}_{2}$ such that

[TABLE]

for any $t,s\in\mathbf{Z}_{2}$. ∎

Proposition 5.13**.**

*Let $A$ and $B$ be unital $C^{*}$-algebras. Let $X$ and $Y$ be an involutive Hilbert $A-A$-bimodule and an involutive Hilbert $B-B$-bimodule, respectively. Let $\mathcal{A}_{X}=\{A_{t}\}_{t\in\mathbf{Z}_{2}}$ and $\mathcal{A}_{Y}=\{B_{t}\}_{t\in\mathbf{Z}_{2}}$ be the $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$ induced by $X$ and $Y$, respectively. Then the following conditions are equivalent:

$(1)$ There is an $\mathcal{A}_{X}-\mathcal{A}_{Y}$-equivalence bundle $\mathcal{M}=\{M_{t}\}_{t\in\mathbf{Z}_{2}}$ over $\mathbf{Z}_{2}$ such that*

[TABLE]

*for any $t,s\in\mathbf{Z}_{2}$, where $C=A\oplus X$ and $D=B\oplus Y$.

$(2)$ There is an $A-B$-equivalence bimodule $M$ such that*

[TABLE]

as involutive Hilbert $B-B$-bimodules.

Proof.

This is immediate by Lemmas 5.10 and 5.12. ∎

Theorem 5.14**.**

*Let $A$ and $B$ be unital $C^{*}$-algebras. Let $X$ and $Y$ be an involutive Hilbert $A-A$-bimodule and an involutive Hilbert $B-B$-bimodule, respectively. Let $A\subset C_{X}$ and $B\subset C_{Y}$ be the unital inclusions of unital $C^{*}$-lgebras induced by $X$ and $Y$, respectively. Then the following hold:

$(1)$ If there is an $A-B$-equivalence bimodule $M$ such that*

[TABLE]

*as involutie Hilbert $B-B$-bimodules, then the unital inclusions $A\subset C_{X}$ and $B\subset C_{Y}$ are strongly Morita equivalent.

$(2)$ We suppose that $X$ and $Y$ are full with the both inner products and that $A^{\prime}\cap C_{X}=\mathbf{C}1$. If the unital inclusions $A\subset C_{X}$ and $B\subset C_{Y}$ are strongly Morita equivalent, then there is an $A-B$-equivalence bimodule $M$ such that*

[TABLE]

as involutive Hilbert $B-B$-bimodules.

Proof.

Let $\mathcal{A}_{X}=\{A_{t}\}_{t\in\mathbf{Z}_{2}}$ and $\mathcal{A}_{Y}=\{B_{t}\}_{t\in\mathbf{Z}_{2}}$ be the $C^{*}$-algebraic bundles over $\mathbf{Z}_{2}$ induced by $X$ and $Y$, respectively. We prove (1). We suppose that there is an $A-B$-equivalence bimodule $M$ such that

[TABLE]

as involutive Hilbert $B-B$-bimodules. Then by Proposition 5.13, there is an $\mathcal{A}_{X}-\mathcal{A}_{Y}$-equivalence bundle $\mathcal{M}=\{M_{t}\}_{t\in\mathbf{Z}_{2}}$ over $\mathbf{Z}_{2}$ such that

[TABLE]

for any $t,s\in\mathbf{Z}_{2}$, where $C=A\oplus X$ and $D=B\oplus Y$. Hence by Proposition 2.1, the unital inclusions of unital $C^{*}$-algebras $A\subset C$ and $B\subset D$ are strongly Morita equivalent. Since we identify $A\subset C$ and $B\subset D$ with $A\subset C_{X}$ and $B\subset C_{Y}$, respectively, $A\subset C_{X}$ and $B\subset C_{Y}$ are strongly Morita equivalent. Next, we prove (2). We suppose that $X$ and $Y$ are full with the both inner products and that $A^{\prime}\cap C_{X}=\mathbf{C}1$. Also, we suppose that $A\subset C_{X}$ and $B\subset C_{Y}$ are strongly Morita equivalent. Then $\mathcal{A}_{X}$ and $\mathcal{A}_{Y}$ are saturated by Lemma 5.4. Since the identity map ${\rm{id}}_{\mathbf{Z}_{2}}$ is the only automorphism of $\mathbf{Z}_{2}$, by Theorem 4.7 there is an $\mathcal{A}_{X}-\mathcal{A}_{Y}$-equivalence bundle $\mathcal{M}=\{M_{t}\}_{t\in\mathbf{Z}_{2}}$ such that

[TABLE]

for any $t,s\in\mathbf{Z}$. Hence Proposition 5.13, there is an $A-B$-equivalence bimodule $M$ such that

[TABLE]

as involutive Hilbert $B-B$-bimodules. ∎