2 Preliminaries

Let $\mathbf K$ be the $C^*$ -algebra of all compact operators on a countably infinite-dimensional Hilbert space and $\{e_{ij}\}_{i, j\in \mathbf N}$ its system of matrix units.

For each $C^*$ -algebra A, we denote by $M(A)$ the multiplier $C^*$ -algebra of A. Let $\pi $ be an isomorphism of A onto a $C^*$ -algebra B. Then, there is the unique strictly continuous isomorphism of $M(A)$ onto $M(B)$ extending $\pi $ by Jensen and Thomsen [Reference Kodaka5, Corollary 1.1.15]. We denote it by $\underline {\pi }$ .

For an algebra A, we denote by ${\mathrm {id}}_A$ the identity map on A. If A is unital, we denote by $1_A$ the unit element of A. If no confusion arises, we denote them by ${\mathrm {id}}$ and $1$ , respectively.

Let A and B be $C^*$ -algebras and X an $A-B$ -bimodule. We denote its left A-action and right B-action on X by $a\cdot x$ and $x\cdot b$ for any $a\in A$ , $b\in B$ , $x\in X$ , respectively. We denote by $\widetilde {X}$ the dual $B-A$ -bimodule of X and let $\widetilde {x}$ denote the element in $\widetilde {X}$ associated to an element $x\in X$ . Furthermore, we regard X as a Hilbert $M(A)-M(B)$ -bimodule in the sense of [Reference Brown, Mingo and Shen4] in the same way as described before [Reference Kodaka8, Definition 2.4].

Let $A\subset C$ and $B\subset D$ be inclusions of $C^*$ -algebras. We give some definitions.

We note that X can be regarded as an $A-B$ -equivalence bimodule. Furthermore, we give the following definition.

Then, we have the following:

$$ \begin{align*}\lambda_t (a\cdot x)=\alpha_t (a)\cdot \lambda_t (x) , \quad \lambda_t (x\cdot b)=\lambda_t (x)\cdot \beta_t (b), \end{align*} $$

for any $a\in A$ , $b\in B$ , $x\in X$ , and $t\in G$ .

Let A and B be $C^*$ -algebras and $\pi $ an isomorphism of B onto A. We construct an $A-B$ -equivalence bimodule $X_{\pi }$ as follows: Let $X_{\pi }=A$ as a $\mathbf C$ -vector space. For any $a\in A$ , $b\in B$ , and $x, y\in X_{\pi }$ ,

$$ \begin{align*} & a\cdot x=ax \, , \quad x\cdot b=x\pi(b), \\ & {}_A \langle x, y \rangle=xy^* \, , \quad \langle x, y \rangle_B=\pi^{-1}(x^* y). \end{align*} $$

By easy computations, we can see that $X_{\pi }$ is an $A-B$ -equivalence bimodule. We call $X_{\pi }$ an $A-B$ -equivalence bimodule induced by $\pi $ . Let $\alpha $ be an automorphism of A. Then, in the same way as above, we construct $X_{\alpha }$ , an $A-A$ -equivalence bimodule. Let $u_{\alpha }$ be a unitary element in $M(A\rtimes _{\alpha }\mathbf Z)$ implementing $\alpha $ . Hence, $\alpha ={\mathrm {Ad}}(u_{\alpha })$ . We regard $Au_{\alpha }$ as an $A-A$ -equivalence bimodule as follows:

$$ \begin{align*} & a\cdot xu_{\alpha}=axu_{\alpha}\, , \quad xu_{\alpha}\cdot a =x\alpha(a), \\ & {}_A \langle xu_{\alpha} \, , \, yu_{\alpha} \rangle =xy^* \, , \quad \langle xu_{\alpha} \, , \, yu_{\alpha} \rangle_A =\alpha^{-1}(x^* y), \end{align*} $$

for any $a, x, y\in A$ .

Let A be a $C^*$ -algebra and X an $A-A$ -equivalence bimodule. Let $A\rtimes _X \mathbf Z$ be the crossed product of A by X defined in [Reference Abadie, Eilers and Exel1]. We regard the $C^*$ -algebra $\mathbf K$ as the trivial $\mathbf K-\mathbf K$ -equivalence bimodule. Then, we obtain an $A\otimes \mathbf K-A\otimes \mathbf K$ -equivalence bimodule $X\otimes \mathbf K$ , and we can also consider the crossed product

$$ \begin{align*}(A\otimes\mathbf K)\rtimes_{X\otimes\mathbf K}\mathbf Z \end{align*} $$

of $A\otimes \mathbf K$ by $X\otimes \mathbf K$ . Hence, we have the following inclusions of $C^*$ -algebras:

$$ \begin{align*}A\subset A\rtimes_X \mathbf Z \, , \quad A\otimes\mathbf K\subset (A\otimes\mathbf K)\rtimes_{X\otimes\mathbf K}\mathbf Z. \end{align*} $$

Because there is an isomorphism $\pi $ of $(A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z$ onto $(A\rtimes _X \mathbf Z)\otimes \mathbf K$ such that $\pi |_{A\rtimes \mathbf K}={\mathrm {id}}$ on $A\otimes \mathbf K$ , we identify $A\otimes \mathbf K \subset (A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z$ with $A\otimes \mathbf K\subset (A\rtimes _X \mathbf Z) \otimes \mathbf K$ . Thus, $A\subset A\rtimes _X \mathbf Z$ and $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z $ are strongly Morita equivalent.

Let $H_A$ be the $A\otimes \mathbf K-A$ -equivalence bimodule defined as follows: Let $H_A =(A\otimes \mathbf K)(1_{M(A)}\otimes e_{11})$ as a $\mathbf C$ -vector space. For any $a\in A$ , $k\in \mathbf K$ , and $x, y\in A\otimes \mathbf K$ ,

$$ \begin{align*} & (a\otimes k)\cdot x(1\otimes e_{11})=(a\otimes k)x(1\otimes e_{11}) \, , \\ & x(1\otimes e_{11})\cdot a =x(a\otimes e_{11}) , \\ & {}_{A\otimes\mathbf K} \langle x(1\otimes e_{11}) \, , \, y(1\otimes e_{11}) \rangle =x(1\otimes e_{11})y^* \, , \\ & \langle x(1\otimes e_{11}) \, , \, y(1\otimes e_{11}) \rangle_A =(1\otimes e_{11})x^* y(1\otimes e_{11}) , \end{align*} $$

where we identify A with $A\otimes e_{11}$ . Let B be a $C^*$ -algebra. Let $H_B$ be as above.

Lemma 2.2 With the above notation, let M be an $A-B$ - equivalence bimodule. Then,

$$ \begin{align*}(1\otimes e_{11})\cdot (M\otimes\mathbf K)\cdot (1\otimes e_{11})\cong M \end{align*} $$

as $A-B$ -equivalence bimodules, where we regard $M\otimes \mathbf K$ as an $A\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodule.

Proof. Because the linear span of the set

$$ \begin{align*}\{x\otimes e_{ij} \, | \, x\in M \, , \, i, j\in \mathbf N \} \end{align*} $$

is dense in $M\otimes \mathbf K$ , $(1\otimes e_{11})\cdot (M\otimes \mathbf K)\cdot (1\otimes e_{11}) \cong M\otimes e_{11}$ as $A-B$ -equivalence bimodules. Hence,

$$ \begin{align*}(1\otimes e_{11})\cdot (M\otimes\mathbf K)\cdot (1\otimes e_{11})\cong M \end{align*} $$

as $A-B$ -equivalence bimodules.▪

Lemma 2.3 With the above notation, let M be an $A-B$ -equivalence bimodule. Then,

$$ \begin{align*}\widetilde{H_A}\otimes_{A\otimes\mathbf K}(M\otimes\mathbf K)\otimes_{B\otimes\mathbf K}H_B \cong M \end{align*} $$

as $A-B$ -equivalence bimodules.

Proof. Let $\pi $ be the map from $\widetilde {H_A}\otimes _{A\otimes \mathbf K}(M\otimes \mathbf K)\otimes _{B\otimes \mathbf K}H_B$ to $(1\otimes e_{11})\cdot (M\otimes \mathbf K)\cdot (1\otimes e_{11})$ defined by

$$ \begin{align*}\pi([a(1\otimes e_{11})]^{\widetilde{}}\otimes x\otimes b(1\otimes e_{11})) =(1\otimes e_{11})\cdot (a^* \cdot x\cdot b)\cdot (1\otimes e_{11}), \end{align*} $$

for any $a\in A\otimes \mathbf K$ , $b\in B\otimes \mathbf K$ , and $x\in M\otimes \mathbf K$ . Then, by easy computations, $\pi $ is an $A-B$ -equivalence bimodule isomorphism of $\widetilde {H_A}\otimes _{A\otimes \mathbf K}(M\otimes \mathbf K)\otimes _{B\otimes \mathbf K}H_B$ onto $(1\otimes e_{11})\cdot (M\otimes \mathbf K)\cdot (1\otimes e_{11})$ . Thus, by Lemma 2.2,

$$ \begin{align*}\widetilde{H_A}\otimes_{A\otimes\mathbf K}(M\otimes\mathbf K)\otimes_{B\otimes\mathbf K}H_B \cong M \end{align*} $$

as $A-B$ -equivalence bimodules.▪

We prepare the following lemma which is applied in the next section.

Proof. Modifying the proof of [Reference Abadie, Eilers and Exel1, Theorem 4.2], we prove this lemma. We suppose that $Y\cong \widetilde {M}\otimes _A X\otimes _A M$ as $B-B$ -equivalence bimodules. Let $L_M$ be the linking $C^*$ -algebra for M defined by

$$ \begin{align*}L_M =\begin{bmatrix} A & M \\ \widetilde{M} & B \end{bmatrix}. \end{align*} $$

Furthermore, let W be the $L_M -L_M$ - equivalence bimodule defined in the proof of [Reference Abadie, Eilers and Exel1, Theorem 4.2], which is defined by

$$ \begin{align*}W=\begin{bmatrix} X & X\otimes_A M \\ Y\otimes_B \widetilde{M} & Y \end{bmatrix}. \end{align*} $$

Let $L_M \rtimes _W \mathbf Z$ be the crossed product of $L_M$ by W, and let

$$ \begin{align*}p=\begin{bmatrix} 1_{M(A)} & 0 \\ 0 & 0 \end{bmatrix} \, , \quad q=\begin{bmatrix} 0 & 0 \\ 0 & 1_{M(B)} \end{bmatrix}. \end{align*} $$

Furthermore, let $N=p(L_M \rtimes _W \mathbf Z)q$ . Then, because $M=pL_M q$ , M is a closed subspace of N. Hence, by the proof of [Reference Abadie, Eilers and Exel1, Theorem 4.2], $A\subset A\rtimes _X \mathbf Z$ and $B\subset B\rtimes _Y \mathbf Z$ are strongly Morita equivalent with respect N and its closed subspace M.

Next, we suppose that $\widetilde {Y}\cong \widetilde {M}\otimes _A X\otimes _A M$ as $B-B$ -equivalence bimodules. Let

$$ \begin{align*}W_0 =\begin{bmatrix} X & X\otimes_A M \\ \widetilde{Y}\otimes_B \widetilde{M} & \widetilde{Y} \end{bmatrix}. \end{align*} $$

Then, $W_0$ is an $L_M -L_M$ -equivalence bimodule. Let $N_0 =p(L_M \rtimes _{W_0}\mathbf Z)q$ . By the above discussions, $A\subset A\rtimes _X \mathbf Z$ and $B\subset B\rtimes _{\widetilde {Y}}\mathbf Z$ are strongly Morita equivalent with respect to $N_0$ and its closed subspace M. On the other hand, there is an isomorphism $\pi $ of $B\rtimes _Y \mathbf Z$ onto $B\rtimes _{\widetilde {Y}}\mathbf Z$ such that $\pi |_{B}={\mathrm {id}}$ on B. Let $X_{\pi }$ be the $B\rtimes _{\widetilde {Y}}\mathbf Z-B\rtimes _Y \mathbf Z$ -equivalence bimodule induced by $\pi $ . Then, B is a closed subspace of $X_{\pi }$ , and we regard B as the trivial $B-B$ -equivalence bimodule, because $\pi |_{B} ={\mathrm {id}}$ on B. Thus, $A\subset A\rtimes _X \mathbf Z$ and $B\subset B\rtimes _Y \mathbf Z$ are strongly Morita equivalent with respect to $N_0 \otimes _{B\rtimes _{\widetilde {Y}}\mathbf Z}X_{\pi }$ and its closed subspace $M\otimes _B B(\cong M)$ . Therefore, we obtain the conclusion.▪

Proof. Because A is $\sigma $ -unital, by [Reference Brown, Green and Rieffel3, Corollary 3.5], there is an automorphism $\alpha $ of $A\otimes \mathbf K$ such that $X\otimes \mathbf K\cong X_{\alpha }$ as $A\otimes \mathbf K -A\otimes \mathbf K$ -equivalence bimodules, where $X_{\alpha }$ is the $A\otimes \mathbf K -A\otimes \mathbf K$ -equivalence bimodule induced by $\alpha $ . Let $u_{\alpha }$ be a unitary element in $M((A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)$ implementing $\alpha $ . We regard $(A\otimes \mathbf K)u_{\alpha }$ as an $A\otimes \mathbf K -A\otimes \mathbf K$ -equivalence bimodule as above. Then, by Lemma 2.1, $X_{\alpha }\cong (A\otimes \mathbf K)u_{\alpha }$ as $A\otimes \mathbf K -A\otimes \mathbf K$ -equivalence bimodules. Let $(A\otimes \mathbf K)\rtimes _{(A\otimes \mathbf K)u_{\alpha }}\mathbf Z$ be the crossed product of $A\otimes \mathbf K$ by $(A\otimes \mathbf K)u_{\alpha }$ . Then, by the definition of the crossed product of a $C^*$ -algebra by an equivalence bimodule, we can see that

$$ \begin{align*}(A\otimes\mathbf K)\rtimes_{\alpha}\mathbf Z\cong (A\otimes\mathbf K)\rtimes_{(A\otimes\mathbf K)u_{\alpha}}\mathbf Z \end{align*} $$

as $C^*$ -algebras. Because $X_{\alpha }\cong X\otimes \mathbf K$ as $A\otimes \mathbf K-A\otimes \mathbf K$ -equivalence bimodules, we obtain that

$$ \begin{align*}(A\otimes\mathbf K)\rtimes_{(A\otimes\mathbf K)u_{\alpha}}\mathbf Z\cong (A\otimes\mathbf K)\rtimes_{X_{\alpha}}\mathbf Z \cong (A\otimes\mathbf K)\rtimes_{X\otimes\mathbf K}\mathbf Z \end{align*} $$

as $C^*$ -algebras. Because the above isomorphisms leave any element in $A\otimes \mathbf K$ invariant, we can see that $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{\alpha } \mathbf Z$ is isomorphic to $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z$ as inclusions of $C^*$ -algebras.▪

3 Strong Morita equivalence

Let A and B be $\sigma $ -unital $C^*$ -algebras and X and Y an $A-A$ -equivalence bimodule and a $B-B$ -equivalence bimodule, respectively. Let $A\subset A\rtimes _X \mathbf Z$ and $B\subset B\rtimes _Y \mathbf Z$ be the inclusions of $C^*$ -algebras induced by X and Y, respectively. We suppose that $A\subset A\rtimes _X \mathbf Z$ and $B\subset B\rtimes _Y \mathbf Z$ are strongly Morita equivalent with respect to an $A\rtimes _X \mathbf Z -B\rtimes _Y \mathbf Z$ -equivalence bimodule N and its closed subspace M. We suppose that $A' \cap M(A\rtimes _X \mathbf Z)=\mathbf C 1$ . Then, because the inclusion $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z$ is isomorphic to the inclusion $A\otimes \mathbf K\subset (A\rtimes _X \mathbf Z)\otimes \mathbf K$ as inclusions of $C^*$ -algebras, by [Reference Kodaka8, Lemma 3.1],

$$ \begin{align*}(A\otimes\mathbf K)' \cap((A\otimes\mathbf K)\rtimes_{X\otimes\mathbf K}\mathbf Z)=\mathbf C 1. \end{align*} $$

Furthermore, by the above assumptions, the inclusion $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z$ is strongly Morita equivalent to the inclusion $B\otimes \mathbf K\subset (B\otimes \mathbf K)\rtimes _{Y\otimes \mathbf K}\mathbf Z$ with respect to the $(A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z-(B\otimes \mathbf K)\rtimes _{Y\otimes \mathbf K}\mathbf Z$ -equivalence bimodule $N\otimes \mathbf K$ and its closed subspace $M\otimes \mathbf K$ . By Lemma 2.5, there are an automorphism $\alpha $ of $A\otimes \mathbf K$ and an automorphism $\beta $ of $B\otimes \mathbf K$ such that $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z$ and $B\otimes \mathbf K\subset (B\otimes \mathbf K)\rtimes _{Y\otimes \mathbf K}\mathbf Z$ are isomorphic to $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z$ and $B\otimes \mathbf K\subset (B\otimes \mathbf K)\rtimes _{\beta }\mathbf Z$ as inclusions of $C^*$ -algebras, respectively. Hence, we can assume that $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z$ and $B\otimes \mathbf K\subset (B\otimes \mathbf K)\rtimes _{\beta }\mathbf Z$ are strongly Morita equivalent with respect to an $(A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z -(B\otimes \mathbf K)\rtimes _{\beta }\mathbf Z$ -equivalence bimodule $N\otimes \mathbf K$ and its closed subspace $M\otimes \mathbf K$ . Because A and B are $\sigma $ -unital, in the same way as in the proof of [Reference Kodaka and Teruya6, Proposition 3.5] or [Reference Brown, Green and Rieffel3, Proposition 3.1], there is an isomorphism $\theta $ of $(B\otimes \mathbf K)\rtimes _{\beta }\mathbf Z$ onto $(A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z$ satisfying the following:

(1) $\theta |_{B\otimes \mathbf K}$ is an isomorphism of $B\otimes \mathbf K$ onto $A\otimes \mathbf K$ ,

(2) There is an $(A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z-(B\otimes \mathbf K)\rtimes _{\beta }\mathbf Z$ - equivalence bimodule isomorphism $\Phi $ of $N\otimes \mathbf K$ onto $Y_{\theta }$ such that $\Phi |_{M\otimes \mathbf K}$ is an $A\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodule isomorphism of $M\otimes \mathbf K$ onto $X_{\theta }$ , where $Y_{\theta }$ is the $(A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z- (B\otimes \mathbf K)\rtimes _{\beta }\mathbf Z$ -equivalence bimodule induced by $\theta $ and $X_{\theta }$ is the $A\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodule induced by $\theta |_{B\otimes \mathbf K}$ .

Let

$$ \begin{align*}\gamma=\theta|_{B\otimes\mathbf K}\circ\beta\circ\theta|_{B\otimes\mathbf K}^{-1}, \end{align*} $$

and let $\lambda $ be the linear automorphism of $X_{\theta }$ defined by $\lambda (x)=\gamma (x)$ for any $x\in X_{\theta }(=A\otimes \mathbf K)$ .

Proof. For any $x, y\in X_{\theta }$ ,

$$ \begin{align*} {}_{A\otimes\mathbf K}\langle \lambda(x) \, , \, \lambda(y) \rangle & =\gamma(xy^* )=\gamma({}_{A\otimes\mathbf K} \langle x \, , \, y \rangle) , \\ \langle \lambda(x) \, ,\, \lambda(y) \rangle _{B\otimes\mathbf K} & =\theta|_{B\otimes\mathbf K}^{-1}(\gamma(x^* y)) =\beta(\theta|_{B\otimes\mathbf K}^{-1}(x^* y))=\beta(\langle x \, \, y \rangle_{B\otimes\mathbf K}). \end{align*} $$

Hence, $\gamma $ and $\beta $ are strongly Morita equivalent with respect to $(X_{\theta }, \lambda )$ .▪

By the proof of [Reference Kodaka8, Theorem 5.5], there is an automorphism $\phi $ of $\mathbf Z$ satisfying that $\gamma ^{\phi }$ and $\alpha $ are exterior equivalent, that is, there is a unitary element $z\in M(A\otimes \mathbf K)$ such that

$$ \begin{align*}\gamma^{\phi}={\mathrm{Ad}}(z)\circ \alpha\, , \quad \underline{\alpha}(z)=z , \end{align*} $$

where $\gamma ^{\phi }$ is the automorphism of $A\otimes \mathbf K$ induced by $\gamma $ and $\phi $ , that is, $\gamma ^{\phi }$ is defined by $\gamma ^{\phi }=\gamma ^{\phi (1)}$ . We note that $\gamma ^{\phi }=\gamma $ or $\gamma ^{\phi }=\gamma ^{-1}$ . We regard $A\otimes \mathbf K$ as the trivial $A\otimes \mathbf K-A\otimes \mathbf K$ -equivalence bimodule. Let $\mu $ be the linear automorphism of $A\otimes \mathbf K$ defined by

$$ \begin{align*}\mu(x)=\alpha(x)z^*, \end{align*} $$

for any $x\in A\otimes \mathbf K$ .

Proof. For any $x, y\in A\otimes \mathbf K$ ,

$$ \begin{align*} {}_{A\otimes\mathbf K} \langle \mu(x) \, , \, \mu(y) \rangle & ={}_{A\otimes\mathbf K} \langle \alpha(x)z^* \, , \, \alpha(y)z^* \rangle =\alpha(xy^* )=\alpha({}_{A\otimes\mathbf K} \langle x \, , \, y \rangle) , \\ \langle \mu(x) \, , \, \mu(y) \rangle_{A\otimes\mathbf K} & =z\alpha(x^* y)z^* =\gamma^{\phi}(x^* y) =\gamma^{\phi}(\langle x \, ,\, y \rangle_{A\otimes\mathbf K}). \end{align*} $$

Therefore, we obtain the conclusion.▪

Let $\nu $ be the linear automorphism of $X_{\theta }$ defined by

$$ \begin{align*}\nu(x)=\gamma^{\phi}(z^* x), \end{align*} $$

for any $x\in X_{\theta }(=A\otimes \mathbf K)$ .

Proof. For any $x, y\in X_{\theta }$ ,

$$ \begin{align*} {}_{A\otimes\mathbf K} \langle \nu(x) \, , \, \nu(y) \rangle & = {}_{A\otimes\mathbf K} \langle \gamma^{\phi}(z^* x) \, , \, \gamma^{\phi}(z^* y) \rangle=\gamma^{\phi}(z^* xy^* z) =z\alpha(z^* xy^* z)z^* \\ & =\alpha(xy^* )=\alpha({}_{A\otimes\mathbf K} \langle x \, , \, y \rangle) , \\ \langle \nu(x) \, , \, \nu(y) \rangle_{B\otimes\mathbf K} & = \langle \gamma^{\phi}(z^* x) \, , \, \gamma^{\phi}(z^* y) \rangle_{B\otimes\mathbf K} =\theta|_{B\otimes\mathbf K}^{-1}(\gamma^{\phi}(x^* y))=\beta^{\phi}(\theta|_{B\otimes\mathbf K}^{-1}(x^* y)) \\ & =\beta^{\phi}(\langle x \, , \, y \rangle_{B\otimes\mathbf K}). \end{align*} $$

Therefore, we obtain the conclusion.▪

Because $\beta ^{\phi }=\beta $ or $\beta ^{\phi }=\beta ^{-1}$ , by Lemma 3.3, $\alpha $ is strongly Morita equivalent to $\beta $ or $\beta ^{-1}$ .

(I) We suppose that $\alpha $ is strongly Morita equivalent to $\beta $ . Then, by Lemma 3.3, there is the linear automorphism $\nu $ of $X_{\theta }$ satisfying the following:

(1) $\nu (a\cdot x)=\alpha (a)\cdot \nu (x)$ ,

(2) $\nu (x\cdot b)=\nu (x)\cdot \beta (b)$ ,

(3) ${}_{A\otimes \mathbf K} \langle \nu (x) \, , \, \nu (y)\rangle =\alpha ({}_{A\otimes \mathbf K} \langle x \, ,\, y \rangle )$ ,

(4) $\langle \nu (x) \, , \, \nu (y)\rangle _{B\otimes \mathbf K}=\beta (\langle x \, ,\, y \rangle _{B\otimes \mathbf K})$ , for any $a\in A\otimes \mathbf K$ , $b\in B\otimes \mathbf K$ , and $x, y\in X_{\theta }$ .

Lemma 3.4 With the above notation and assumptions, let $X_{\alpha }$ and $X_{\beta }$ be the $A\otimes \mathbf K-A\otimes \mathbf K$ -equivalence bimodule and the $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodule induced by $\alpha $ and $\beta $ , respectively. Then,

$$ \begin{align*}X_{\beta}\cong\widetilde{X_{\theta}}\otimes_{A\otimes\mathbf K}X_{\alpha}\otimes_{A\otimes\mathbf K}X_{\theta} \end{align*} $$

as $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodules.

Proof. Let $\Psi $ be the map from $\widetilde {X_{\theta }}\otimes _{A\otimes \mathbf K}X_{\alpha }\otimes _{A\otimes \mathbf K}X_{\theta }$ to $X_{\beta }$ defined by

$$ \begin{align*}\Psi(\widetilde{x}\otimes a\otimes y)=\langle x\, , \, a\cdot \nu(y) \rangle_{B\otimes\mathbf K}, \end{align*} $$

for any $x, y\in X_{\theta }$ and $a\in X_{\alpha }$ . Then, for any $x, x_1 , y, y_1 \in X_{\theta }$ and $a, a_1 \in X_{\alpha }$ ,

$$ \begin{align*} {}_{B\otimes\mathbf K} \langle \widetilde{x}\otimes a\otimes y \, , \, \widetilde{x_1}\otimes a_1 \otimes y_1 \rangle & ={}_{B\otimes\mathbf K} \langle \widetilde{x}\cdot {}_{A\otimes\mathbf K} \langle a\otimes y \, , \, a_1 \otimes y_1 \rangle \, , \, \widetilde{x_1} \rangle \\ & =\langle \, {}_{A\otimes\mathbf K} \langle a_1 \otimes y_1 \, , \, a \otimes y \rangle \cdot x \, , \, x_1 \rangle_{B\otimes\mathbf K} \\ & =\langle \, {}_{A\otimes\mathbf K} \langle a_1 \cdot {}_{A\otimes\mathbf K} \langle y_ 1 \, , \, y \rangle \, , \, a \rangle \cdot x \, , \, x_1 \rangle_{B\otimes\mathbf K} \\ & =\langle \, {}_{A\otimes\mathbf K} \langle a_1 \alpha({}_{A\otimes\mathbf K} \langle y_1 \, , \, y \rangle) \, , \, a \rangle \cdot x \, , \, x_1 \rangle_{B\otimes\mathbf K} \\ & =\langle a_1 \alpha({}_{A\otimes\mathbf K} \langle y_1 \, ,\, y \rangle)a^* \cdot x \, , \, x_1 \rangle_{B\otimes\mathbf K}. \end{align*} $$

On the other hand,

$$ \begin{align*} {}_{B\otimes\mathbf K} \langle \Psi(\widetilde{x}\otimes a\otimes y) \, , \, \Psi(\widetilde{x_1}\otimes a_1 \otimes y_1 )\rangle & ={}_{B\otimes\mathbf K} \langle \langle x\, , \, a\cdot \nu(y) \rangle_{B\otimes\mathbf K} \, , \, \langle x_1 \, , \, a_1 \cdot \nu(y_1 ) \rangle_{B\otimes\mathbf K} \rangle \\ & =\langle x\, , \, a\cdot\nu(y) \rangle_{B\otimes\mathbf K}\, \, \langle a_1 \cdot \nu(y_1 ) \, , \, x_1 \rangle_{B\otimes\mathbf K} \\ & =\langle x\, , \, a\cdot \nu(y)\cdot \langle a_1 \cdot \nu(y_1 ) \, , \, x_1 \rangle_{B\otimes\mathbf K} \, \, \rangle_{B\otimes\mathbf K} \\ & =\langle x \, , \, {}_{A\otimes\mathbf K} \langle a\cdot \nu(y) \, , \, a_1 \cdot \nu(y_1 ) \rangle \cdot x_1 \rangle_{B\otimes\mathbf K} \\ & =\langle x\, , \, a \, {}_{A\otimes\mathbf K} \langle \nu(y) \, , \, \nu(y_1 ) \rangle a_1^* \cdot x_1 \rangle_{B\otimes\mathbf K} \\ & =\langle a_1 \, {}_{A\otimes\mathbf K} \langle \nu(y_1 ) \, , \, \nu(y) \rangle a^* \cdot x\, , \, x_1 \rangle_{B\otimes\mathbf K} \\ & =\langle a_1 \alpha({}_{A\otimes\mathbf K} \langle y_1 \, , \, y \rangle)a^* \cdot x \, , \, x_1 \rangle_{B\otimes\mathbf K}. \end{align*} $$

Hence, $\Psi $ preserves the left $B\otimes \mathbf K$ -valued inner products. Furthermore,

$$ \begin{align*} \langle \widetilde{x}\otimes a\otimes y \, , \, \widetilde{x_1}\otimes a_1 \otimes y_1 \rangle_{B\otimes\mathbf K} & =\langle y \, , \, \langle \widetilde{x}\otimes a \, , \, \widetilde{x_1 }\otimes a_1 \rangle_{A\otimes\mathbf K}\cdot y_1 \rangle_{B\otimes\mathbf K} \\ & =\langle y \, , \, \langle a \, , \, {}_{A\otimes\mathbf K} \langle x\, , \, x_1 \rangle \cdot a_1 \rangle_{A\otimes\mathbf K}\cdot y_1 \rangle_{B\otimes\mathbf K} \\ & =\langle y \, , \, \langle a \, , \, {}_{A\otimes\mathbf K} \langle x\, , \, x_1 \rangle a_1 \rangle_{A\otimes\mathbf K}\cdot y_1 \rangle_{B\otimes\mathbf K} \\ & =\langle y\, , \, \alpha^{-1}(a^* \, {}_{A\otimes\mathbf K} \langle x\, , \, x_1 \rangle a_1 )\cdot y_1 \rangle_{B\otimes\mathbf K}. \end{align*} $$

On the other hand,

$$ \begin{align*} \langle \Psi(\widetilde{x}\otimes a\otimes y)\, , \, \Psi(\widetilde{x_1}\otimes a_1 \otimes y_1 ) \rangle_{B\otimes\mathbf K} & =\langle \langle x\, , \, a\cdot\nu(y) \rangle_{B\otimes\mathbf K}\, , \, \langle x_1 \, , \, a_1 \cdot\nu(y_1 ) \rangle_{B\otimes\mathbf K} \rangle_{B\otimes\mathbf K} \\ & =\beta^{-1}(\langle a\cdot\nu(y) \, , \, x \rangle_{B\otimes\mathbf K}\, \langle x_1 \, , \, a_1 \cdot \nu(y_1 )\rangle_{B\otimes\mathbf K}) \\ & =\beta^{-1}(\langle a\cdot\nu(y) \, , \, x\cdot\langle x_1 \, , \, a_1 \cdot \nu(y_1 ) \rangle_{B\otimes\mathbf K} \, \rangle_{B\otimes\mathbf K}) \\ & =\beta^{-1}(\langle a\cdot\nu(y) \, , \, {}_{A\otimes\mathbf K} \langle x \, , \, x_1 \rangle a_1 \cdot\nu(y_1 ) \rangle_{B\otimes\mathbf K}) \\ & =\beta^{-1}(\langle \nu(y) \, , \, a^* {}_{A\otimes\mathbf K} \langle x \, ,\, x_1 \rangle a_1 \cdot \nu(y_1 ) \rangle_{B\otimes\mathbf K}) \\ & =\langle y \, , \, \alpha^{-1}(a^* \, {}_{A\otimes\mathbf K} \langle x \, , , x_1 \rangle a_1 )\cdot y_1 \rangle_{B\otimes\mathbf K}. \end{align*} $$

Hence, $\Psi $ preserves the right $B\otimes \mathbf K$ -valued inner products. Therefore, we obtain the conclusion.▪

(II) We suppose that $\alpha $ is strongly Morita equivalent to $\beta ^{-1}$ . Then, by Lemma 3.4,

$$ \begin{align*}X_{\beta^{-1}}\cong \widetilde{X_{\theta}}\otimes_{A\otimes\mathbf K}X_{\alpha}\otimes_{A\otimes\mathbf K}X_{\theta} \end{align*} $$

as $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodules. Thus, we obtain the following lemma.

Lemma 3.5 With the above notation and assumptions,

$$ \begin{align*}\widetilde{X_{\beta}}\cong X_{\beta^{-1}} \cong\widetilde{X_{\theta}}\otimes_{A\otimes\mathbf K} X_{\alpha}\otimes_{A\otimes\mathbf K}X_{\theta} \end{align*} $$

as $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodules.

We recall that there is an $(A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z- (B\otimes \mathbf K)\rtimes _{\beta }\mathbf Z$ - equivalence bimodule isomorphism $\Phi $ of $N\otimes \mathbf K$ onto $Y_{\theta }$ such that $\Phi |_{M\otimes \mathbf K}$ is an $A\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodule isomorphism of $M\otimes \mathbf K$ onto $X_{\theta }$ . We identify $M\otimes \mathbf K$ with $X_{\theta }$ by $\Phi |_{M\otimes \mathbf K}$ . Then, by Lemmas 3.4 and 3.5,

$$ \begin{align*}X_{\beta} \cong \widetilde{(M\otimes\mathbf K)}\otimes_{A\otimes\mathbf K}X_{\alpha}\otimes_{A\otimes\mathbf K} (M\otimes\mathbf K) \end{align*} $$

or

$$ \begin{align*}\widetilde{X_{\beta}}\cong(\widetilde{M\otimes\mathbf K)}\otimes_{A\otimes\mathbf K}X_{\alpha} \otimes_{A\otimes\mathbf K}(M\otimes\mathbf K) \end{align*} $$

as $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodules. Furthermore, we recall that $X_{\alpha }\cong X\otimes \mathbf K$ as $A\otimes \mathbf K-A\otimes \mathbf K$ -equivalence bimodules and that $X_{\beta }\cong Y\otimes \mathbf K$ as $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodules. Thus,

$$ \begin{align*}Y\otimes\mathbf K\cong \widetilde{(M\otimes\mathbf K)}\otimes_{A\otimes\mathbf K}(X\otimes\mathbf K) \otimes_{A\otimes\mathbf K}(M\otimes\mathbf K) \end{align*} $$

or

$$ \begin{align*}\widetilde{Y\otimes\mathbf K}\cong(\widetilde{M\otimes\mathbf K)}\otimes_{A\otimes\mathbf K}(X\otimes\mathbf K) \otimes_{A\otimes\mathbf K}(M\otimes\mathbf K) \end{align*} $$

as $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodules. Furthermore, by Lemma 2.3,

$$ \begin{align*}X\otimes\mathbf K\cong H_A \otimes_A X \otimes_A \widetilde{H_A} \end{align*} $$

as $A\otimes \mathbf K-A\otimes \mathbf K$ -equivalence bimodules and

$$ \begin{align*}Y\otimes\mathbf K\cong H_B \otimes_B Y\otimes_B \widetilde{H_B} \end{align*} $$

as $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodules. Thus,

$$ \begin{align*} Y & \cong\widetilde{H_B}\otimes_{B\otimes\mathbf K}(Y\otimes\mathbf K)\otimes_{B\otimes\mathbf K}H_B \\ & \cong\widetilde{H_B}\otimes_{B\otimes\mathbf K}(\widetilde{M\otimes\mathbf K})\otimes_{A\otimes\mathbf K} (X\otimes\mathbf K)\otimes_{A\otimes\mathbf K}(M\otimes\mathbf K)\otimes_{B\otimes\mathbf K}H_B \\ & \cong\widetilde{H_B}\otimes_{B\otimes\mathbf K}(\widetilde{M\otimes\mathbf K})\otimes_{A\otimes\mathbf K} H_A \otimes_A X\otimes_A \widetilde{H_A}\otimes_{A\otimes\mathbf K}(M\otimes\mathbf K)\otimes_{B\otimes\mathbf K}H_B \end{align*} $$

or similarly

$$ \begin{align*}\widetilde{Y}\cong\widetilde{H_B}\otimes_{B\otimes\mathbf K}(\widetilde{M\otimes\mathbf K}) \otimes_{A\otimes\mathbf K} H_A \otimes_A X\otimes_A \widetilde{H_A}\otimes_{A\otimes\mathbf K}(M\otimes\mathbf K)\otimes_{B\otimes\mathbf K}H_B \end{align*} $$

as $B\otimes \mathbf K-B\otimes \mathbf K$ -equivalence bimodules. Furthermore, by Lemma 2.3,

$$ \begin{align*}\widetilde{H_A}\otimes_{A\otimes\mathbf K}(M\otimes\mathbf K)\otimes_{B\otimes\mathbf K} H_B \cong M \end{align*} $$

as $A-B$ -equivalence bimodules. Hence,

$$ \begin{align*}Y\cong\widetilde{M}\otimes_A X\otimes_A M \quad \text{or}\quad \widetilde{Y}\cong\widetilde{M}\otimes_A X\otimes_A M \end{align*} $$

as $B-B$ -equivalence bimodules. Therefore, we obtain the following theorem.

4 The Picard groups

Let A be a unital $C^*$ -algebra and X an $A-A$ -equivalence bimodule. Let $A\subset A\rtimes _X \mathbf Z$ be the inclusion of unital $C^*$ -algebras induced by X. We suppose that $A' \cap (A\rtimes _X \mathbf Z)=\mathbf C 1$ . In this section, we shall compute ${\mathrm {Pic}}(A, A\rtimes _X \mathbf Z)$ , the Picard group of the inclusion $A\subset A\rtimes _X \mathbf Z$ (See [Reference Kodaka and Teruya6]).

Let G be the subgroup of ${\mathrm {Pic}}(A)$ defined by

$$ \begin{align*} G=\{[M]\in{\mathrm{Pic}}(A) \, | \, X\cong\widetilde{M}\otimes_A X\otimes_A M \, \text{or} \, \widetilde{X} & \cong\widetilde{M}\otimes_A X\otimes_A M \\ & \text{as }A-A\text{-equivalence bimodules}\}. \end{align*} $$

Let $f_A$ be the homomorphism of ${\mathrm {Pic}}(A, A\rtimes _X \mathbf Z)$ to ${\mathrm {Pic}}(A)$ defined by

$$ \begin{align*}f_A ([M, N])=[M] \end{align*} $$

for any $[M, N]\in {\mathrm {Pic}}(A, A\rtimes _X \mathbf Z)$ . First, we show ${\mathrm {Im}} f_A=G$ , where ${\mathrm {Im}} f_A$ is the image of $f_A$ .

Next, we compute ${\mathrm {Ker}} f_A$ , the kernel of $f_A$ . Let ${\mathrm {Aut}} (A, A\rtimes _X \mathbf Z)$ be the group of all automorphisms $\alpha $ of $A\rtimes _X \mathbf Z$ such that $\alpha |_A$ is an automorphism of A. Let ${\mathrm {Aut}}_0 (A, A\rtimes _X \mathbf Z)$ be the group of all automorphisms $\alpha $ of $A\rtimes _X \mathbf Z$ such that $\alpha |_A ={\mathrm {id}}$ on A. It is clear that ${\mathrm {Aut}}_0 (A, A\rtimes _X \mathbf Z)$ is a normal subgroup of ${\mathrm {Aut}} (A, A\rtimes _X \mathbf Z)$ . Let $\pi $ be the homomorphism of ${\mathrm {Aut}}(A, A\rtimes _X \mathbf Z)$ to ${\mathrm {Pic}}(A, A\rtimes _X \mathbf Z)$ defined by

$$ \begin{align*}\pi(\alpha)=[M_{\alpha}, N_{\alpha}], \end{align*} $$

for any $\alpha \in {\mathrm {Aut}}(A, A\rtimes _X \mathbf Z)$ , where $[M_{\alpha }, N_{\alpha }]$ is an element in ${\mathrm {Pic}}(A, A\rtimes _X \mathbf Z)$ induced by $\alpha $ (See [Reference Kodaka and Teruya6, Section 3]).

Lemma 4.2 With the above notation,

$$ \begin{align*}{\mathrm{Ker}} f_A =\{[A, N_{\beta}]\in{\mathrm{Pic}}(A, A\rtimes_X \mathbf Z) \, | \, \beta\in{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z) \}. \end{align*} $$

Proof. Let $[M, N]\in {\mathrm {Ker}} f_A$ . Then, $[M]=[A]$ in ${\mathrm {Pic}}(A)$ , and by [Reference Kodaka and Teruya6, Lemma 7.5], there is a $\beta \in {\mathrm {Aut}}_0 (A, A\rtimes _X \mathbf Z)$ such that

$$ \begin{align*}[M, N]=[A, N_{\beta}] \end{align*} $$

in ${\mathrm {Pic}}(A, A\rtimes _X \mathbf Z)$ , where $N_{\beta }$ is the $A\rtimes _X \mathbf Z -A\rtimes _X \mathbf Z$ -equivalence bimodule induced by $\beta $ . Therefore, we obtain the conclusion.▪

Let ${\mathrm {Int}}(A, A\rtimes _X \mathbf Z)$ be the group of all ${\mathrm {Ad}}(u)$ such that u is a unitary element in A. By [Reference Kodaka and Teruya6, Lemma 3.4],

$$ \begin{align*}{\mathrm{Ker}} \, \pi \cap{\mathrm{Aut}}_0 (A, A\rtimes_X\mathbf Z)={\mathrm{Int}}(A, A\rtimes_X \mathbf Z)\cap {\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z). \end{align*} $$

Hence,

$$ \begin{align*} & {\mathrm{Ker}} \, \pi\cap{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z) \\ &\quad =\{{\mathrm{Ad}}(u)\in{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z) \, | \, u \text{ is a unitary element in }A\} \\ &\quad =\{{\mathrm{Ad}}(u)\in{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z) \, | \, u \text{ is a unitary element in }A' \cap A\}. \end{align*} $$

Because $A' \cap (A\rtimes _X \mathbf Z)=\mathbf C1$ , $A' \cap A=\mathbf C1$ . Thus,

$$ \begin{align*}{\mathrm{Ker}} \, \pi \cap{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z)=\{1\}. \end{align*} $$

It follows that we can obtain the following lemma.

Proof. Because ${\mathrm {Ker}} \, \pi \cap {\mathrm {Aut}}_0 (A, A\rtimes _X \mathbf Z)=\{1\}$ , by Lemma 4.2,

$$ \begin{align*} {\mathrm{Ker}} f_A & =\pi({\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z))\cong{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z)/ ({\mathrm{Ker}} \, \pi\cap{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z) )\\ & ={\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z). \end{align*} $$

Therefore, we obtain the conclusion.▪

We recall that the inclusions of $C^*$ -algebras $A\otimes \mathbf K\subset (A\rtimes _X \mathbf Z)\otimes \mathbf K$ and $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z$ are isomorphic as inclusions of $C^*$ -algebras. Furthermore, there is an automorphism $\alpha $ of $A\otimes \mathbf K$ such that $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{X\otimes \mathbf K}\mathbf Z$ and $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z$ are isomorphic as inclusions of $C^*$ -algebras.

Proof. Because $A' \cap (A\rtimes _X \mathbf Z)=\mathbf C1$ , by [Reference Kodaka8, Lemma 3.1], $(A\otimes \mathbf K)' \cap M((A\rtimes _X \mathbf Z)\otimes \mathbf K)=\mathbf C1$ . Hence, because $A\otimes \mathbf K\subset (A\rtimes _X \mathbf Z)\otimes \mathbf K$ is isomorphic to $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z$ as inclusions of $C^*$ -algebras,

$$ \begin{align*}(A\otimes\mathbf K)' \cap M((A\otimes\mathbf K)\rtimes_{\alpha}\mathbf Z)=\mathbf C1. \end{align*} $$

Thus, by [Reference Kodaka8, Corollary 4.2], the action $\alpha $ is free.▪

For any $n\in \mathbf Z$ , let $\delta _n$ be the function on $\mathbf Z$ defined by

$$ \begin{align*}\delta_n (m) =\begin{cases} 1 & m=n \\ 0 & m\ne n \end{cases}. \end{align*} $$

We regard $\delta _n$ as an element in $M((A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)$ .

Let $E^{M(A\otimes \mathbf K)}$ be the canonical faithful conditional expectation from $M(A\otimes \mathbf K)\rtimes _{\underline {\alpha }}\mathbf Z$ onto $M(A\otimes \mathbf K)$ defined in [Reference Bédos and Conti2, Section 3]. Then, we may let $E^{A\otimes \mathbf K}$ be the restriction of $E^{M(A\otimes \mathbf K)}$ to $(A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z$ , that is, $E^{A\otimes \mathbf K}=E^{M(A\otimes \mathbf K)}|_{(A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z}$ . Let $\{u_i \}_{i\in I}$ be an approximate unit of $A\otimes \mathbf K$ . We fix the approximate unit $\{u_i \}_{i\in I}$ of $A\otimes \mathbf K$ . For any $x\in M((A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)$ , we define the Fourier coefficient of x at $n\in \mathbf Z$ as in the same way as in [Reference Kodaka8, Section 2]. We show that ${\mathrm {Aut}}_0 (A\otimes \mathbf K, (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)\cong \mathbf T$ .

Let $\beta \in {\mathrm {Aut}}_0 (A\otimes \mathbf K, (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)$ . For any $a\in A\otimes \mathbf K$ ,

$$ \begin{align*}\underline{\beta}(\delta_1 )a\underline{\beta}(\delta_1^* )=\beta(\delta_1 a \delta_1^* )=\beta(\alpha(a)) =\alpha(a). \end{align*} $$

Hence, $\underline {\beta }(\delta _1 )a=\alpha (a)\underline {\beta }(\delta _1 )$ for any $a\in A\otimes \mathbf K$ .

Lemma 4.5 With the above notation, let $a_n$ be the Fourier coefficient of $\underline {\beta }(\delta _1 )$ at $n\in \mathbf Z$ . Then, for any $a\in A\otimes \mathbf K$ ,

$$ \begin{align*}a_n \alpha^{n-1}(a)=aa_n. \end{align*} $$

Proof. Let $a\in A\otimes \mathbf K$ . Then, because $||au_i -u_i a||\to 0 (i\to \infty )$ , the Fourier coefficient of $\underline {\beta }(\delta _1 )a$ at $n\in \mathbf Z$ is given by

$$ \begin{align*} \lim_i E^{A\otimes\mathbf K}(\underline{\beta}(\delta_1 )au_i \delta_n ) & = \lim_i E^{A\otimes\mathbf K}(\underline{\beta}(\delta_1 )u_i a\delta_n ) \\ & =\lim_i E^{A\otimes\mathbf K}(\underline{\beta}(\delta_1 )u_i \delta_n \alpha^n (a)) \\ & =\lim_i E^{A\otimes\mathbf K}(\underline{\beta}(\delta_1 )u_i \delta_n ) \alpha^n (a) \\ & =a_n \alpha^n (a). \end{align*} $$

Furthermore, the Fourier coefficient of $\alpha (a)\underline {\beta }(\delta _1 )$ at $n\in \mathbf Z$ is given by

$$ \begin{align*}\lim_i E^{A\otimes\mathbf K}(\alpha(a)\underline{\beta}(\delta_1 )u_i \delta_n )=\alpha(a)\lim_i E^{A\otimes\mathbf K} (\underline{\beta}(\delta_1 )u_i \delta_n )=\alpha(a)a_n. \end{align*} $$

Because $\underline {\beta }(\delta _1 )a=\alpha (a)\underline {\beta }(\delta _1 )$ , we get that

$$ \begin{align*}a_n \alpha^n (a)=\alpha(a)a_n, \end{align*} $$

for any $a\in A\otimes \mathbf K$ . Because a is an arbitrary element in $A\otimes \mathbf K$ , replacing a by $\alpha ^{-1}(a)$ , we obtain the conclusion.▪

Lemma 4.6 With the above notation,

$$ \begin{align*}{\mathrm{Aut}}_0 (A\otimes\mathbf K, (A\otimes\mathbf K)\rtimes_{\alpha}\mathbf Z)\cong\mathbf T. \end{align*} $$

Proof. Let $\beta \in {\mathrm {Aut}}_0 (A\otimes \mathbf K, (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)$ , and let $a_n$ be the Fourier coefficient of $\underline {\beta }(\delta _1 )$ at $n\in \mathbf Z$ . Then, by Lemma 4.5, $a_n \alpha ^{n-1}(a)=aa_n$ for any $a\in A\otimes \mathbf K$ . Because the automorphism $\alpha ^{n-1}$ is free for any $n\in \mathbf Z\setminus \{1\}$ by Lemma 4.4, $a_n =0$ for any $n\in \mathbf Z\setminus \{1\}$ . Thus, $\underline {\beta }(\delta _1 )=a_1 \delta _1$ . Because $\underline {\beta }(\delta _1 )a\underline {\beta }(\delta _1^* )=\alpha (a)$ , for any $a\in A\otimes \mathbf K$ ,

$$ \begin{align*}a_1 \delta_1 a \delta_1^* a_1^* =\alpha(a). \end{align*} $$

Because $\delta _1 a\delta _1^* =\alpha (a)$ ,

$$ \begin{align*}a_1 \alpha(a)a_1^* =\alpha(a), \end{align*} $$

for any $a\in A\otimes \mathbf K$ . Because $\delta _1$ and $\underline {\beta }(\delta _1 )$ are unitary elements in $M((A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)$ , $a_1$ is a unitary element in $M(A\otimes \mathbf K)$ . Thus,

$$ \begin{align*}a_1 \alpha(a)=\alpha(a)a_1, \end{align*} $$

for any $a\in A\otimes \mathbf K$ . Because $(A\otimes \mathbf K)' \cap M(A\otimes \mathbf K)=\mathbf C1$ , $a_1 \in \mathbf C1$ . Because $a_1$ is a unitary element in $M(A\otimes \mathbf K)$ , there is the unique element $c_{\beta }\in \mathbf T$ such that $a_1 =c_{\beta }1$ . Let $\epsilon $ be the map from ${\mathrm {Aut}}_0 (A\otimes \mathbf K, (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)$ onto $\mathbf T$ defined by $\epsilon (\beta )=c_{\beta }$ . By routine computations, we can see that $\epsilon $ is an isomorphism of ${\mathrm {Aut}}_0 (A\otimes \mathbf K, (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z)$ onto $\mathbf T$ .▪

Lemma 4.7 With the above notation,

$$ \begin{align*}{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z)\cong{\mathrm{Aut}}_0 (A\otimes\mathbf K, (A\otimes\mathbf K)\rtimes_{\alpha}\mathbf Z). \end{align*} $$

Proof. Because $A\otimes \mathbf K\subset (A\otimes \mathbf K)\rtimes _{\alpha }\mathbf Z$ and $A\otimes \mathbf K\subset (A\rtimes _X\mathbf Z)\otimes \mathbf K$ are isomorphic as inclusions of $C^*$ -algebras, it suffices to show that

$$ \begin{align*}{\mathrm{Aut}}_0 (A, A\rtimes_X \mathbf Z)\cong {\mathrm{Aut}}_0 (A\otimes\mathbf K, (A\rtimes_X \mathbf Z)\otimes\mathbf K). \end{align*} $$

Let $\kappa $ be the homomorphism of ${\mathrm {Aut}}_0 (A, A\rtimes _X \mathbf Z)$ to ${\mathrm {Aut}}_0 (A\otimes \mathbf K, (A\rtimes _X \mathbf Z)\otimes \mathbf K)$ defined by

$$ \begin{align*}\kappa(\beta)=\beta\otimes{\mathrm{id}}_{\mathbf K}, \end{align*} $$

for any $\beta \in {\mathrm {Aut}}_0 (A, A\rtimes _X \mathbf Z)$ . Then, it is clear that $\kappa $ is a monomorphism of ${\mathrm {Aut}}_0 (A, A\rtimes _X \mathbf Z)$ to ${\mathrm {Aut}}_0 (A\otimes \mathbf K, (A\rtimes _X \mathbf Z)\otimes \mathbf K)$ . We show that $\kappa $ is surjective. Let $\gamma \in {\mathrm {Aut}}_0 (A\otimes \mathbf K, (A\rtimes _X \mathbf Z)\otimes \mathbf K)$ . Then,

$$ \begin{align*}\gamma(a\otimes e_{ij})=a\otimes e_{ij}, \end{align*} $$

for any $a\in A$ , $i, j \in \mathbf N$ . Thus,

$$ \begin{align*}\gamma(x\otimes e_{11})=(1\otimes e_{11})\gamma(x\otimes e_{11})(1\otimes e_{11}), \end{align*} $$

for any $x\in A\rtimes _X \mathbf Z$ . Hence, there is an automorphism $\beta $ of $A\rtimes _X \mathbf Z$ such that

$$ \begin{align*}\gamma(x\otimes e_{11})=\beta(x)\otimes e_{11}, \end{align*} $$

for any $x\in A\rtimes _X \mathbf Z$ . For any $i, j\in \mathbf N$ and $x\in A\rtimes _X \mathbf Z$ ,

$$ \begin{align*} \gamma(x\otimes e_{ij}) & =\gamma((1\otimes e_{i1})(x\otimes e_{11})(1\otimes e_{1j})) =(1\otimes e_{i1})(\beta(x)\otimes e_{11})(1\otimes e_{1j}) \\ & =\beta(x)\otimes e_{ij}. \end{align*} $$

Especially, if $a\in A$ , $\beta (a)\otimes e_{ij}=\gamma (a\otimes e_{ij})=a\otimes e_{ij}$ , for any $i, j\in \mathbf N$ . Thus, $\beta (a)=a$ , for any $a\in A$ . Therefore, $\gamma =\beta \otimes {\mathrm {id}}_{\mathbf K}$ and $\beta \in {\mathrm {Aut}}_0 (A, A\rtimes _X \mathbf K)$ . Hence, we have shown that $\kappa $ is surjective.▪

By Lemmas 4.1 and 4.8, we have the following exact sequence:

$$ \begin{align*}1\longrightarrow\mathbf T\longrightarrow{\mathrm{Pic}}(A, A\rtimes_X \mathbf Z)\longrightarrow G\longrightarrow 1, \end{align*} $$

where

$$ \begin{align*} G=\{[M]\in{\mathrm{Pic}}(A) \, | \, X\cong\widetilde{M}\otimes_A X\otimes_A M \, \text{or} \, \widetilde{X} & \cong\widetilde{M}\otimes_A X\otimes_A M \\ & \text{as }A-A\text{-equivalence bimodules}\}. \end{align*} $$

Let g be the map from G to ${\mathrm {Pic}}(A, A\rtimes _X \mathbf Z)$ defined by

$$ \begin{align*}g([M])=[M, N], \end{align*} $$

where N is the $A\rtimes _X \mathbf Z -A\rtimes _X \mathbf Z$ -equivalence bimodule defined in the proof of Lemma 2.4. Then, g is a homomorphism of G to ${\mathrm {Pic}}(A, A\rtimes _X \mathbf Z)$ such that

$$ \begin{align*}f_A \circ g={\mathrm{id}} \end{align*} $$

on G. Thus, we obtain the following theorem.