## Idea

For rings, the classical Morita theorem, which says that for Morita equivalent rings $R,S$ there is an $R-S$-bimodule ${}_R X_S$ that takes ${}_S M$ to ${}_R(X\otimes_S M)$. For $C^\ast$-algebras, if one uses the Hilbert space representations as the category of modules, there is no such theorem.

Given a right Hilbert $B$-module $X_B$, a left action of $A$ by adjointable operators converts representations $\pi$ of $B$ to representations $X-\operatorname{Ind}\pi$ of $A$ on $X\otimes_{B} \mathcal{H}_\pi$. Rieffel’s notion of strong Morita equivalence extends this to a equivalence relation on $C^\ast$-algebras implemented by a suitable Hilbert module.

## Definition

*imprimitivity bimodule*is an full $A-B$-Hilbert bimodule such that for $x,y\in X$, $a\in A$, $b\in B$, $$ \begin{gather} \langle a\cdot x,y \rangle_B = \langle x, a^\ast\cdot y\rangle_B\\\ {}_A\langle x\cdot b,y\rangle = {}_A\langle x, y\cdot b^\ast \rangle\\\ {}_A\langle x,y\rangle\cdot z = x\cdot\langle y,z\rangle_B \end{gather} $$

Some points to be made:

- The conditions say that $A$ acts by adjointable operators on $X_{B}$ and $B$ acts by adjointable operators on ${}_{A}X$ and two algebra-valued inner products are compatible.
- A Hilbert $A$-module $X$ is
*full*if the ideal $I=\operatorname{span}\{\langle x,y\rangle_A | x,y\in X\}$ is dense in $A$. The fullness makes left and right multiplications adjointable automatically, so what is really needed is the third condition. - The adjoints of a operator on a Hilbert module does not automatically exist. Every adjointable map $rho:X\to Y$ between Hilbert $A$-modues is a bounded linear $A$-module map.

*strongly Morita equivalence*if there is an $A-B$-imprimitivity bimodule ${}_A X_B$.

## Internal Tensor Product and Equivalence

The *internal tensor product* ${}_A X_B\otimes {}_B Y_C$, of two imprimitivity bimodules $X,Y$, is defined in such a way that it is in turn an $A-C$-imprimitivity bimodule, thus providing transitivity for strong Morita equivalence.

*internal tensor product*$X\otimes_B Y$ is the completion of the $B$-balanced algebraic tensor product $X\odot_{B} Y$ with respect to the $A$- and $C$-valued pre-inner products $$ \begin{gather} \langle x\otimes_B y, z\otimes_B w\rangle_C = \langle \langle z,x\rangle_B\cdot y, w\rangle_C\\\ {}_A\langle x\otimes_B y, z\otimes_B w\rangle = {}_A\langle x,z\cdot {}_B\langle w \cdot\ y\rangle\rangle. \end{gather} $$

$B$-balanced means $(x\cdot b) \otimes_B y = x\otimes_B (b\cdot y)$, that is, $X\odot_B Y = X\odot Y / \sim$ where $\sim$ is the subspace spanned by $\{(x\cdot b)\otimes y - x\otimes(b\cdot y)| x\in X, y\in Y, b\in B\}$.

The internal tensor product is in turn an $A-C$-imprimitivity bimodule.

## Some Results

- Two unital $C^\ast$-algebras are Strongly Morita equivalent iff they are Morita equivalent as abstract algebras.
- Two $\sigma$-unital $C^\ast$-algebras (have countable approximate identity) are strongly Morita equivalent if they are stably isomorphic.
- Two separable $C^\ast$-algebras (the spectrums are separable,
*i.e.*have countable dense subset) are strongly Morita equivalent iff they are stably isomorphic.

There is another characterzation of Morita equivalence by corners. A *corner* is a $C^\ast$-subalgebra of the form $pAp$ where $p$ is a self-adjoit idempotent in the multiplier algebra of $A$. If $ApA$ is dense in $A$ then it is a full corner. The opposite corner to $pAp$ is $(1-p)A(1-p)$, now

The bicategory $\mathsf{C}^\ast-\operatorname{Alg}$ formed by $C^\ast$-algebras with morphisms imprimitivity bimodules and 2-morphisms intertwiners form a $(2,1)$-category, *i.e.*, all the intertwiners are isomorphisms.