I'm trying to get an understanding of Hilbert $C^\*$-bi-modules from a geometric point of view. As is well-known, we have that


i) Commutative unital $C^\*$-algebras correspond to compact Hausdorff spaces through the Gelfand--Naimark theorem and the identification $X \mapsto C(X)$.


ii) In this setting, Hermtian vector bundles are generalized as Hilbert $C^\*$-modules.


Does the notion of a Hilbert $C^\*$-bimodule, over a pair of $C^\*$-algebras $(A,B)$, have a corresponding geometric interpretation in the classical case?


My guess, since we are now dealing with *two* $C^\*$-algebras, there must a second space whose functions can multiply the sections of the bundle. The only way I can think of doing this is if the new space is simply a subspace of the first. But this doesn't feel like the "correct picture" here.



Geometric Motivation for Hilbert $C^*$-Bimodules
Since an $(A,B)$-module is an $A\otimes B^{\text{op}}$ module and since in case $A\simeq C(X)$, $B\simeq C(Y)$ we have that $A\otimes B^{\text{op}}\simeq C(X\times Y)$, we get that an $(A,B)$-module is an $C(X\times Y)$-module.
Thus your geometric example would be an Hermtian vector bundle (over a space $Z$ endowed with maps $Z\to X$, $Z\to Y$).


Let me detail the trivial case. 
Fix measures on $X$ and $Y$. Take the trivial line bundle on $X\times Y$. The space of $L^2$-sections is merely $L^2(X\times Y)$. A nice way of interpreting this space as a bi-module is by thinking of it as the space of Hilbert-Schmidt operators from $L^2(X)$ to $L^2(Y)$, acted by post- and pre-compositions of the multipliers $C(X)$ and $C(Y)$ on these spaces.


