TITLE: When are countably generated Hilbert modules generated by c.p.c. order zero maps?
QUESTION [5 upvotes]: Throughout let $B$ be a stable C*-algebra, i.e. $B\cong B\otimes K$, where $K$ is the C*-algebra of compact operators on an infinite dimensional separable Hilbert space. It is well-known that any countably generated Hilbert $B$-module $X$ is singly generated, i.e. there exists a positive element $b\in B$ such that $X\cong \overline{bB}$.
Assume the following as the definition of c.p.c. order zero maps between C*-algebras.
Definition A completely positive linear map $\phi:A\to B$ between C*-algebras has order zero if there exist a positive element $h\in\mathcal M(C)\cap C'$ and a $*$-homomorphism $\pi:A\to\mathcal M(C)\cap\{h\}'$ such that $\Vert h\Vert = 1$ and
$$\phi(a) = h\pi(a)=\pi(a)h$$
for any $a\in A$, where $C = C^*(\phi(A))\subset B$, i.e. the C*-algebra generated by the image of $\phi$ and $\mathcal M(C)$ is the multiplier algebra of $C$.
Let $A$ be a separable C*-algebra and let $\phi:A\to B$ be a c.p.c. order zero map. One can construct a Hilbert $B$-module $X_\phi$ out of $\phi$ by setting
$$X_\phi:=\overline{\phi(A)B}.$$
Conversely, given a countably generated Hilbert $B$-module $X$ and a separable C*-algebra $A$, when is that there exists a c.p.c. order zero map $\phi:A\to B$ such that $X\cong\overline{\phi(A)B}$?
As a special case, suppose that $X$ is the inductive limit of a sequence of isometric inclusions of modules $\overline{\phi_n(A)B}\hookrightarrow\overline{\phi_{n+1}(A)B}$, where $\{\phi_n\}_{n\in\mathbb N}$ is a sequence of c.p.c. order zero maps. Is there a c.p.c. order zero map $\phi$ such that $X\cong\overline{\phi(A)B}$? I believe this last question has a positive answer when the connecting maps "commute" with the c.p.c. order zero maps, so I would rather be interested in the case where there are no a priori connections between the $\phi_n$s.

REPLY [4 votes]: If $A=\mathbb C$ then the answer to both questions is yes.
If $A=M_2(\mathbb C)$, then the modules $H=\overline{\phi(A)B}$ are those that have a direct sum decomposition $H\cong E\oplus E$ (where $E= \overline{\phi(e_{1,1})B}$).
It is clear that not all modules need to have this property. The answer to the second question is also negative in this case. One can arrange for this: a locally compact space $X$ covered by compactly contained open sets $\bigcup_n U_n=X$
and a dimension 2 vector bundle over $X$ that is trivial on all the sets $U_n$
but non-trivial on $X$ (a ``phantom" vector bundle). ($X$ can be obtained by a telescoping construction. See Example 5.6 of http://arxiv.org/abs/0910.2967). Viewing the vector bundle as a Hilbert module $H$ over $C_0(X)$, the Hilbert modules $HC_0(U_n)$ have the desired direct sum decomposition but $H$ itself does not.
If $A=\mathcal K$, then the modules in question have the form $\bigoplus_{n=1}^\infty E$ (a.k.a., a stable module). By Kasparov's stabilization  $H$ is isomorphic to $\ell^2(I)$, where $I$ is a closed two-sided ideal of $B$. Again this need not exhaust all possible modules, but the  second question has a positive answer in this case. If $H=\overline{\bigcup_n H_n}$ and all the modules $H_n$ are stable (and countably generated) then $H$ is stable.