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.


1 Answer 1


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.

  • $\begingroup$ Many thanks for your answer! I was wondering if the case of $A=\mathcal K$ generalises to any stable C*-algebra just by considering $E_A:=\overline{\phi(A\otimes e)B}$, where $e\in\mathcal K$ is any minimal projection; and if there is the possibility of explicitly constructing the c.p.c. order zero map associated to the limit. I was thinking along the lines of a representation of $A$ on $\ell^2(\mathbb N)$ tensor with some positive element in $I$, but I'm not sure to what extent this intuition is correct. $\endgroup$
    – Phoenix87
    Apr 17, 2015 at 13:55
  • $\begingroup$ In that case there are still obstructions. Take the case $B=\mathcal K$. If there is a non-zero order zero map $\phi\colon A\otimes \mathcal K\to \mathcal K$ then $A\otimes\mathcal K$ has a non-trivial densely finite trace (by functional calculus on $\phi$ you can make sure that it has finite rank operators in its range). But $A\otimes \mathcal K$ may not have non-zero densely finite traces. $\endgroup$ Apr 17, 2015 at 21:40
  • $\begingroup$ Perhaps I'm overlooking something, but if there are no non-trivial c.p.c. order zero maps between $A\otimes\mathcal K$ and $B$ then the only sequence one can construct out of a countable family of c.p.c. order zero maps is the constant sequence given by the trivial module, which has limit in the trivial module itself. $\endgroup$
    – Phoenix87
    Apr 18, 2015 at 15:22
  • 1
    $\begingroup$ Oh OK, that's right. My comment was more relevant for the first question: not all stable modules will arise in this construction when $A$ is an arbitrary stable algebra. Regarding the second question, you are right that the answer is yes when $A$ is stable. For these reasons: (1) the range of modules arising by the construction is closed under countable orthogonal sums, (2) the limit of an increasing sequence of stable modules is isomorphic to their direct sum. $\endgroup$ Apr 20, 2015 at 2:38
  • $\begingroup$ OK many thanks. That was really helpful! $\endgroup$
    – Phoenix87
    Apr 20, 2015 at 7:29

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