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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.