This question is about mapping finitely generated projective modules over $C^*$ algebras (or K-theory classes - I don't mind much) under maps between algebras. If we have an idempotent matrix $P\in M_n(A)$ and a star algebra map $\phi:A\to B$ then taking $\phi(P)$ (applying to each entry) will give another idempotent. However suppose that we only have a completely positive map $\psi:A\to B$ -- what do we do then?

One possible way would be to use the KSGNS theorem to say that $\psi$ is given by a Hilbert bimodule structure on some $M$ together with an element $m\in M$. The module given by the applying the map to an $A$-module $E$ might be taken as $M\otimes_A E$ (or rather the conjugate of $M$ to get the sides right...). However this seems to be a very non-unique construction, and likely far too big.

The reason I am interested in this is the problem of calculating characteristic classes (a la Connes) for algebras which simply do not have any nicely calculable $n$-cycles. It may be simplest to transfer the bundle to an easier to deal with algebra. (Just as classically we can define a cycle on a manifold by using an embedded submanifold). However then we run into the problem that we often only have CP maps, not algebra maps... I would be grateful for any assistance.


1 Answer 1


This is not really an answer, but an idea. I am not sure if it works, plus you need an extra assumption on the completely positive map.

Definition: We call a completely positive map $\psi \colon A \to B$ of order zero if it is orthogonality preserving, i.e. if $ab = 0$ implies that $\psi(a)\,\psi(b) = 0$ for any two self-adjoint elements $a,b \in A$.

Let $A$ and $B$ be unital $C^*$-algebras and let $\psi \colon A \to B$ be a completely positive order zero map. There is a powerful structure theorem for these maps, which can be found in Section 3.3 of the paper "Completely positive maps of order zero" by Wilhelm Winter and Joachim Zacharias: Let $C = C^*(\psi(A))$ be the $C^*$-algebra generated by the image of $\psi$, then $h = \psi(1_A) \in C$ commutes with $C$ and there is a $*$-homomorphism $\pi_{\psi} \colon A \to M(C) \cap \{h\}'$ such that $$ h^{1/2}\,\pi_{\psi}(a)\,h^{1/2} = h\,\pi_{\psi}(a) = \pi_{\psi}(a)h = \psi(a) $$ for all $a \in A$. This says that completely positive order zero maps can always be written as compressions of a $*$-homomorphism into a bigger algebra by a positive element.

If you now start with a projection $p \in A$ (or in a matrix algebra over $A$), then you can consider the sequence $$ q_n = \left(h +\frac{1}{n}\right)^{-1}\psi(p) . $$ This should get arbitrarily close to a projection in $B$ and hence can be cut down to an honest projection for $n$ big enough.

  • $\begingroup$ There is a problem with this: You have to take the strong limit of the sequence in a representation and end up with a projection in some $B(H)$. I don't know if it really lies in $B$ in the end... $\endgroup$ Jan 17, 2017 at 17:15
  • $\begingroup$ You have to add that $a, b$ are self-adjoint in the definition of "order zero". $\endgroup$ Jan 21, 2017 at 20:06
  • $\begingroup$ @SabrinaGemsa Thank you! I fixed that. $\endgroup$ Jan 22, 2017 at 13:56

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