4
$\begingroup$

I am looking for a reference for the following result:

If $A$ and $B$ are C* algebras, $H$ is a right Hilbert $A$-modules, $\phi :A \rightarrow B$ is a morphism, and assume that there is a map $\eta : H \rightarrow B$ such that:

$\eta$ is $\mathbb{C}$-linear, $\eta(h) \phi(a)= \eta(ha)$ and $\eta(h)^* \eta(h') = \phi(\langle h | h' \rangle)$.

Then there is a unique morphism $\tilde{\phi}$ of $C^*$-algebras from the algebras $K$ of compact endomorphisms of $H$ to $B$ which send $|h\rangle \langle h' |$ to $\phi(h)\phi(h')^*$ and moreover, if $\phi$ is injective then $\tilde{\phi}$ is injective too.

Unless I'm mistaken, it is true in this form, but maybe I'm wrong and one also need to assume that $\eta(h) \eta(h')=0$, but I think this condition is only necessary if one wants a morphism from the matrix algebra with diagonal coefficient in $A$ and $K$ and off diagonal coefficients in $H$ and $H^*$.

Edit: So the Key lemma behind this results (at least for the proof I know) is that in any $C^*$ algebra the norm of an element of the form $\sum_{i=1}^n u_i (v_i)^*$ can be expressed by something that only involve the algebra generated by the $(v_i)^* u_i$ (the norm of some matrix with coefficient in this algebra typically), so any reference to a formula of this kind would be good too.

$\endgroup$
  • $\begingroup$ This looks close to Proposition 1.1.28 in the book "Elements of KK-theory" by Jensen and Thomsen, though it seems like one would need further assumptions on $\eta$ (closed range?) to make that proposition directly applicable (by viewing $\text{ran}(\eta)$ as a $\text{ran}(\phi)$-module). $\endgroup$ – Mike Jury Jan 27 '16 at 14:46
  • $\begingroup$ This looks close indeed, but I think that the additional conditions in the book by Jensen and Thomsen are needed to go from a morphism at the level of compact operators to a morphisms on the algebra of all endomorphisms (you need some non-degenratedness assumption for this...) $\endgroup$ – Simon Henry Jan 27 '16 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.