A C-correspondence over a C-algebra $A$ is a (right) Hilbert $A$-module (so a Hilbert space) $H$ together with a faithful representation $A\rightarrow B(H)$. Am I right in understanding that a C*-correspondence is just a Hilbert space that you get from the Gelfand-Naimark theorem?

Just to know more, is it possible to have multiple C-correspondences over a single C-algebra? I think yes. I would be happy to see some examples. I am also interested in knowing about any literature where they consider a family of C*-correspondences for whatever purposes.

  • $\begingroup$ A GNS representation doesn’t have to be faithful. And a single representation can be a direct sum of multiple GNS representations. But it is certainly true that every nondegenerate representation of a $C^\ast$-algebra is a direct sum of GNS representations. $\endgroup$
    – David Gao
    Feb 2 at 22:37
  • $\begingroup$ A $C^\ast$-algebra can, of course, have multiple non-equivalent faithful representations. Say, a faithful cyclic representation cannot be equivalent to a faithful non-cyclic one. $\endgroup$
    – David Gao
    Feb 2 at 22:40
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    $\begingroup$ If $A$ is a $C^*$-algebra of dimension $\ge2$, a Hilbert $A$ module is not going to have a (compatible) structure of a Hilbert space. $\endgroup$
    – Aweygan
    Feb 3 at 3:45
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    $\begingroup$ A Hilbert $A$-module is not a Hilbert space per se. You should check your definitions again! $\endgroup$ Feb 4 at 10:09
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    $\begingroup$ Awygan and Just dropped in raised a good point. I was following the description in your question statement and assumed we are talking about a Hilbert space, but a Hilbert $A$-modules is not a Hilbert space. While it does have an inner product, the inner product does not take values in $\mathbb{C}$ but instead in $A$. Which definition do you actually want to use? $\endgroup$
    – David Gao
    Feb 4 at 21:03

1 Answer 1


Let $A,B$ be $C^*$-algebras. An $A$-$B$-$C^*$-correspondence consists of a right Hilbert $B$-module $\mathcal{E}$ together with a (non-degenerate) $*$-homomorphism $$\pi: A \to \mathcal{L}_B(\mathcal{E}).$$ Thus, $A$-$\mathbb{C}$-correspondences correspond to (non-degenerate) $*$-representations of $A$ on Hilbert spaces. This motivates the notion of $C^*$-correspondences, as it is natural to replace Hilbert spaces by Hilbert modules (and sometimes this is really necessary, as the representation theory of $C^*$-algebras on Hilbert spaces is not always strong enough to capture relevant information).

The underlying idea here is simple: $\mathcal{E}$ has (by definition) a right $B$-action. The existence of the $*$-morphism $\pi$ means that $\mathcal{E}$ also carries a left $A$-action, namely $$a.\xi :=\pi(a)\xi, \quad a \in A, \xi \in \mathcal{E}$$ that is compatible with the right $B$-module structure. In other words, an $A$-$B$-$C^*$-correspondence should be thought of some kind of $C^*$-algebraic version of the notion of $A$-$B$-bimodule. Note however the asymmetry that $\mathcal{E}$ does not come with an $A$-valued inner product.

  • $\begingroup$ It probably wasnt your intention, but reading this answer just made the connection between the various definitions of KK modules click in my mind! $\endgroup$ 19 hours ago
  • $\begingroup$ @Opisthokont Not in my mind when writing the answer, but glad to help :) $\endgroup$
    – J. De Ro
    18 hours ago

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