Is an $A$-$B$—$C^*$-correspondence a representation of a $G$-$C^*$-algebra, $\rho \colon A \otimes_{ \alpha } B \to \mathcal{L} ( \mathcal{H} )$?

Question

Let $R$ and $S$ be two rings.

It is known that an $R$-$S$-bimodule is actually the same thing as a left module over the ring $R \otimes_{\mathbb{Z}} S^{\mathrm{op}}$, where $S^{\mathrm{op}}$ is the opposite ring of $S$, and the multiplication is defined with the arguments exchanged.

Now, by definition, if $A$ and $B$ are two $G$-$C^*$-algebras, then, an $A$-$B$–$C^*$-correspondence is a pair $( \mathcal{H}_1 , \varphi )$, where, $\mathcal{H}_1$ is a right $G$-Hilbert $B$-module, and $\varphi \colon A \to \mathcal{L} ( \mathcal{H}_1 )$ is a $*$-homomorphism with, $\varphi (ga) = g \varphi (a)$, for all $g \in G$ and $a \in A$.

Is it true that, an $A$-$B$—$C^*$-correspondence is the same thing as a representation of a $G$-$C^*$-algebra, $\rho \colon A \otimes_{ \alpha } B^{\mathrm{op}} \to \mathcal{L} ( \mathcal{H}_2 )$, or something like that, with, $\mathcal{H}_2$, a $\mathbb{C}$-Hilbert space, and, $\alpha$, relative to a $C^*$-norm $\| \cdot \|_{ \alpha }$, which makes the algebraic tensor product $A \odot B^{\mathrm{op}}$, complete?

How are $\mathcal{H}_2$, and the $C^*$-norm $\| \cdot \|_{ \alpha }$, defined?

$B^{\mathrm{op}}$ is the opposite algebra of $B$.

Thanks in advance for your help.

Answer 1

No.
The underlying vector space of a $C^*$-correspdondence is typically not a Hilbert space.

For example, the trivial $A$-$A$-$C^*$-correspdondence has underlying vector space given by the $C^*$-algebra $A$ itself.

So the data of a $C^*$-correspdondence is not equivalent to the data of a Hilbert space equipped with the action of some $C^*$-algebra.