I am currently learning about the theory of C*-Algebras, and came across the idea of quasi-equivalence of representations, defined as follows:
Two representations $\pi: A \rightarrow \mathcal{B}(\mathcal{H})$ and $\rho: A \rightarrow \mathcal{B}(\mathcal{K})$ of a unital C*-algebra $A$ are quasi-equivalent if there is an isomorphism $\Phi: \pi(A)'' \rightarrow \rho(A)''$, where $A''$ denotes the double commutant, such that $\Phi(\pi(a)) = \rho(a)$ for all $a \in A$.
Now, my problem is this: Given two unital representations which are quasi-equivalent, it should be clear that they are not necessarily unitarily equivalent. But how does one show this? Is there an easy counter-example?
I honestly don't really know how to start thinking about an example, since I haven't really seen any concrete examples of representations apart from the GNS construction and the corresponding universal representation. Any help would be appreciated!
Edit: I was given the hint to consider direct products of representations. If I find an appropriate unital representation $\pi$, I think one could maybe show $\pi$ and $\pi \oplus \pi$ are quasi-equivalent, but not equivalent. Not really sure how to approach showing this though.
Best Answer
An equivalent formulation of your definition is that $\Phi:\pi(A)\to\rho(A)$ is a normal isomorphism.
The hint you were given is a good way to do the counterexample. If you take $A=\mathbb C$, $\pi:A\to\mathbb C$ and $\rho:A\to M_2(\mathbb C)$ be $$ \pi(a)=a,\qquad\qquad \rho(a)=\begin{bmatrix} a&0\\0&a\end{bmatrix}, $$ then $\Phi=\rho$ realizes the quasiequivalence of $\pi$ and $\rho$ but $\pi$ and $\rho$ cannot be approximately unitarily equivalent because the operators $\pi(a)$ and $\rho(a)$ have different rank.