TITLE: Why need two directions to make $\sim_{\rm wa}$ an equivalence relation?
QUESTION [7 upvotes]: Let $\pi$ and $\sigma$ be representations of a $C^*$-algebra $\mathcal{A}$. They are weak approximately equivalent ($\pi\mathbin{\sim_{\rm wa}}\sigma$) if there are sequences of unitary operators $\{U_n\}$ and $\{V_n\}$ such that \begin{equation}
\sigma(A)=\operatorname{WOT-lim} U_n\pi(A) U_n^*, 
\end{equation} \begin{equation}
\pi(A)=\operatorname{WOT-lim} V_n\sigma(A) V_n^*
\end{equation} for all $A\in\mathcal{A}$.
Many books point out that both directions are needed to obtain an equivalence relation but no clue is given why. Since for approximate equivalence ($\mathbin{\sim_{\rm a}}$), only one direction is needed, I wonder why for $\mathbin{\sim_{\rm wa}}$ we need two.
Thanks!

REPLY [2 votes]: I think I have an example.  The representations are degenerate, but I don't see any assumption in Davidson that they shouldn't be.
Let $A=B(\ell^2(\mathbb Z_{\geq 0}))$ (or any nonzero $C^*$-subalgebra).  Let $P:\ell^2(\mathbb Z)\to\ell^2(\mathbb Z_{\geq 0})$ be orthogonal projection, and define $\pi:A\to B(\ell^2(\mathbb Z))$ by $\pi(a)=P^*aP$ (essentially embedding $A$ in the lower right corner of $B(\ell^2(\mathbb Z))$).  Let $U$ be the right shift on $\ell^2(\mathbb Z)$, and define the sequence $(U_n)_{n\geq 1}$ of unitary operators on $\ell^2(\mathbb Z)$ by $U_n=U^n$.  Then for all $a\in A$, $\text{WOT-}\lim_n U_n \pi(a) U_n^*=0$.  Thus $\pi$ is "halfway" weak approximately equivalent to the $0$ representation $\sigma(a)=0$ on $\ell^2(\mathbb Z)$, but not weak approximately equivalent to the $0$ representation.