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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!

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"Many books": Examples? –  Jonas Meyer Jul 30 '12 at 21:16
    
@JonasMeyer C^* algebras by example, C^* algebras and operator theory. –  Hui Yu Jul 30 '12 at 21:24
    
Davidson, Murphy? –  Jonas Meyer Jul 30 '12 at 21:24
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I probably wouldn't have used the word "direction" myself, but it is very clear what it means in this context. If you define a relation by using only one of the equalities above, you don't get an equivalence relation (and a concrete example of that is what this question is requesting); if you use both equalities ("back and forth", and hence a vague notion of "direction") you do get an equivalence relation. By contrast, if you use the same definition but with norm-limits instead of WOT limits, one equality already defines an equivalence relation. –  Martin Argerami Jul 30 '12 at 23:59
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@JonasMeyer I agree. I found another reference talking about this weak approximate equivalent in Arveson's Notes on Extenstions of C^*-algebras. But he did not give an explanation either. –  Hui Yu Jul 31 '12 at 15:23

1 Answer 1

up vote 2 down vote accepted

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.

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My reflex is to introduce a preorder $\pi \mathbin{\preceq_{\rm wa}} \sigma$ instead of "halfway weak approximate equivalence". w.a. equivalence would then be the equivalence relation associated to the preorder, i.e., $\pi \mathbin{\sim_{\rm wa}} \sigma$ if and only if $\pi \preceq \sigma$ and $\sigma \preceq \pi$. At least this is what people do in related circumstances (weak containment of unitary representations of groups)---I didn't check too closely how strong the relation really is –  t.b. Jul 31 '12 at 23:21
    
@t.b.: I'm not trying to coin terminology or notation, although I do agree that introducing notation such as yours would be useful if referring to it more than once. I hope it is clear from context what is meant in my answer. Thanks for the link. –  Jonas Meyer Jul 31 '12 at 23:48
    
Me neither. It is perfectly clear what is meant. Since I do not know the background of this notion I can't tell how meaningful it is, but it seems to be a natural way of looking at things: the pre-order is there and how you write it or call it is of secondary importance (I wrote this before seeing the new version of the comment) –  t.b. Jul 31 '12 at 23:48

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