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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?2012-07-30
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    @JonasMeyer C^* algebras by example, C^* algebras and operator theory.2012-07-30
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    Davidson, Murphy?2012-07-30
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    @JonasMeyer Yes.2012-07-30
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    What do you mean by 'direction'? But even without knowing what 'direction' means, your question sounds odd: 'no clue is given why we need two directions to obtain an equivalence relation'. Apparently you can prove it to be an equivalence relation using 'one direction'?2012-07-30
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    @wildildildlife: verbatim from Davidson's book: "Both directions are needed to obtain an equivalence relation." Hui's question sounds perfectly reasonable to me.2012-07-30
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    @MartinArgerami: then perhaps you can explain what 'direction' means? (Still, that Davidson statement would only puzzle me if I could convince myself that I only need one direction.)2012-07-30
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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.2012-07-30
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    @MartinArgerami Could you explain a little bit? Why we do not get an equivalence relation using only one direction? Is there a concrete example with one direction that violates reflexivity? transitivity? symmetry?2012-07-31
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    HuiYu: I do not know whether Martin had an answer ready, but I'm guessing that he would have posted it if he did. Martin's last comment is just explaining to wildildildlife what the question means. (Note that reflexivity is obviously not violated with only one direction. Symmetry and transitivity are not so clear.)2012-07-31
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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.2012-07-31
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    Hui, as Jonas said, I have no idea about how one could prove that. If you look at Davidson's, you see that he uses both conditions to show that they imply something stronger (i.e. the same but with $*$-SOT). And I cannot even see how to prove symmetry or transitivity in the presence of both relations, or the *-SOT condition. Not that I spent a lot of time with it, but I was surprised because most often these things are more or less obvious.2012-07-31
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    @Martin: Isn't symmetry obvious when you have both relations? It is after all an intersection of a relation and its inverse.2012-07-31
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    @Jones: yes, absolutely, it is by definition. I still don't know how to do transitivity, though (as multiplication is not WOT-jointly continuous).2012-07-31
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    @Jonas: sorry for the typo in your name, I saw it after the 5 minutes had elapsed.2012-07-31

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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](http://www.math.vanderbilt.edu/~peters10/teaching/Spring2011/Lecture%2010%20-%20Weak%20containment.pdf) of unitary representations of groups)---I didn't check too closely how strong the relation really is2012-07-31
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    @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.2012-07-31
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    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)2012-07-31