Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
"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
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
@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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.