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This is a somewhat vague question, but I think it is not too open-ended and should admit well-circumscribed answers by specialists in operator algebras.$\newcommand{\Cst}{{\rm C}^*}$ It arises from some background reading I was doing while working on a different area/problem.

For sake of simplicity I'll restrict to unital separable $\Cst$-algebras, although I think everything can be discussed more generally with appropriate modifications. $\otimes$ will denote the completed minimal tensor product of $\Cst$-algebras.

Definition. A (unital, separable) $\Cst$-algebra $A$ is said to have approximately inner flip if there is a sequence of unitaries $(u_n)\subset A\otimes A$ such that $u_n(x\otimes y)u_n^* \to y\otimes x$ for every $x,y\in A$. It is said to have approximately inner half-flip if $u_n(x\otimes 1_A)u_n^* \to 1_A\otimes x$ for all $x\in A$.

It's known that having a.i. half-flip already imposes some fairly strong restrictions on $A$, such as being simple and nuclear. We also know that ${\mathcal O}_2$ and ${\mathcal O}_\infty$ have a.i. flips.

From a quick look at the 1978 paper of Effros and Rosenberg, I gather that these properties are analogues of von Neumann algebraic properties of the hyperfinite ${\rm II}_1$ factor. What I would like to know is: why is it interesting/important to characterize those $\Cst$-algebras with a.i. (half-)flip? and if we can show that a given simple, nuclear $\Cst$-algebra has a.i. (half-)flip, what further structural consequences does one typically hope to deduce?

I guess part of what I would like to find out is whether we can think of these properties in any of the following ways:

1) "goes unseen by many of the usual invariants"

2) "has some kind of homogeneity not shared by general unital nuclear simple $\Cst$-algebras"

3) "looks like one of several standard examples on a short list".

Of course, if any or all of these three claims are wide of the mark, I'd welcome any clarifications or corrections.

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  • $\begingroup$ Not sure whether you already know this reference, but you should check out this paper by Aaron Tikuisis: arxiv.org/abs/1505.00572 $\endgroup$ – Gabor Szabo Mar 23 '16 at 9:58
  • $\begingroup$ @GaborSzabo Thanks - I did have a quick look at Aaron's paper but it wasn't very clear to me what the implications of his characterization were. (This is is not intended as any slight on his paper, it's just that I come from a different are of FA and so don't really have a feel for what the conclusions tell us.) $\endgroup$ – Yemon Choi Mar 23 '16 at 18:21
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If $A$ is a simple AF algebra, then having an approximately inner flip entails more than that it has unique trace—because $K_0(A \otimes A)$ is order isomorphic to $K_0(A) \otimes_{Z} K_0(A)$, the latter as partially ordered abelian groups—it also says that the infinitesimals are trivial, so in fact, it forces $K_0(A)$ to be a totally ordered subgroup of the rationals, i.e., $A$ is UHF. (The key and perhaps only point is that approximately inner automorphisms induce the identity on taking $K$-groups.) This is drastic, and the result presumably extends to much larger classes of C*-algebras. This was surely in Effros-Rosenberg? (I am too tired to check, and it's fairly late for me.) So maybe (2) and (3) apply?

More generally, if $K_1(A)$ is trivial and UCT applies (whatever that is), then $K_0(A \otimes A)$ is isomorphic to the group tensor product, but I don't know whether anyone has checked that it is a pre-order isomorphism; however, if traces behave well (meaning, are determined by their effects on projections), ai flip should imply at most one trace.

What's an example of an ai half-flip that is not an ai flip?

Edit: It's clear (without using K-theory) that only C*-algebras with zero or one trace can have an ai flip ... .

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    $\begingroup$ The article that I referenced in my other comment shows exactly what K-theory is allowed for C*-algebras with approximately inner flip. In contrast, any pair of abelian groups is allowed as the K-theory of something with approximately inner half-flip. For Kirchberg algebras, a.i.h.f. is already satisfied if the algebra is in Cuntz standard form, i.e. the unit represents the zero in K_0. This follows from Kirchberg-Phillips classification. $\endgroup$ – Gabor Szabo Mar 23 '16 at 10:06

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