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I don't expect to find an explicit counterexample to my question, because any example which was known to have the Haagerup property yet not have AP would have given an exact group without AP, and the existence of such groups the existence of exact groups without AP was open until the recent(ish) work of Lafforgue--de la Salle and de Laat--Haagerup. My understanding is that all these examples of groups without AP all have Kazhdan's property (T), so can't have the Haagerup property.

On the other hand, it is not obvious to me why the Haagerup property should imply AP. But perhaps some MO readers will know of an argument that proves such an implication? If not, what is an example of a group which is known to have the Haagerup property yet which is not known to have AP?

(One characterization of the Haagerup property for discrete groups is that there should exist a net of normalized, positive-definite functions on the group, each of which vanishes at infinity, such that the net converges pointwise to the constant function 1. The definition of the AP is given here.)

(Correction 2015-04-15: the original version mistakenly claimed, tacitly, that the Haagerup property implies exactness. My thanks to N. Ozawa for pointing out that this is not known either.)

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    $\begingroup$ Not known. Thompson's group $F$ does have the Haagerup property (D. S. Farley), but it's not known whether it is amenable, exact, or neither. Thompson's group $F$ aside, I'd bet there's a non-exact group with the Haagerup property, because the analogous fact is true for coarse metric spaces with bounded geometry (Arzhantseva--Guentner--Spakula). $\endgroup$
    – Narutaka OZAWA
    Commented Apr 13, 2015 at 23:40
  • $\begingroup$ @NarutakaOZAWA Thanks -- for some reason I did not think of F. If you post your comment as an answer then I'd be happy to accept (since you have answered my second question, and hence impliictly my first one). $\endgroup$
    – Yemon Choi
    Commented Apr 14, 2015 at 16:07

1 Answer 1

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The answer is no. It's solved in the following paper by Osajda. http://arxiv.org/abs/1406.5015

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  • $\begingroup$ Thanks! Just to make sure I have understood (and for others reading): Osajda's examples have the Haagerup property but do not have Property A, and for discrete groups the AP of Haagerup--Kraus implies Property A. Is that so? $\endgroup$
    – Yemon Choi
    Commented Apr 16, 2015 at 14:04
  • $\begingroup$ Yes, that's the case. $\endgroup$
    – Narutaka OZAWA
    Commented Apr 16, 2015 at 21:36

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