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Let $\Gamma$ be a discrete group.

Q: If $l^\infty(\Gamma)\rtimes \Gamma=l^\infty(\Gamma)\rtimes_r \Gamma$ canonically, can we conclude that $\Gamma$ is an exact group?

The converse implication is well-known and if we replace $l^\infty(\Gamma)$ by $l^\infty(\Gamma)/C_0(\Gamma)$ we have indeed an affirmative answer to the question. The latter fact can be found in Section 5 of "Ghostbusting and property A" by Roe and Willett or Theorem 5.6 in http://arxiv.org/pdf/1603.01829.pdf

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  • $\begingroup$ By "canonically isomorphic", do you mean that "the canonical map from the full to the reduced cross product is an isomorphism"? $\endgroup$
    – Yemon Choi
    Commented Mar 12, 2016 at 0:56
  • $\begingroup$ @m07kl Yes. See Theorem 4.4.3 in Brown and Ozawa's book, C*-algebras and finite dimensional approximation properties $\endgroup$
    – Caleb Eckhardt
    Commented Mar 12, 2016 at 4:53
  • $\begingroup$ @YemonChoi Yes. :) $\endgroup$
    – m07kl
    Commented Mar 12, 2016 at 12:14
  • $\begingroup$ @CalebEckhardt: I don't think you can apply Theorem 4.4.3, because here you only use one $C^*$-algebra namely $l^\infty(\Gamma)$. Please see arxiv.org/abs/1204.3050 for a partial solution. $\endgroup$
    – m07kl
    Commented Mar 12, 2016 at 12:21
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    $\begingroup$ The answer is affirmative and it follows from a very recent paper by Anantharaman-Delaroche see arXiv:1604.01724 $\endgroup$
    – m07kl
    Commented Apr 21, 2016 at 18:08

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See final remark of "SOME REMARKS ABOUT THE WEAK CONTAINMENT PROPERTY FOR GROUPOIDS AND SEMIGROUPS" by Claire Anantharaman-Delaroche for an affirmative answer of the question.

https://arxiv.org/pdf/1604.01724.pdf

edit (by AlexE): the cited argument is flawed, see my comment below.

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  • $\begingroup$ It looks like a nice argument. Why not give an outline of the argument here? $\endgroup$
    – Yemon Choi
    Commented May 18, 2016 at 16:34
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    $\begingroup$ The final remark that you refer to is flawed. In fact, in the new version of the paper you don't find it anymore. Concretely, I was told by Kang Li that Cor. 2.12 in version 3 of the paper is wrong, which invalidates the final step in the final remark you mention. As far as I know, your question is currently still open. $\endgroup$
    – AlexE
    Commented Nov 19, 2020 at 11:11

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