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>



Uniform Roe algebras and exact groups
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.


