Does the braid group $B\_n, n\ge 3$, act properly by isometries on a CAT(0) cube complex?

 **Update 1.**  During a recent talk of Nigel Higson in Pennstate Dmitri Burago asked whether the braid groups are a-T-menable. I seem to remember that somebody proved that they do act properly by isometries on CAT(0) cube complexes. That would imply a-T-menability by Niblo and Roller or Cherix, Martin and Valette. Hence the question. Note: no co-compactness required.

 **Update 2.**  It looks like my question is still an open problem for $n\gt 3$. I think my confusion came from terminology. Dan Farley proved that all braided diagram groups (including the R. Thompson group $V$) act property by isometries on CAT(0)-cube complexes. But braided diagram groups (defined by Victor Guba and myself) are not related to braid groups, at least not explicitly because wires there intersect and do not form braids. One can define the notion of "really braided" diagram groups where wires form braids, but I do not think Farley's method will work. So I got confused by my own terminology. By the way, I do not see an obvious reason that $B\_n$ does not embed into $V$. $V$ is a big group with lots of complicated subgroups.

 **Update 3.**  As Bruce Hughes pointed out to me, even though the Haagerup property (a-T-menability) is unknown for $B\_n, n\ge 4$, all forms of Baum-Connes conjecture have been proved for it by Thomas Schick in [*Finite group extensions and the Baum-Connes conjecture*](https://arxiv.org/abs/math/0209165).

 **Update 4**  Concerning the question from Update 2. Collin Bleak and Olga Salazar-Diaz proved in [*Free products in R. Thompson's group V*](https://arxiv.org/abs/0911.0979) that $V$ does not contain subgroups isomorphic to ${\mathbb Z}^2\ast {\mathbb Z}$. Does $B\_n$ contain such subgroups?