Do all right orderable groups have the Haagerup property?
Recall that a group is right orderable if there exists a total order $\leq$ on it such that $a\leq b\Rightarrow ac\leq bc$. This property is important for its connection to conjectures regarding group rings. Recall also that a group has the Haagerup property if it has a proper 1-cocycle. This property is e.g. relevant to Sorin Popa's deformation/rigidity techniques and to the Baum-Connes conjecture.
No. Consider $\Gamma\subset\mathrm{GL}_n(\mathbf{Z})$. Assume that the Zariski closure of $\Gamma$ in $\mathrm{GL}_n(\mathbf{R})$ contains $\mathrm{SL}_n(\mathbf{R})$, or more generally is irreducible and non-compact. Then $(\Gamma\ltimes\mathbf{Z}^n,\mathbf{Z}^n)$ has Kazhdan's relative Property T (by an argument due to Kazhdan, Burger, Shalom, see Theorem 1.4.5 in the Bekka-Harpe-Valette book), and in particular $\Gamma\ltimes\mathbf{Z}^n$ fails to be Haagerup. If moreover $\Gamma$ is right-orderable (e.g., is free), then $\Gamma\ltimes\mathbf{Z}^n$ is right-orderable as well. For every $n\ge 2$ some such free subgroup can indeed be found (e.g., of finite index when $n=2$). [With no irreducibility assumption, it is more generally true that $\Gamma\ltimes\mathbf{Z}^n$ fails to be Haagerup whenever $\Gamma\subset\mathrm{GL}_n(\mathbf{Z})$ is non-virtally-solvable, by a slight refinement of the argument).
