Commensurability of groups quasi-isometric to RAAG's

Let $G$ be a right-angled Artin group with defining graph $\mathrm{\Gamma }$ and let $H$ be a finitely generated group quasi-isometric to $G(\mathrm{\Gamma })$. We show if $G$ satisfies (1) its outer automorphism group is finite; (2) $\mathrm{\Gamma }$ does not have induced 4-cycle; (3) $\mathrm{\Gamma }$ is star-rigid; then $H$ is commensurable to $G$. We show condition (2) is sharp in the sense that if $\mathrm{\Gamma }$ contains an induced 4-cycle, then there exists an $H$ quasi-isometric to $G(\mathrm{\Gamma })$ but not commensurable to $G(\mathrm{\Gamma })$. Moreover, one can drop condition (1) if $H$ is a uniform lattice acting on the universal cover of the Salvetti complex of $G(\mathrm{\Gamma })$. As a consequence, we obtain a conjugation theorem for such uniform lattices. The ingredients of the proof include a blow-up building construction in \cite{cubulation} and a Haglund-Wise style combination theorem for certain class of special cube complexes. However, in most of our cases, relative hyperbolicity is absent, so we need new ingredients for the combination theorem.