TITLE: Is $\mathbb{H}^n$ quasi-isometric to a leaf of a codimension 1 foliation of a compact manifold?
QUESTION [13 upvotes]: If we extend the action of $\pi_1(\Sigma_g), g\geq 2,$ from $\mathbb{H}^2$ to its boundary $\partial_{\infty}\mathbb{H}^2=S^1$, the surface bundle corresponding to this action of $\pi_1(\Sigma_g)$ on $S^1$ has a leaf that is isometric to $\mathbb{H}^2$.  Hence $\mathbb{H}^2$ is even isometric to a leaf in a codimension 1 foliation of a compact manifold.
For $n>2$ the above argument fails, and hence my question in the title:  Is it known whether for every $n$ $\mathbb{H}^n$ is quasi-isometric (or even isometric) to a leaf of a foliation of a compact (n+1)-manifold?

REPLY [2 votes]: Let me add a comment. One knows that a codimension-one foliation without leafwise holonomy must be given by a Z^n action, by a result of Sacksteder
Sacksteder, Richard
Foliations and pseudogroups.
Amer. J. Math. 87 1965 79–102. 
So, if the foliation is defined by a group action on the circle, and the orbits are QI to a hyperbolic space, then there must be leaves with holonomy. This means there is some point on the circle for which there is a non-trivial element of the group that fixes the point. Maybe this can be used to prove that no such actions exist. In any case, it proves that not all leaves are QI to H^n.