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?



Is $\mathbb{H}^n$ quasi-isometric to a leaf of a codimension 1 foliation of a compact manifold?
One may generalize your 3-dimensional example to all dimensions in some sense. 


There are compact hyperbolic $n$-manifolds which have orderable fundamental group, since the fundamental groups embed in a right-angled Artin group (which is in fact [bi-orderable](http://en.wikipedia.org/wiki/Linearly_ordered_group)), by a [result of Haglund and Wise](http://link.springer.com/article/10.1007/s00039-007-0629-4). Let $M=\mathbb{H}^n/\Gamma$ be such a manifold, with faithful action $\Gamma \to Homeo^+(\mathbb{R})$ (this is equivalent to being orderable), and consider the twisted product $(\mathbb{H}^n \times \mathbb{R})/ \Gamma$, where $\Gamma$ acts diagonally. The foliation $\mathbb{H}^n\times \{x\}$ naturally descends to a foliation on the quotient. This manifold is homeomorphic to $M\times \mathbb{R}$, so embeds in $M\times [-\infty,\infty]$, where the boundary $M\times \{\pm\infty\}$ are leaves of the foliation. There is a point $x\in\mathbb{R}$ on which $\Gamma$ acts faithfully (in particular, the orbit induces the total ordering on $\Gamma$) if the action is well-chosen (see the proof of Theorem 6.8 in [Ghys' paper](http://www.math.ethz.ch/~bgabi/ghys%20groups%20acting%20on%20the%20circle.pdf)). Thus, $(\mathbb{H}^n \times \{\Gamma x\})/\Gamma \cong \mathbb{H}^n$, so there is a leaf of the foliation which is a copy of $\mathbb{H}^n$. Since it is uniformly close to $\tilde{M}\times \{\infty\}$ in the universal cover, it will be quasisometric to $\tilde{M}=\mathbb{H}^n$. Glue $M\times{-\infty}$ to $M\times \{\infty\}$ to obtain a closed manifold $M\times S^1$ with the desired property (really, we are embedding $Homeo^+(\mathbb{R})\subset Homeo^+(S^1)$ by the 1-point compactification, and taking the corresponding twisted product). 


