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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?

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  • $\begingroup$ Presumably you want something other than taking a compact hyperbolic $n$-manifold $\times S^1$? $\endgroup$
    – Ian Agol
    Commented Nov 5, 2013 at 19:42
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    $\begingroup$ @Ian: The question is about leaves which are qi to hyperbolic space. A compact leaf will not qualify for this. $\endgroup$
    – Misha
    Commented Nov 5, 2013 at 20:52
  • $\begingroup$ Ah, right, I was thinking the universal cover... $\endgroup$
    – Ian Agol
    Commented Nov 6, 2013 at 8:26

2 Answers 2

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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), by a result of Haglund and Wise. 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). 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).

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  • $\begingroup$ Nice example but not smooth as far as I can tell. In the usual setting of smooth foliations one takes a Riemannian metric on the compact manifold and restricts it to the leaves in order to define a geometric structure leaf-wise. In this example one can use a lead-wise metric which is continuous in the transversal direction, so it is sort if ok. Still, I would prefer to see a smooth example. $\endgroup$
    – Misha
    Commented Nov 6, 2013 at 19:12
  • $\begingroup$ I guess your question about which RAAGs embed in $Diff(S^1)$ is relevant. Is it possible that there are compact hyperbolic $n$-manifolds that embed in $F^k$ for some $k$ ($F$ a free group)? $\endgroup$
    – Ian Agol
    Commented Nov 6, 2013 at 19:50
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    $\begingroup$ One more thing: Since all RAAGs embed in $Diff(S^2)$, it follows that there are codimension 2 examples. $\endgroup$
    – Misha
    Commented Nov 6, 2013 at 20:15
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    $\begingroup$ One more-more thing: As an alternative to orderability, one can use the fact that all RAAGs embed in some mapping class group of an open surface and the latter all embed in $Homeo(S^1)$ (but maybe not in $Diff(S^1)$ as far as we know). $\endgroup$
    – Misha
    Commented Nov 6, 2013 at 20:20
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    $\begingroup$ The question of embedding RAAGs in $Diff(\mathbb{R})$ is now resolved: front.math.ucdavis.edu/1404.5559 However, I don't believe that the embedding they obtain admits a free orbit in general, so this leaves the answer to the question in the smooth category still open. $\endgroup$
    – Ian Agol
    Commented Jul 17, 2014 at 16:55
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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.

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