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Recall a topological space is called planar if it can be embedded in $S^2$. I'm interested in understanding hyperbolic groups with planar boundaries.

In [1], it is shown that if a one-ended hyperbolic group has 1-dimensional planar boundary, then this boundary is either a circle or a Sierpinski carpet. It is also well known that there exists hyperbolic groups with boundary homeomorphic to $S^2$, and moreover the only 0-dimensional such space is the Cantor set.

My question is: What other planar spaces arise as boundaries of hyperbolic groups? Does a nice "list" of these spaces exist, or is this a hopeless problem? What if we restrict ourselves to just one-ended groups?

Any references would be appreciated. Thanks!


References

[1] Kapovich, Michael, and Bruce Kleiner. "Hyperbolic groups with low-dimensional boundary." Annales scientifiques de l'Ecole normale supérieure. Vol. 33. No. 5. 2000.

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    $\begingroup$ A free product of two one surface groups with $\mathbf{Z}$? The boundary is a compact space whose connected components are countably many circle, and continuum many points (if I'm correct). I'd even wonder whether it occurs as undistorted subgroup of $\mathrm{SL}_2(\mathbf{C})$. $\endgroup$
    – YCor
    Dec 19, 2021 at 15:31

1 Answer 1

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There are many further examples with local cut points, which can be obtained by amalgams over $\mathbb{Z}$, as @YCor suggests in his comment.

Perhaps the easiest example is obtained by gluing three one-holed tori along their boundary. The resulting group $\Gamma$ is hyperbolic, and its boundary cannot be a Sierpinski carpet or $S^2$, since it has a local cut point, coming from either of the limit points of the amalgamating copy of $\mathbb{Z}$). On the other hand, that local cut point has valence 3, so the boundary also cannot be a circle.

An even richer family of examples can be constructed via the following construction. Take any simple closed curve $\gamma$ on the boundary of a handlebody $U$, and take the double

$\Gamma=\pi_1(U)*_{\langle\gamma\rangle} \pi_1(U)$ .

This will be a 3-manifold group, and hence its boundary will be planar.

By the Convergence Group Theorem of Tukia--Casson--Jungreis--Gabai, the boundary of $\Gamma$ is only a circle if $\Gamma$ is a surface group, which in turn only occurs if $\gamma$ was a ``surface element" of $\pi_1(U)$. And the boundary won't be a Sierpinski carpet, because it has a local cut point.

It's complicated to describe the $\partial\Gamma$ for this construction in general, but you can make a start by looking at the papers of Cashen--Macura and Cashen, who explain how to construct the decomposition space $D(\gamma)$. The cut-point structure of $D(\gamma)$ gives information about the cut-point structure of $\partial\Gamma$.

As far as I know, the problem of characterising all planar compacta that arise as boundaries hasn't been worked out, but I suspect it is tractable, and would make a very nice thesis problem, say. One would also want to make use of Bowditch's theorem, which explains how local cut points in the boundary correspond to cyclic splittings, and the Strong Accessibiltiy Theorem of Louder--Touikan, which asserts (under mild hypotheses) that hierarchies of cyclic splittings terminate.

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  • $\begingroup$ Wonderful answer, thank you! $\endgroup$
    – jpmacmanus
    Dec 20, 2021 at 11:11

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