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Let $G$ be a semisimple Lie group of rank one and let $\Gamma$ be a convex-cocompact, Zariski dense subgroup. Let $X=G/K$ denote the symmetric space and $\partial X$ its visibility boundary. Let $\Lambda(\Gamma)\subset\partial X$ be the limit set and let $C(\Gamma)\subset X$ be the convex hull of the limit set (intersected with $X$).

My question is about the shape that $C(\Gamma)$ can take. Does it have interior points? Is it the closure of its interior points? Is it a smooth Riemmannian manifold with boundary (or corners)? If not, is this true if $\Gamma$ is a subgroup of an arithmetic group?

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  • $\begingroup$ In what category do you consider manifolds with boundary/corners? If purely topological then you should clarify what you mean by a manifold with corners (since in TOP, this is just an extra structure one can impose on manifolds with boundary). If smooth, did you think of any examples, e.g. the case $G=SL(2,C)$? D you know about pleated surfaces? All in all, apart from lack of clarity, this is a nice convex geometry exercise. $\endgroup$
    – Moishe Kohan
    Commented Sep 10 at 10:58
  • $\begingroup$ Sorry, I meant as smooth Riemann manifolds, or at least as smooth manifolds. $\endgroup$
    – Antonius
    Commented Sep 10 at 13:03
  • $\begingroup$ The surface case is clear to me. My question refers to higher dimensions. I also think the question is not si much about convex geometry, as for the answers the shape of the limit set is decisive, which in the surface case can be quite nasty. $\endgroup$
    – Antonius
    Commented Sep 10 at 13:05
  • $\begingroup$ I do not think you understand the surface case. Have you read about convex hulls of limit sets of finitely generated discrete groups in hyperbolic 3-space? They are never smooth unless the subgroup is not Zariski dense. What sources did you consult? $\endgroup$
    – Moishe Kohan
    Commented Sep 10 at 14:32
  • $\begingroup$ Sorry, what I meant was the hyperbolic 2 space. But 3 dimensional space doesn't sound like surface to me. But indeed, where do I find this result? $\endgroup$
    – Antonius
    Commented Sep 10 at 15:17

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To convert my comments to an answer. Consider the case of convex-cocompact nonelementary subgroups $\Gamma< PSL(2,\mathbb C)$ whose domain of discontinuity in $S^2$ is nonempty (i.e. $\mathbb H^3/\Gamma$ is noncompact). Then the boundary $S$ of the convex hull of the limit set of $\Gamma$ is never a smooth surface unless it is a union of pairwise disjoint totally geodesic hyperbolic planes (which is very rarely the case). It is a "pleated surface" with $\Gamma$-invariant bending geodesic lamination $L\subset S$. The lamination $L$ is where smoothness fails. The complement $S\setminus L$ is the union of countably many convex totally geodesic regions. You can find this discussed in many details in

Epstein, D. B. A. and Marden, A. "Convex hulls in hyperbolic space, a theorem of Sullivan, and measured pleated surfaces". In: Fundamentals of hyperbolic geometry: selected expositions, 117–266. London Math. Soc. Lecture Note Ser., 328. Cambridge University Press, Cambridge, 2006

Another place where it is discussed is

Kamishima, Yoshinobu; Tan, Ser P., Deformation spaces on geometric structures, Matsumoto, Y. (ed.) et al., Aspects of low dimensional manifolds. Tokyo: Kinokuniya Company Ltd.. Adv. Stud. Pure Math. 20, 263-299 (1992). ZBL0798.53030.

and also in higher dimensions:

Kulkarni, Ravi S.; Pinkall, Ulrich, A canonical metric for Möbius structures and its applications, Math. Z. 216, No. 1, 89-129 (1994). ZBL0813.53022.

As far as I know the only solid proof of nonsmoothness (in general) is in the case of the hyperbolic 3-space. In the higher-dimensional case, examples given by "bending" of hyperbolic metrics give rise to nonsmooth boundaries of convex hulls. I do not think anything was worked out in the case of rank 1 symmetric spaces of nonconstant curvature, but I do not think there is a reason to believe they are any better than in the case of the hyperbolic space.

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