Let $G$ be a the fundamental group of a hyperbolic knot complement. Then $G$ is hyperbolic relative to a subgroup $P\cong \mathbb Z \oplus \mathbb Z$. The knot complement has a $2$-dimensional spine with contractible universal cover illustrating that $G$ has geometric dimension $2$.

Question: Does there exist a $2$-dimensional $G$-complex $X$ with the property that the fixed point set $X^Q$ is contractible (in particular non-empty) for every subgroup $Q$ conjugate to a subgroup of $P$ (in particular $X$ is contractible)? EDIT: Add the condition that the $G$-stabilizer of every cell is a parabolic subgroup -- this is part of the definition of $E_{\mathcal F}G$.

The question can be phrased as whether there exist a $2$-dimensional model for $E_{\mathcal F}G$ where $\mathcal F$ is the full family of parabolic subgroups of $G$.

One obtains a $3$-dimensional $G$-space with this property by coning-off the flats in the universal cover of the $2$-dimensional spine. The flats are isolated and it is conceivable that one can push them towards the cone points.

The motivation of the question is that I have recently proved an extension for toral relatively hyperbolic groups of Gersten's result that finitely presented subgroups of hyperbolic groups of cohomological dimension $2$ are hyperbolic. If the answer to the question is known, then $G$ illustrates the result since finitely generated subgroups of $G$ are either a virtual fibers or geometrically finite. ---Apologies, for the many references of important work that I am skipping here---


1 Answer 1


Proposition 9.6 of M. Kapovich's 'Homological dimension and critical exponent of Kleinian groups' (it's Proposition 9.5 of the arXiv version) asserts that the cohomological dimension of the pair $(G,\mathcal{F})$ is equal to $\mathrm{dim}~\Lambda(G)+1$, where $\Lambda(G)$ is as usual the limit set of $G$. For cofinite $G$, $\Lambda(G)$ is the full 2-sphere and so $cd(G,\mathcal{F})=3$. In particular, no 2-dimensional model exists.

  • $\begingroup$ Henry, thanks for the reference, this settles the question. $\endgroup$ Jun 30, 2015 at 22:33
  • $\begingroup$ No problem. Nice theorem, by the way! $\endgroup$
    – HJRW
    Jul 1, 2015 at 9:51

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