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Let $P$ be a compact convex polyhedron in $\mathbb{H}^3$. Let $G$ be a group generated by side-pairing isometries of $P$. Is there an algorithm to find the rank of $G$?

edited Mar 4 2012 at 17:57

asked Mar 4 2012 at 17:44

bb
1,191●5●9

Misha · Accepted Answer · 2012-03-04 18:21:50Z UTC

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I assume that you want $P$ to be a fundamental domain for $G$. Then the answer is positive, see:

I. Kapovich, R. Weidmann, Kleinian groups and the rank problem. Geom. Topol. 9 (2005), 375-402.

answered Mar 4 2012 at 18:21

Misha
12.6k●1●26●48

Do you know of a practical algorithm? – bb Mar 4 2012 at 18:29

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This algorithm is as practical as any in 3d topology and geometric group theory. None of these you would want to implement on your computer. – Misha Mar 4 2012 at 19:10

The obvious question is: what is the application? In practice, there are often heuristics which work quite fast. – Igor Rivin Mar 4 2012 at 21:39

I was curious about whether one could determine some properties of the polyhedron that would imply the difference between the rank of the group and the number of side-pairings was bounded. – bb Mar 5 2012 at 2:53

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Even for cyclic groups, the number of faces of convex (Dirichlet) fundamental polyhedron could be arbitrarily high, see T. Drumm, J. Poritz, Ford and Dirichlet fundamental domains for cyclic subgroups of $PSL(2; C)$, Conformal Geometry and Dynamics, 3 (1999) 116-150. Short of taking a group generated by reflections, I do not think there are nice reasonably general conditions to ensure a relation between rank and the number of faces. – Misha Mar 5 2012 at 3:38

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Rank of a group generated by side-pairing isometries of a polyhedron

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