Past exam of the mathematics course of the University of Cambridge 2006 ii Paper 3 3F Solution Created 2026-09-24 Updated 2026-10-14
Extend the Möbius transformations to isometries of three-dimensional hyperbolic space. For any interior point , the limit set of a Kleinian group is the set of boundary accumulation points of . Its definition is independent of : two interior orbits stay at a bounded hyperbolic distance, so have the same boundary limits. The Kleinian limit set is closed and -invariant. Iterating a loxodromic Möbius transformation in the positive and negative directions shows that its two fixed points belong to the Kleinian limit set.
The two fixed-point sets in this problem cannot have exactly one common point. Otherwise conjugate that point to infinity and the other fixed point of the first transformation to zero. The transformations then have the forms , with , and , with . But is a sequence of distinct group elements converging in the Möbius group. Taking quotients of consecutive elements gives nonidentity elements converging to the identity, contrary to discrete subgroup. Consequently the Kleinian limit set contains at least three points.
Let be any point of the Kleinian limit set, and choose distinct with . The rank-one convergence of divergent Möbius transformations supplies a subsequence converging to uniformly away from one exceptional boundary point . To see the lemma directly, normalize determinant-one representing matrices by their norms. The norms tend to infinity, and a subsequential limit has rank one: its kernel is and its image is the attracting point. In the positive-Hermitian-matrix model of hyperbolic space, has this same image limit, so the attracting point is indeed .
Choose distinct in the Kleinian limit set other than . Then and both tend to , belong to the Kleinian limit set, and remain distinct because is injective. If were isolated, both would eventually equal , an impossibility. The limit set has no isolated points.