2008/01/01 by Marc Lackenby, D. D. Long, Alan W. Reid · 23 citations
Mathematics · #Geometric and Algebraic Topology #Advanced Algebra and Geometry #Geometry and complex manifolds #Mathematics #Conjecture #Congruence subgroup #Normal subgroup #Betti number #Kleinian group #Abelian group #Combinatorics #Group (periodic table) #Pure mathematics #Discrete mathematics
paper · doi:10.1093/imrn/rnn036
published in International Mathematics Research Notices 2008 (Oxford University Press)
openalex publication_date 2008/01/01 · openalex created_date 2025/10/10 · openalex updated_date 2026/07/15
Let G be an arithmetic Kleinian group, and let O be the associated hyperbolic 3-orbifold or 3-manifold. In this paper, we prove that, in many cases, G is large, which means that some finite index subgroup admits a surjective homomorphism onto a non-abelian free group. This has many consequences, including that O has infinite virtual first Betti number and G has super-exponential subgroup growth. Our first result, which forms the basis for the entire paper, is that G is commensurable with a lattice containing the Klein group of order 4. Our second result is that if G has a finite index subgroup with first Betti number at least 4, then G is large. This is known to hold for many arithmetic lattices. In particular, we show that it is always the case when G contains A4, S4 or A5. Our third main result is that the Lubotzky-Sarnak conjecture and the geometrisation conjecture together imply that any arithmetic Kleinian group G is large. We also give a new ‘elementary’ proof that arithmetic Kleinian groups do not have the congruence subgroup property, which avoids the use of the Golod-Shafarevich inequality.