Type: Article
Publication Date: 2008-01-01
Citations: 5
DOI: https://doi.org/10.1515/crelle.2008.072
Let k be a global field and let k υ be the completion of k with respect to υ a non-archimedean place of k . Let G be a connected, simply-connected algebraic group over k , which is absolutely almost simple of k υ -rank 1. Let G = G ( k υ ). Let Г be an arithmetic lattice in G and let C = C (Г) be its congruence kernel . Lubotzky has shown that C is infinite, confirming an earlier conjecture of Serre. Here we provide complete solution of the congruence subgroup problem for Г by determining the structure of C . It is shown that C is a free profinite product, one of whose factors is , the free profinite group on countably many generators. The most surprising conclusion from our results is that the structure of C depends only on the characteristic of k . The structure of C is already known for a number of special cases. Perhaps the most important of these is the ( non-uniform ) example , where is the ring of S-integers in k , with S = { υ }, which plays a central role in the theory of Drinfeld modules. The proof makes use of a decomposition theorem of Lubotzky, arising from the action of Г on the Bruhat-Tits tree associated with G .