Density of the commensurability groups of uniform tree lattices
Let \(G\) be a group and \(\Gamma_ 0\), \(\Gamma_ 1\) subgroups of \(G\). The groups \(\Gamma_ 0\) and \(\Gamma_ 1\) are called commensurable if the intersection \(\Gamma_ 0 \cap \Gamma_ 1\) has finite index in both. The commensurator \(C_ G(\Gamma)\) of a subgroup \(\Gamma\) is the group of all \(g \in G\) such that \(\Gamma\) and \(g\Gamma g^{-1}\) are commensurable. If for example \(G\) is a Lie group and \(\Gamma\) an arithmetic subgroup, then the group \(C_ G(\Gamma)\) is essentially the group of rational points of the corresponding arithmetic structure. Now let \(X\) be a uniform tree, i.e. the universal cover of a finite graph. The automorphism group \(G\) of \(X\) is equipped with the compact open topology. Let \(\Gamma\) be a discrete cocompact subgroup of \(G\). The main result of the paper under consideration is that the commensurator \(C_ G(\Gamma)\) of \(\Gamma\) is dense in the topological group \(G\). Under the hypothesis that also the quotient graph \(G\setminus X\) is a tree this was already proven by \textit{H. Bass} and \textit{R. Kulkarni} [J. Am. Math. Soc. 3, No. 4, 843-902 (1990; Zbl 0734.05052)]. For the proof the claim is reduced to the fact that every \(g \in G\) is interpolated on every finite subset of \(X\) by elements of \(C_ G(\Gamma)\). This then is shown by a careful analysis of the combinatorial situation.
scientific article; zbMATH DE number 589944