Commensurators of normal subgroups of lattices (Q6621563)

Let \(\Lambda\) be an irreducible lattice in a real or p-adic semsimple algebraic group \(S\) which has no compact factors. Let \(C\) denote its ``commensurator'', i.e., the set of all \(g\in S\) such that \(\Lambda\cap (g\Lambda g^{-1})\) is of finite index in both \(\Lambda\) and \(g\Lambda g^{-1}\). By a theorem of Margulis \(\Lambda\) is arithmetic if and only if \(C\) is dense in \(S\).\N\NGreenberg and Shalom asked whether the condition ``lattice'' may be weakened. May the condition of ``finite volume'' be replaced by ``Zariski dense''?\N\NGreenberg proved this for finitely generated subgroups of \(SL(2,\mathbb{R})\). Later, some other special cases have been dealt with.\N\NThis paper gives a positive answer for the following special case: There is an irreducible lattice \(\Gamma\) such that \(\Lambda\) is a normal subgroup of \(\Gamma\).\N\NThe result is also proved for algebraic groups over local fields with positive characteristic.\N\NMethods used include a precise study of relative profinite completions.\N\NIn \S1.3 some related questions are discussed. The following question is raised: If \(\Lambda\) is a discrete subgroup in a product \(G_1\times G_2\) of semisimple Lie groups such that both projections are dense, must it be an irreducible lattice?\N\NThis is an intriguing question. However, it warrants some minor reformulation.\N\NAs stated, the answer is clearly negativ, as seen by the following counterexample: Let \(S\) be a simple Lie group, \(\Gamma\) an irreducible lattice in \(S\times S\) with embedding \(j=(j_1,j_2):\Gamma\to S\times S\) and \(G_1=G_2=S\times S\). Then take\N\[\N\Lambda=\{(j_1(\gamma_1),j_1(\gamma_2), j_2(\gamma_1),j_2(\gamma_2)):\gamma_1,\gamma_2\in\Gamma\} \subset \underbrace{S\times S}_{G_1}\times\underbrace{S\times S}_{G_2}\N\]\N\NOne might ask this question with ``lattice'' instead of ``irreducible lattice''.