On the profinite rigidity of lattices in higher rank Lie groups

Abstract

We study the existence of lattices in a higher rank simple Lie group which are profinitely but not abstractly commensurable. We show that no such examples exist for the complex forms of type E 8 , F 4 , and G 2 . In contrast, there are arbitrarily many such examples for all other higher rank Lie groups, except possibly SL 2n+1 (R), SL 2n+1 (C), SL n (H), and any form of type E 6 .

Adelic superrigidity and profinitely solitary lattices

“…A special case of this theorem has geometrical relevance. It was no coincidence that all examples of profinitely commensurable but noncommensurable lattices that we constructed in [9] were cocompact.…”

“…p. 15. In [9], we recently showed that for most (but not all) groups G, profinite commensurability is a strictly weaker equivalence relation on the set of lattices than commensurability. So the problem suggests itself to find the profinite commensurability classification of lattices in terms of G-arithmetic pairs.…”

“…The conjecture remains open only for certain anisortopic forms of type A n and E 6 . A recent survey with the precise definitions and statements can be found in [9,Appendix A]. Two G-arithmetic pairs are locally isomorphic if the groups are isomorphic over a local isomorphism of the number fields of definition by which we mean an isomorphism of the finite adele rings; cf.…”

“…We saw in [9] that for K = C, all lattices in G(C), cocompact or not, are profinitely solitary if the complex group G has type E 8 , F 4 , or G 2 . In contrast, if the complex group G has any other type except possibly E 6 and A n where CSP is open, then a non-cocompact lattice Γ ⊂ G(C) is not necessarily profinitely solitary as we shall explain in Remark 5.2.…”

On absolutely profinitely solitary lattices in higher rank Lie groups

“…Recent breakthroughs are Yi Liu's result [10] that for G = PSL 2 (C), only finitely many lattices Γ ≤ G can have the same profinite completion and M. Stover's negative answer [25] for G = SU(n, 1). For most higher rank Lie groups G, the question can be answered in the negative by means of the congruence subgroup property (CSP) as we recalled in [7,Theorem 1.3]. These non-isomorphic but profinitely isomorphic lattices are however commensurable and hence not genuinely distinct.…”