Gel′fand’s problem on unitary representations associated with discrete subgroups of {{𝑃𝑆𝐿}}_2({{𝐑}})

Abstract

In 1978 M.-F. Vigneras [10] gave a negative answer to the question posed by I. M. Gel'fand [2], who asked if the induced representation of PSL2(R) on L(r\PSL2(R)) determines a discrete subgroup up to conjugation. She constructed explicitly two nonconjugate discrete groups arising from indefinite quaternion algebras defined over number fields giving rise to isomorphic induced representations. Such examples are necessarily quite sporadic since there are only finitely many conjugacy classes of arithmetic groups with a fixed signature; see K. Takeuchi [9]. The purpose of this note is to give, in a rather simple way, a large family of (nonarithmetic) discrete groups that are not determined by their induced representations. The key idea is to reduce the problem to the case of finite groups where the situations are simple and well understood. We should point out that a similar idea can be applied to constructions of various isospectral Riemannian manifolds [8]. Our construction is based on the following proposition, which follows from standard facts about induced representations. PROPOSITION. Let G be a locally compact topological group, and let C Ti, T2 C To be discrete subgroups, with normal and of finite index in IV Then if the subgroups U% = t%/T, i = 1,2, of Q = r 0 / r meet each conjugacy class of Q in the same number of elements, the representations of G on the spaces L (r \G), i ~ 1,2, are unitarily equivalent To construct nonconjugate T and I2 in PSL2(R), we first choose an appropriate triple ($, #i, #2) with the same induced representation Ind$ (1) = IndjJ (1). For instance, we let Q be the semidirect product (Z/8Z) • (Z/8Z) and set ^ = {(1,0), (3,0), (5,0), (7,0)}, X2 = {(1,0), (3,4), (5,4), (7,0)}. It is easy to check that M and #2 are not conjugate in C, and each conjugacy class of Q meets Mi and #2 in the same number of elements. We then take a torsion-free discrete subgroup To C PSL/2(R) satisfying the following conditions: (i) the genus of the Riemann surface F0\PSL2(R)/SO(2) is greater than two; (ii) To is maximal in PGL2(R); and (iii) To is nonarithmetic. Since the fundamental group of a Riemann surface of genus k has a free group on k generators as quotient, we may always, if k is large enough, find a sur-