Dependence of Kazhdan constants on generating subsets (Q1601464)

Let \(\Gamma\) be a discrete group generated by a finite subset \(S\). \(\varepsilon(S)>0\) is a Kazhdan constant for \(S\) if there is \(s \in S\) such that \(||\pi(s)u-u||\geq \varepsilon(S)||u||\) whenever \((\pi, {\mathcal H})\) is a unitary representation of \(\Gamma\) without invariant vectors and \(u \in {\mathcal H}\). \(\Gamma\) is said to be a Kazhdan group if it has a generating set \(S\) with a Kazhdan constant. If this is the case and \(S'\) is another finite generating set, then \(S'\) has a Kazhdan constant \(\varepsilon(S')\), i.e., being Kazhdan does not depend on the set of generators. In page 126 of [Discrete Groups, Expanding Graphs and Measures, Prog. Math. 125 (Basel 1994; Zbl 0826.22012)] \textit{A. Lubotzky} asks whether one can choose a Kazhdan constant which is independent of \(S\). The authors answer this question negatively by proving that if a Kazhdan group \(\Gamma\) has a dense homomorphic image contained in a connected topological group \(G\) which has a continuous unitary representation without invariant vectors, then for every \(\varepsilon>0\) there is a generating set \(S\) of \(\Gamma\) such that \(\varepsilon\) is not a Kazhdan constant for \(S\). The authors also point out that there are concrete examples of families of Kazhdan groups densely embedded into connected simple Lie groups.