Exercise 15, § 3

Let G be a locally compact group, C a subset of G and $\varepsilon >0$. One says that $(C, \varepsilon )$ is a Kazhdan pair for G if, for every unitary representation $\pi$ of G in a Hilbert space E, the existence of a vector $x\in E$ such that

sup$_{g\in C}\|\pi (g)x-x\|< \varepsilon \|x\|$,

implies the existence of a nonzero trivial subrepresentation of $\pi$. One says that C is a Kazhdan set if there exists $\varepsilon >0$ such that $(C, \varepsilon )$ is a Kazhdan pair.

a) Let $(C, \varepsilon )$ be a Kazhdan pair for G. Let $\pi$ be a unitary representation of G in a Hilbert space E. Let $x\in E$ be nonzero and let $\delta >0$ be such that

sup$_{g\in C}\|\pi (g)x-x\|< \delta \varepsilon \|x\|$.

Show that then $\|p(x)-x\|\leqslant \delta \|x\|$, where $p$ is the orthogonal projector of E whose image is the space of G-invariant vectors.

b) Prove that a locally compact group G has property ]T[ if and only if there exists a Kazhdan pair $(C, \varepsilon )$. (To prove that property ]T[ implies the assertion, proceed by contraposition by considering the unitary representation which is the Hilbert sum of representations $\pi_{C,\varepsilon}$ parametrized by a compact subset C of G and $\varepsilon >0$ in such a way that $\pi_{C,\varepsilon}$ contains no trivial subrepresentation but that there exists $x_{C,\varepsilon}$ of norm 1 with sup$_{g\in\mathbf{C}}\|\pi (g)x_{C,\varepsilon}-x_{C,\varepsilon}\|< \varepsilon .$)

c) Let G be a locally compact group having property ]T[. Show that every compact subset C of G which generates G is a Kazhdan set.

d) Let G be a locally compact group having property ]T[, and let C be a compact subset of G with nonempty interior which is a Kazhdan set. Show that then C generates G. In particular, if G is discrete, then the Kazhdan sets are the finite generating subsets of G. (The subgroup H of G generated by C is open; consider the quasi-regular representation $\pi$ of G on $G/H$ and show that the projection of the characteristic function $\varphi$ of $H\in G/H$ onto the space of invariant vectors of $\pi$ is equal to $\varphi .$)