Property (T) is equivalent to the existence of a compact Kazhdan pair

Assume (ii), and let $\pi$ have almost invariant vectors. Choose the compact $Q$ and $\epsilon >0$ from (ii). By [F1], $\pi$ has a $(Q,\epsilon )$-invariant unit vector, so the pair property [F2] supplies a nonzero invariant vector. Hence $G$ has property (T).

Assume (i) and suppose, for contradiction, that no compact Kazhdan pair exists. Let $I$ be the set of pairs $(Q,\eta )$ with $Q$ compact in $G$ and $\eta >0$. For each $i=(Q,\eta )\in I$, let $C_i$ be the subset of the set ${\mathbb{C}}^G$ consisting of normalized diagonal coefficients of unit vectors in strongly continuous representations with no nonzero invariant vector that are $(Q,\eta )$-invariant. Failure of $(Q,\eta )$ to be a pair makes $C_i$ nonempty. AC chooses one function ${\phi }_i\in C_i$ for each $i\in I$; this is a choice from subsets of the set ${\mathbb{C}}^G$, not from the class of all representations.

If (ii) holds for a compact $Q$ and $\epsilon >0$, [F2] gives $\kappa (G,Q)\ge \epsilon >0$, so (iii) holds. Conversely, if (iii) holds for a compact $Q$, choose $\alpha >0$ strictly below $\kappa (G,Q)$ (choose $\alpha =1$ when the threshold is $+\mathrm{\infty }$); [F2] says $(Q,\alpha )$ is a Kazhdan pair. Hence (ii) and (iii) are equivalent.

For the quantitative clause put $M=H^G$. By [F5] it is closed and invariant, its orthogonal complement is invariant, and the restricted representation on $M^{\perp }$ has no nonzero invariant vector. By [F6], write $\xi =P\xi +{\xi }^{\mathrm{\prime }\mathrm{\prime }}$ with ${\xi }^{\mathrm{\prime }\mathrm{\prime }}=\xi -P\xi \in M^{\perp }$. If ${\xi }^{\mathrm{\prime }\mathrm{\prime }}=0$, the claimed estimate is immediate. If ${\xi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\ne }0$ and $Q=\mathrm{\varnothing }$, then ${\xi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}\parallel {\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel$ is vacuously $(Q,\epsilon )$-invariant, so the pair would force an invariant vector in the restricted representation, a contradiction.

For each $i$, let $({\rho }_i,K_i,{\zeta }_i)$ be the canonical pointed cyclic GNS representation of ${\phi }_i$ from [F3]. This representation has no nonzero invariant vector: a witness in the definition of $C_i$ has a closed cyclic subspace generated by its unit vector; that subspace has no invariant vector, and its pointed cyclic representation has coefficient ${\phi }_i$, so [F3] identifies it unitarily with the GNS representation. In particular, $\mathrm{\parallel }{\zeta }_i\mathrm{\parallel }=1$ and ${\zeta }_i$ is $(Q,\eta )$-invariant in ${\rho }_i$.

Thus in the remaining case $Q\mathrm{\ne }\mathrm{\varnothing }$ and ${\xi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\ne }0$. The unit vector ${\xi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}\parallel {\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel$ cannot be $(Q,\epsilon )$-invariant, because the restricted representation has no nonzero invariant vector and $(Q,\epsilon )$ is a Kazhdan pair. Hence some $q\in Q$ satisfies $\parallel \pi (q){\xi }^{\mathrm{\prime }\mathrm{\prime }}-{\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel \ge \epsilon \parallel {\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel$. Since $P\xi \in H^G$, this displacement equals $\parallel \pi (q)\xi -\xi \parallel$, which is strictly less than $\delta \epsilon \parallel \xi \parallel$ by the assumed invariance of $\xi$. Canceling $\epsilon >0$ gives $\parallel \xi -P\xi \parallel =\parallel {\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel <\delta \parallel \xi \parallel$, and therefore the stated weak inequality.

Form the set-indexed Hilbert direct sum $\rho ={\widehat{⨁}}_{i\in I}{\rho }_i$. Given any compact $Q$ and $ϵ>0$, the coordinate indexed by $(Q,ϵ\mathrm{/}2)$ contains a unit vector that is $(Q,ϵ\mathrm{/}2)$-invariant; its image under the coordinate inclusion is $(Q,ϵ)$-invariant in $\rho$. Thus $\rho$ has almost invariant vectors. By (i) it has a nonzero invariant vector, but each coordinate of an invariant vector is invariant in its summand by [F4], and every ${\rho }_i$ has no such vector by step 2.1. All coordinates must therefore vanish, a contradiction. This proves (i)$\Rightarrow$(ii).

Steps 1.1 and 3.1 prove (i)$⟺$(ii), step 1.3 proves (ii)$⟺$(iii), and steps 1.4 and 2.2 prove the quantitative clause, including the zero vector, zero representation and empty-$Q$ cases. AC is used for the set-indexed coefficient selection, GNS/direct-sum constructions, and through Countable Choice in the orthogonal-decomposition and projection suppliers; no proper-class selection is used.