Property (T) is a uniform spectral gap over all representations

Suppose $G$ has property (T). By [F1] choose a compact Kazhdan pair $(Q_0,\epsilon )$. Put $Q=Q_0\cup \{e\}$; this is compact because a cover has a finite subcover on $Q_0$ and one additional member covering $e$, and it is still a Kazhdan pair because $(Q,\epsilon )$-invariance implies $(Q_0,\epsilon )$-invariance. Let $\pi$ be any strongly continuous unitary representation and $\xi \in (H^G{)}^{\perp }$. The assertion is immediate for $\xi =0$. If $\xi \mathrm{\ne }0$ and the displayed supremum were less than $\epsilon \parallel \xi \parallel$, then $\xi \mathrm{/}\parallel \xi \parallel$ would be a $(Q,\epsilon )$-invariant unit vector in the restriction to $(H^G{)}^{\perp }$. By [F3] that restriction has no nonzero invariant vector, contradicting the pair property. Thus the uniform lower bound holds for every $\pi$ and $\xi$.

Conversely, assume a compact $Q$ and $\epsilon >0$ satisfy the uniform bound. Let $(\pi ,H)$ have a $(Q,\epsilon )$-invariant unit vector $\xi$. By [F4], write $\xi =P\xi +{\xi }^{\mathrm{\prime }\mathrm{\prime }}$ with ${\xi }^{\mathrm{\prime }\mathrm{\prime }}\in (H^G{)}^{\perp }$. If ${\xi }^{\mathrm{\prime }\mathrm{\prime }}=0$, then $P\xi =\xi$ is already a nonzero invariant vector. If ${\xi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\ne }0$ and $Q=\mathrm{\varnothing }$, the uniform inequality reads $0\ge \epsilon \parallel {\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel$, a contradiction; so $Q$ is nonempty. Since $P\xi$ is fixed, $\parallel \pi (x){\xi }^{\mathrm{\prime }\mathrm{\prime }}-{\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel =\parallel \pi (x)\xi -\xi \parallel <\epsilon$ for every $x\in Q$. By [F5] the compact-set supremum is a maximum $D<\epsilon$. Applying the uniform bound to ${\xi }^{\mathrm{\prime }\mathrm{\prime }}$ gives $\epsilon \parallel {\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel \le D<\epsilon$, so $\parallel {\xi }^{\mathrm{\prime }\mathrm{\prime }}\parallel <1$. Orthogonality in [F4] now gives $\parallel P\xi {\parallel }^2=1-\parallel {\xi }^{\mathrm{\prime }\mathrm{\prime }}{\parallel }^2>0$. Thus $P\xi$ is a nonzero invariant vector, and $(Q,\epsilon )$ is a Kazhdan pair.

Let $\pi$ be any strongly continuous unitary representation of $G$ with almost invariant vectors. Since the Kazhdan pair from step 1.2 has compact $Q$, almost invariance supplies a $(Q,\epsilon )$-invariant unit vector; step 1.2 then gives a nonzero invariant vector. This is property (T) by [F2].

Steps 1.1–2.1 prove the equivalence and show that a compact Kazhdan pair is itself a uniform spectral-gap witness. For the locally compact Hausdorff clause, apply [F6] to each representation. If a uniform witness has empty $Q$, the bound forces every fixed-space complement to be zero and $\{e\}$ is then a common witness; otherwise adjoining $e$ preserves the lower bound by [F7]. Thus a single compact set and constant witness spectral gap simultaneously for all representations exactly when $G$ has property (T). AC is used through the pair-equivalence theorem and, via Countable Choice, by the orthogonal-decomposition/projection suppliers.