Kazhdan's Property T and Spectral Gap — Examples

If $\xi$ is a unit vector and $M:={\sup }_{x\in K}\parallel \pi (x)\xi -\xi \parallel <\epsilon \le 1$, then $M<1$. By [F1], the Haar average $\eta ={\int }_K\pi (x)\xi d\mu (x)$ is nonzero and $K$-invariant, and $\parallel \eta -\xi \parallel \le M$.

By [F1], $(K,\epsilon )$ is a Kazhdan pair for every $0<\epsilon \le 1$. Taking $\epsilon =1$ makes $K$ a Kazhdan set; its pair property applied to almost invariant unit vectors gives property (T) by [F2]. A representation on the zero Hilbert space has no unit vector, so the pair implication is vacuous there.

The finite discrete space has a finite subcover for every open cover, since a choice of one covering set for each of its finitely many points gives a finite subcover; distinct points are separated by open singletons, and the compact whole group is a neighbourhood of each point. Singleton rectangles make the product topology on $\mathrm{\Gamma }\times \mathrm{\Gamma }$ discrete, so multiplication and inversion are continuous; hence $\mathrm{\Gamma }$ is a compact Hausdorff locally compact topological group by [F1]. Since $m\ge 1$, define $\mu (A):=\mathrm{\mid }A\mathrm{\mid }\mathrm{/}m$ for $A\subseteq \mathrm{\Gamma }$. By [F2] and positive rescaling it is a Borel measure on the discrete topology. Every subset is open and compact. If $E$ is Borel and $V$ is open with $E\subseteq V$, finite additivity gives $\mu (V)=\mu (E)+\mu (V\setminus E)\ge \mu (E)$, and the open set $V=E$ attains this lower bound; if $U$ is open and $K$ is compact with $K\subseteq U$, then $\mu (U)=\mu (K)+\mu (U\setminus K)\ge \mu (K)$, and $K=U$ attains this upper bound. Every compact set has finite measure. For each $h\in \mathrm{\Gamma }$, left translation $g\mapsto hg$ bijects $\mathrm{\Gamma }$ and preserves cardinality, hence $\mu (hA)=\mu (A)$, while $\mu (\mathrm{\Gamma })=m\mathrm{/}m=1$. Thus $\mu$ is a normalized left Haar probability; by uniqueness it is the normalized Haar probability of [F3].

Applying the compact-group theorem [F4] to $\mathrm{\Gamma }$ and this $\mu$ proves that every $(\mathrm{\Gamma },\epsilon )$ with $0<\epsilon \le 1$ is a Kazhdan pair and that $\mathrm{\Gamma }$ has property (T).

Let $M:=H^{\mathrm{\Gamma }}$. It is a linear subspace because each $\pi (g)$ is linear. For each $g$, the map $T_g(\xi ):=\pi (g)\xi -\xi$ is continuous, since $\mathrm{\parallel }{T}_g(\xi )-T_g(\eta )\mathrm{\parallel }\le 2\mathrm{\parallel }\xi -\eta \mathrm{\parallel }$; [F7] makes $\ker T_g=T_g^{-1}(\{0\})$ closed. Therefore $M={\bigcap }_{g\in \mathrm{\Gamma }}\ker T_g$ is closed.

Define $P\xi :=m^{-1}{\sum }_{g\in \mathrm{\Gamma }}\pi (g)\xi$. For $h\in \mathrm{\Gamma }$, $\pi (h)P\xi =m^{-1}{\sum }_g\pi (hg)\xi =P\xi$ by the bijective reindexing $g\mapsto hg$, so $P\xi \in M$. If $\zeta \in M$, then every summand in $P\zeta$ equals $\zeta$, whence $P\zeta =\zeta$.

For $\xi \in H$ and $\zeta \in M$, inner-product preservation gives $⟨\pi (g)\xi ,\zeta ⟩=⟨\xi ,\pi (g^{-1})\zeta ⟩=⟨\xi ,\zeta ⟩$ because $\pi (g^{-1})\zeta =\zeta$. Summing yields $⟨P\xi ,\zeta ⟩=⟨\xi ,\zeta ⟩$, so $\xi -P\xi \in M^{\perp }$. Since $M$ is a closed linear subspace, [F9] identifies $P\xi$ with its Hilbert orthogonal projection component; [F9] also gives $\mathrm{\parallel }P\xi \mathrm{\parallel }\le \mathrm{\parallel }\xi \mathrm{\parallel }$.

The additive group $\mathbb{Z}$ with the discrete topology is a topological group: every subset of $\mathbb{Z}$ and of $\mathbb{Z}\times \mathbb{Z}$ is open, so addition and negation are continuous by [F4].

The compact set $Q$ is finite by [F1]. Let $M:=\max (\{0\}\cup \{\mathrm{\mid }n\mathrm{\mid }:n\in Q\})$ in $\mathbb{R}$, using the order-preserving embeddings in [F5]. This maximum exists because the displayed set is finite and linearly ordered; it also gives $M=0$ when $Q=\mathrm{\varnothing }$. Set $t:=\epsilon \mathrm{/}(4(M+1))>0$ and $z:=((1-t^2)+2it)\mathrm{/}(1+t^2)$. Then $\mathrm{\mid }z{\mathrm{\mid }}^2=((1-t^2{)}^2+4t^2)\mathrm{/}(1+t^2{)}^2=1$, so $\mathrm{\mid }z\mathrm{\mid }=1$ by nonnegativity of the modulus; the imaginary part $2t\mathrm{/}(1+t^2)$ is nonzero, so $z\mathrm{\ne }1$. Moreover, $\mathrm{\mid }z-1{\mathrm{\mid }}^2=(4t^4+4t^2)\mathrm{/}(1+t^2{)}^2=4t^2\mathrm{/}(1+t^2)\le 4t^2$, and hence $\mathrm{\mid }z-1\mathrm{\mid }\le 2t=\epsilon \mathrm{/}(2(M+1))$.

For every integer $n$, the geometric-sum identity and [F3] give $\mathrm{\mid }{z}^n-1\mathrm{\mid }\le \mathrm{\mid }n\mathrm{\mid }\mathrm{\mid }z-1\mathrm{\mid }$: for $n=k\ge 1$ factor $z^k-1=(z-1)(1+z+\cdots +z^{k-1})$ and use $\mathrm{\mid }{z}^j\mathrm{\mid }=1$; for $n=-k<0$, $\mathrm{\mid }{z}^{-k}-1\mathrm{\mid }=\mathrm{\mid }{z}^{-k}(1-z^k)\mathrm{\mid }=\mathrm{\mid }1-z^k\mathrm{\mid }$; for $n=0$ both sides are zero. Therefore, for every $n\in Q$, $\mathrm{\mid }{z}^n-1\mathrm{\mid }\le \mathrm{\mid }n\mathrm{\mid }\epsilon \mathrm{/}(2(M+1))\le M\epsilon \mathrm{/}(2(M+1))<\epsilon$. The last strict inequality also holds when $M=0$.

Since $\mathrm{\mid }z\mathrm{\mid }=1$, the map ${\chi }_z(n)=z^n$ is a homomorphism from the additive group $\mathbb{Z}$ into the unit scalars by [F3], and it is continuous because its domain is discrete by [F4]. On $H=\mathbb{C}$ with the inner product from [F2], define ${\pi }_z(n)u:={\chi }_z(n)u$. Each ${\pi }_z(n)$ is complex linear by the field laws and preserves the norm since $\mathrm{\mid }{z}^{n}u\mathrm{\mid }=\mathrm{\mid }{z}^n\mathrm{\mid }\mathrm{\mid }u\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }$; its inverse is ${\pi }_z(-n)$. The homomorphism law for ${\chi }_z$ gives the representation law, and its orbit maps are continuous by [F4], so ${\pi }_z$ is a strongly continuous unitary representation.

The vector $1\in \mathbb{C}$ has norm one by [F2] and satisfies $\mathrm{\parallel }{\pi }_z(n)1-1\mathrm{\parallel }=\mathrm{\mid }{z}^n-1\mathrm{\mid }<\epsilon$ for each $n\in Q$ by step 2.1. If $u\in \mathbb{C}$ were invariant under ${\pi }_z$, invariance under $1\in \mathbb{Z}$ would give $zu=u$; since $z\mathrm{\ne }1$ and $\mathbb{C}$ is a field, this forces $u=0$. Thus $1$ is a $(Q,\epsilon )$-invariant unit vector in a representation with no nonzero invariant vector.

Since $Q$ and $\epsilon >0$ were arbitrary, the witness in step 3.1 shows that no compact $Q$ and positive tolerance form a Kazhdan pair. Hence $\mathbb{Z}$ has no compact Kazhdan set. By [F6], this rules out property (T). The character construction itself uses no Choice; AC is used only in this final implication.

By [F1], each fixed $I_{0,\nu }$ is irreducible and strongly continuous unitary; [F2] gives a nonzero vector $f_2$ with ${\mathrm{\Pi }}_{\nu }(k_{\pi \mathrm{/}2})f_2=-f_2$, so it is not the trivial representation. Its fixed subspace is closed and invariant, hence irreducibility implies that it is zero.

Fix any compact $Q\subseteq G$ and $\epsilon >0$. If $Q=\mathrm{\varnothing }$, every unit vector is $(Q,\epsilon )$-invariant. If $Q$ is nonempty and compact, [F3] lets us choose $\nu$ close enough to $1$ that ${\sup }_{g\in Q}\mathrm{\mid }{\phi }_{\nu }(g)-1\mathrm{\mid }<{\epsilon }^2\mathrm{/}2$; [F1] supplies the unit vector $f_0$, and [F5] gives $\mathrm{\parallel }{\mathrm{\Pi }}_{\nu }(g)f_0-f_0\mathrm{\parallel }<\epsilon$ for every $g\in Q$. Thus the same fixed-parameter family has nontrivial representations with near-invariant vectors for every compact test and tolerance.

By [F4], the nontrivial classes from step 1.1 converge to the trivial class in the Fell topology as $\nu \uparrow 1$. No fixed $I_{0,\nu }$ is asserted to weakly contain the trivial representation; the convergence is a parameter-family statement.

The A-page proposition in [F6] constructs the direct sum over an explicit cofinal sequence ${\nu }_j\uparrow 1$, which has almost invariant vectors and no invariant vector; hence $G$ fails property (T), refuting the statement above. AC is inherited as in [A1].