Property (T) and isolation of the trivial representation in the Fell dual

Suppose $[1_G]$ is not isolated in $\widehat{G}$, and let $S=\widehat{G}\setminus \{[1_G]\}$. Then $[1_G]\in \bar{S}$, so [F3] gives $1_G\prec {\mathrm{\Pi }}_S:={\widehat{⨁}}_{[\sigma ]\in S}\sigma$, with representatives selected by AC; if $S$ is empty its closure is empty, so this case cannot occur.

Now assume $[1_G]$ is isolated in $\widehat{G}$. Let $A=C^{\ast }(G)$, let $\chi$ be the character corresponding to $1_G$, and set $I={\bigcap }_{[\sigma ]\in \widehat{G}\setminus \{[1_G]\}}\ker \sigma$, with the empty intersection interpreted as $A$. By [F3], isolation is equivalent to $I⊈\ker \chi$. The set $I$ is a closed two-sided star-ideal as an intersection of representation kernels.

Assume property (T) and let $R\supseteq \widehat{G}$ satisfy the stated no-invariants condition outside $[1_G]$. If $[1_G]$ were not isolated in $R$, each Fell neighborhood $W(1_G;1,Q,ϵ)$ would contain a class $[\sigma ]\in R\setminus \{[1_G]\}$. By [F9], the constant coefficient $1$ is then approximated on $Q$ by a finite sum of coefficients of $\sigma$; embedding those vectors in its coordinate of ${\mathrm{\Pi }}_R:={\widehat{⨁}}_{[\tau ]\in R\setminus \{[1_G]\}}\tau$ gives the same finite-sum approximation in ${\mathrm{\Pi }}_R$. Every diagonal coefficient of $1_G$ is a nonnegative constant, so scaling the approximating vectors by its square root (with the zero coefficient handled by the zero vector) gives $1_G\prec {\mathrm{\Pi }}_R$. Representatives and the direct sum exist by [F10].

Every summand of ${\mathrm{\Pi }}_S$ is irreducible and nontrivial, so [F2] gives no nonzero invariant vector in any summand and hence none in the direct sum. By [F4], ${\mathrm{\Pi }}_S$ has almost invariant vectors, contradicting property (T) in [F1]. Therefore property (T) implies isolation of $[1_G]$ in $\widehat{G}$.

If $a\in I\cap \ker \chi$, then every irreducible nondegenerate star-representation of $A$ kills $a$: by [F6] each corresponds to an irreducible group representation, and [F2] says that a class with invariant vectors is trivial, while all other classes occur in the intersection defining $I$. Thus [F5] gives $a=0$, so $\chi {\mathrm{\mid }}_I$ is injective. Since $I⊈\ker \chi$, choose $b\in I$ with $\chi (b)\mathrm{\ne }0$ and set $p=b\mathrm{/}\chi (b)\in I$; then $\chi (p)=1$.

The elements $p^{\ast }-p$ and $p^2-p$ lie in $I$ and have $\chi$-value zero, so injectivity of $\chi {\mathrm{\mid }}_I$ gives $p^{\ast }=p=p^2$. For every $a\in A$, both $ap-\chi (a)p$ and $pa-\chi (a)p$ lie in $I$ and have $\chi$-value zero; hence $ap=pa=\chi (a)p$. Thus $p$ is a central projection.

Let $\rho$ be any strongly continuous unitary representation of $G$ with almost invariant vectors, and let $\pi$ be its nondegenerate representation of $A$ from [F6]. By [F4], $1_G\prec \rho$. If $P:=\pi (p)$ were zero, then $p\in \ker \pi$ and kernel inclusion in [F4] would give $p\in \ker \chi$, contradicting $\chi (p)=1$. Thus $P\mathrm{\ne }0$.

The operator $P$ is a bounded self-adjoint idempotent by [F7]. Its range $M=PH=\ker (I-P)$ is closed, and $\ker P=M^{\perp }$: if $x\in \ker P$ and $y=Pz$, then $⟨x,y⟩=⟨x,Pz⟩=⟨P^{\ast }x,z⟩=0$; conversely, if $x\perp M$, then $⟨Px,z⟩=⟨x,Pz⟩=0$ for every $z$, so $Px=0$. Every $x\in H$ decomposes as $x=Px+(I-P)x$ with the two terms in $M$ and $M^{\perp }$. Centrality of $p$ makes $P$ commute with $\pi (A)$, so these two closed subspaces reduce $\pi (A)$.

The restrictions ${\pi }_1=\pi {\mathrm{\mid }}_M$ and ${\pi }_2=\pi {\mathrm{\mid }}_{M^{\perp }}$ are nondegenerate: applying the commuting projections $P$ and $I-P$ to finite sums from the dense span of $\pi (A)H$ shows that $\pi (A)M$ spans $M$ and $\pi (A)M^{\perp }$ spans $M^{\perp }$. They are star-representations, and $\pi ={\pi }_1\oplus {\pi }_2$.

On $M$, for $y=Pz$ and $a\in A$, ${\pi }_1(a)y=\pi (a)\pi (p)z=\pi (ap)z=\chi (a)y$ by step 3.1. Thus ${\pi }_1$ is the amplification of $\chi$, which is the integrated form of the trivial group representation amplified on the nonzero space $M$. The direct sum of the group representations corresponding to ${\pi }_1$ and ${\pi }_2$ integrates to ${\pi }_1\oplus {\pi }_2=\pi$; uniqueness in [F6] identifies it with $\rho$. Hence $\rho$ has the nonzero fixed subspace $M$, proving that isolation of $[1_G]$ implies property (T).

By [F4], ${\mathrm{\Pi }}_R$ has almost invariant vectors, so property (T) gives a nonzero invariant vector. But each summand indexed by $R\setminus \{[1_G]\}$ has no nonzero invariant vector by hypothesis, hence neither does the direct sum; contradiction. Conversely, the universal $R$ condition includes $R=\widehat{G}$, and every nontrivial irreducible class has no invariant vector by [F2], so it implies isolation in $\widehat{G}$ and then property (T) by step 5.2. This proves the stated equivalence.