Factor (primary) representations

For a discrete group $\mathrm{\Gamma }$, let ${\mathrm{\#}}_{\mathrm{\Gamma }}$ be counting measure. Every function on $\mathrm{\Gamma }$ is measurable and a ${\mathrm{\#}}_{\mathrm{\Gamma }}$-null set is empty. By [F5], ${\int }_{\mathrm{\Gamma }}\mathrm{\mid }f{\mathrm{\mid }}^{2}d{\mathrm{\#}}_{\mathrm{\Gamma }}={\sum }_{\gamma \in \mathrm{\Gamma }}\mathrm{\mid }f(\gamma ){\mathrm{\mid }}^2$, so the identity on functions identifies ${\mathrm{\ell }}^2(\mathrm{\Gamma },\mathbb{C})$ isometrically with $L^2(\mathrm{\Gamma },{\mathrm{\#}}_{\mathrm{\Gamma }};\mathbb{C})$; the pairings agree as well, since [F4] makes $f\bar{g}$ absolutely summable for $f,g\in {\mathrm{\ell }}^2$. Counting measure is Radon on the locally compact discrete space by [F5], so [F6] makes ${\mathrm{\ell }}^2(\mathrm{\Gamma },\mathbb{C})$ complete and hence a Hilbert space. Its standard vectors ${\delta }_{\gamma }$ form an orthonormal basis: given $f\in {\mathrm{\ell }}^2$ and $\epsilon >0$, the finite-subset supremum defining its square sum gives a finite $F\subseteq \mathrm{\Gamma }$ with ${\sum }_{\gamma \mathrm{\notin }F}\mathrm{\mid }f(\gamma ){\mathrm{\mid }}^2<{\epsilon }^2$, and truncation to $F$ approximates $f$ within $\epsilon$.

Put $S=\pi (G)$. Since $\pi (g{)}^{\ast }=\pi (g^{-1})$ by [F2], $S$ is self-adjoint. By [F1], $S^{\mathrm{\prime }}$ is a weak-operator-closed unital $\ast$-subalgebra, and $M=S^{\mathrm{\prime }\mathrm{\prime }}=(S^{\mathrm{\prime }}{)}^{\mathrm{\prime }}$ is also one; hence $M$ is a concrete von Neumann algebra. The commutant identity is algebraic: $S\subseteq S^{\mathrm{\prime }\mathrm{\prime }}$, so every operator in $(S^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\prime }}$ belongs to $S^{\mathrm{\prime }}$; conversely every $T\in S^{\mathrm{\prime }}$ commutes with every element of $S^{\mathrm{\prime }\mathrm{\prime }}=(S^{\mathrm{\prime }}{)}^{\mathrm{\prime }}$, hence $T\in (S^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\prime }}$. Therefore $\pi (G{)}^{\mathrm{\prime }}=(\pi (G{)}^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\prime }}$.

If $\pi$ is irreducible, [F3] gives $\pi (G{)}^{\mathrm{\prime }}=\mathbb{C}{I}_H$. Step 1.2 identifies this with $M^{\mathrm{\prime }}$, so $Z(M)=M\cap M^{\mathrm{\prime }}=\mathbb{C}{I}_H$ because $I_H\in M$. Thus $\pi$ is factorial.

Let $\mathrm{\Gamma }$ be nontrivial and ICC, and let $\lambda (g)f(h)=f(g^{-1}h)$ and $\rho (g)f(h)=f(hg)$ on the Hilbert space ${\mathrm{\ell }}^2(\mathrm{\Gamma })$ of step 1.1. The left translations form a unitary representation, and each right translation is unitary; direct substitution shows $\lambda (a)\rho (g)=\rho (g)\lambda (a)$. Thus $\rho (g)\in \lambda (\mathrm{\Gamma }{)}^{\mathrm{\prime }}=(\lambda (\mathrm{\Gamma }{)}^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\prime }}$ by step 1.2. Put $M_{\mathrm{\Gamma }}=\lambda (\mathrm{\Gamma }{)}^{\mathrm{\prime }\mathrm{\prime }}$. For $z\in Z(M_{\mathrm{\Gamma }})$ let $\xi =z{\delta }_e$. Since $z$ commutes with both $\lambda (g)$ and $\rho (g^{-1})$, and $\rho (g^{-1}){\delta }_e={\delta }_g$, we have $\lambda (g)\xi =z{\delta }_g=z\rho (g^{-1}){\delta }_e=\rho (g^{-1})\xi$. In coordinates this is $\xi (g^{-1}h)=\xi (h{g}^{-1})$; setting $h=gk$ shows $\xi (k)=\xi (gk{g}^{-1})$. Hence $\xi$ is constant on conjugacy classes. Every nonidentity conjugacy class is infinite, so square summability forces $\xi$ to vanish off $e$: $\xi =c{\delta }_e$. For every $h\in \mathrm{\Gamma }$, $z{\delta }_h=z\lambda (h){\delta }_e=\lambda (h)z{\delta }_e=c{\delta }_h$. The standard vectors span densely by step 1.1, hence $z=cI$ and $M_{\mathrm{\Gamma }}$ is a factor.

Choose $g\mathrm{\ne }e$ in $\mathrm{\Gamma }$. The right translation $\rho (g)$ is a bounded self-intertwiner of $\lambda$ but is not scalar, since $\rho (g){\delta }_e={\delta }_{g^{-1}}\mathrm{\ne }{\delta }_e$. If $\lambda$ were irreducible, [F3] would force this self-intertwiner to be scalar. Therefore $\lambda$ is not irreducible. To see the example class is nonempty, let $\mathrm{\Gamma }$ be the finitary symmetric group on $\mathbb{N}$. For any nonidentity finite permutation $\sigma$, let $a$ be the least moved point; for each distinct $b$ outside the finite support, conjugation by the transposition $(ab)$ gives support $(\mathrm{supp}\sigma \setminus \{a\})\cup \{b\}$. These supports are distinct, so the conjugacy class is infinite. Thus this group is nontrivial ICC and supplies the promised example.