Showing that a von neumann algebra in the bounded operators of $l^2(S_\infty)$ is a factor

Question

Consider $S_\infty$ i.e. the group of permutations with functions $\sigma:\mathbb{N} \rightarrow \mathbb{N}$ such that $\sigma(n) = n$ for all but finitely many. The left regular represenation defined on $S_\infty$ into $B(l^2(S_\infty))$ as $$\lambda_\sigma(\delta_\psi)) := \delta_{\sigma \circ \psi}$$where $\{\delta_\sigma\}$ is the orthnormal basis. Not obvious, but I know that if I show that weak operator closure of $\text{span} \lambda(S_\infty)$(which is the weak operator closure is clearly a von Neumann algebra)is a factor i.e. the center of this von Neumann algebra is equal to $\mathbb{C}I$, then $S_\infty$ has infinite conjugacy classes(this implication is not obvious). I am having trouble understanding the weak operator closure of $\text{span} \lambda(S_\infty)$ is a factor. I am not used to dealing with $l^2(G)$ spaces.

Answer 1

When $G$ is a countable group and $\lambda:G\to B(\ell^2(G))$ is the left regular representation, $\lambda(G)''$ is a factor if and only if $G$ is icc. Below is the standard argument.

Indeed, suppose first that $G$ is not icc, so there exists $g\in G$ such that $\{hgh^{-1}: h\in G\}=\{g_1,\ldots,g_n\}$. Consider the element $T=\sum_{j=1}^n \lambda(g_j)\in\lambda(G)''$. For any $h\in G$, $$ \lambda(h)T\lambda(h)^*=\sum_{j=1}^n\lambda(hg_jh^{-1})=\sum_{j=1}^n\lambda(g_j)=T. $$ So $\lambda(h)T=T\lambda(h)$ for all $h\in G$. So $T\in\lambda(G)'\cap\lambda(G)''=Z(\lambda(G)'')$. This shows that if $G$ is not icc, then $\lambda(G)''$ is not a factor.

Conversely, suppose that $G$ is icc. Let $T\in Z(\lambda(G)'')$. Here we use that $\lambda(G)'=\rho(G)''$. So $T\in \lambda(G)''\cap\rho(G)''$. Write $T\delta_e=\sum_{g\in G}\alpha_g\delta_g$. Then, for $h\in G$,
$$ \sum_{g\in G}\alpha_g\delta_g=T\delta_e=T\lambda(h)\rho(h)\delta_e=\lambda(h)\rho(h)T\delta_e =\sum_{g\in G}\alpha_g\delta_{hgh^{-1}}=\sum_{g\in G}\alpha_{h^{-1}gh}\delta_g. $$ It follows that $\alpha_g=\alpha_{hgh^{-1}}$ for all $h\in G$. That is, $\alpha$ is contant on the conjugacy class of $g$, which is infinite. In other words, for each $g\ne e$ the number $\alpha_g$ appears infinitely many times in the expansion $\sum_{g\in G}\alpha_g\delta_g$. As $\sum_g|\alpha_g|^2<\infty$, this implies that $\alpha_g=0$ for all $g\ne e$. Thus $T\delta_e=\alpha_e\delta_e$. Using that $T$ is central, $$ T\delta_g=T\lambda(g)\delta_e=\lambda(g)T\delta_e=\alpha_e\,\lambda(g)\delta_e=\alpha_e\,\delta_e, $$ and by linearity and continuity this gives us that $T=\alpha_e\,I$. So $Z(\lambda(G)'')=\mathbb C\,I$ and $\lambda(G)''$ is a factor.