Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 / 7 / b

Write $M=\lambda (\Gamma {)}^{\prime \prime }$, the group von Neumann algebra, and define right translations by $\rho (g){\delta }_h={\delta }_{h{g}^{-1}}$. Left and right translations commute, so every $T\in M$ commutes with every $\rho (g)$. A Von Neumann factor means that the center of $M$ is exactly $\mathbb{C}I$.

Suppose every nonidentity conjugacy class is infinite and $T$ is central. Put $\eta =T{\delta }_e$. Since $T$ commutes with both regular actions and $\lambda (g)\rho (g){\delta }_e={\delta }_e$,The action on a basis vector is ${\delta }_h\mapsto {\delta }_{gh{g}^{-1}}$, so the coefficients of $\eta \in {\ell }^2(\Gamma )$ are constant on conjugacy classes. A nonzero constant on an infinite class would have infinite squared sum. Thus all coefficients away from the identity vanish and $\eta =c{\delta }_e$.

Commutation with the right action now gives $T{\delta }_h=c{\delta }_h$ for every $h$, because those vectors form the right orbit of ${\delta }_e$. Hence $T=cI$ and $M$ is a factor. This argument also shows that ${\delta }_e$ is separating for $M$.

Conversely, if a nonidentity class $C$ is finite, thencommutes with every $\lambda (g)$, since conjugation permutes $C$. Therefore $Z\in M\cap \lambda (\Gamma {)}^{\prime }=Z(M)$. It is not scalar: $Z{\delta }_e={\sum }_{h\in C}{\delta }_h$ is nonzero and orthogonal to ${\delta }_e$. Consequently $M$ is not a factor. We have provedThe condition is vacuous for the trivial group, whose algebra $\mathbb{C}I$ is indeed a factor.