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For a discrete group, its group von Neumann algebra is generated on ${\ell }^2(\Gamma )$ by left translations $\lambda (g){\delta }_h={\delta }_{gh}$. The vector ${\delta }_e$ is cyclic and separating, and its vector functional is tracial. Its modular conjugation acts by $J{\delta }_g={\delta }_{g^{-1}}$ with conjugated scalar coefficients. The commutant is the algebra generated by right translations $\rho (g){\delta }_h={\delta }_{h{g}^{-1}}$.

A discrete group is ICC when every nonidentity element has an infinite conjugacy class. The trivial group satisfies this condition vacuously. It characterizes when the group von Neumann algebra is a Von Neumann factor.

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.

The dense commutant orbit makes $\Omega$ separating for $M$, so $S_0(a\Omega )=a^{\ast }\Omega$ is well defined on the dense subspace $M\Omega$. Put $\tau (a)=⟨a\Omega ,\Omega ⟩$. The assumed identity says that this positive vector functional is a tracial positive functional. HenceIt extends to an antiunitary involution $J$ on $H$, with $J\Omega =\Omega$. This is the modular conjugation, and the modular operator in this tracial case is $\Delta =I$.

On the dense subspace $M\Omega$,Thus it acts by right multiplication and commutes with all left multiplications; $JMJ\subseteq M^{\prime }$. We prove the reverse inclusion rather than assuming the general Tomita commutation theorem.

Let $T\in M^{\prime }$, $C=\Vert T\Vert$, and $\eta =T\Omega$. The setis weakly compact and convex. Indeed, trace-class duality and the Banach-Alaoglu theorem make the bounded-operator ball compact in a topology stronger than the weak operator topology. The algebra $M$ is weak-operator closed, and $a\mapsto a\Omega$ is weakly continuous, so its image is weakly compact.

For $b\in M$ use the polar decomposition of a bounded operator $b=uh$, $h=|b|$. Both $h$ and $u$ belong to $M$: continuous functional calculus gives $h$, and $b(h+\epsilon I{)}^{-1}$ tends strongly to $u$. The polar decomposition is proved independently in Question 8. Since $T$ commutes with $M$,The trace identity givesTherefore $|⟨\eta ,b\Omega ⟩|\le C\tau (h)$. Meanwhilewith the supremum attained by $a=Cu$. For the upper bound, traciality gives $\tau (b^{\ast }a)=\tau (h^{1/2}{u}^{\ast }a{h}^{1/2})$; evaluating this on $\Omega$ bounds its absolute value by $\Vert a\Vert \tau (h)$.

These support inequalities imply $\eta \in \mathcal{B}$. Otherwise the Hahn-Banach separation theorem would separate $\eta$ from this closed convex set by a real inner-product functional. Since $M\Omega$ is dense, its separating vector could be approximated by some $b\Omega$, preserving strict separation, contrary to the displayed inequality. Thus $T\Omega =a\Omega$ for an $a\in M$ of norm at most $C$.

The bounded operator $R_a=J{a}^{\ast }J$ commutes with $M$ and sends $\Omega$ to $a\Omega$. It agrees with $T$ on the cyclic vector and therefore on all of $M\Omega$, hence on $H$. This proves $T\in JMJ$, and so

For the regular representation, take $\Omega ={\delta }_e$. It is cyclic for the left action and for the commuting right action. On finite sums of left translations, its vector functional is the identity coefficient, so $\tau (ab)=\tau (ba)$. This extends to the whole group von Neumann algebra by the bounded strong-star approximations of Question 6, taking limits of the vector functional and products. The conjugation is explicitlyConsequentlyThus the commutant in (b) is exactly the Von Neumann algebra of right regular translations, with the inverse in $\rho (g)$ fixing the representation convention.

A Von Neumann algebra whose center consists exactly of scalar multiples of the identity. A group von Neumann algebra is a factor exactly when every nonidentity conjugacy class is infinite. A central operator has a square-summable coefficient vector constant on those classes; an infinite class must therefore have coefficient zero. Conversely, the sum of left translations over a finite nonidentity class is a nonscalar central operator.