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For a finite-rank operator $T$, define $|T|=(T^{\ast }T{)}^{1/2}$ and its trace norm bywhere $s_j>0$ are its nonzero singular values, counted with multiplicity. The finite-rank spectral theorem gives orthonormal systems $(u_j),(v_j)$ with $T={\sum }_j{s}_j{R}_{u_j,v_j}$, where $R_{x,y}z=⟨z,y⟩x$.

For every bounded operator $B$,Choose the contraction $B$ taking $u_j$ to $v_j$ and vanishing on their orthogonal complement. It attains equality, soThis formula proves the triangle inequality and absolute homogeneity. Also $\Vert T\Vert \le \Vert T{\Vert }_1$, so vanishing of the trace norm forces $T=0$. Thus it is a norm on the vector space of finite-rank operators.

The map $B\mapsto F_B$, where $F_B(T)=\mathrm{Tr}(BT)$, is a bounded linear map into the dual of this normed space. Its norm is at most $\Vert B\Vert$. Rank-one tests give the reverse inequality: $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, so taking unit vectors $x,y$ gives $\Vert F_B\Vert =\Vert B\Vert$.

Conversely, let $F$ be a bounded linear functional for the trace norm. The sesquilinear form $\beta (x,y)=F(R_{x,y})$ satisfiesThe Riesz representation theorem gives a unique bounded operator $B$ with $\beta (x,y)=⟨Bx,y⟩$, and $\Vert B\Vert \le \Vert F\Vert$. Since every finite-rank operator is a finite sum of rank-one operators, linearity implies $F(T)=\mathrm{Tr}(BT)$ for every $T$. HenceThe same identification extends to the completion, the trace-class operators. This is trace-class duality; completeness of the original finite-rank space is not required to identify its bounded dual.

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.