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Interpret the involution as the adjoint for the Hilbert-space structure on $V$. The hypotheses make $\Omega$ a cyclic vector for an operator algebra for both $A$ and $A^{\prime }$. They also make it a separating vector for each: if $a\Omega =0$, then $a$ vanishes on $A^{\prime }\Omega =V$, and similarly for $A^{\prime }$. Thus the Tomita operatoris well defined, antilinear, invertible and satisfies $S^2=I$. Define the modular operator and modular conjugation byWe calculate them explicitly to prove the commutant assertion.

The finite-dimensional structure of a unital star-algebra gives an orthogonal decompositionThis is the usual matrix-block structure: minimal central projections separate the simple summands, and each simple matrix algebra acts as its defining representation with a multiplicity space. Identify each vector block with an $n_j$-by-$m_j$ matrix $D_j$ with its Hilbert-Schmidt inner product. The two algebras act by left and right multiplication. If $D_j$ has rank $r_j$, the left orbit has dimension $n_j{r}_j$ and the right orbit dimension $m_j{r}_j$. Cyclicity on both sides therefore forces $r_j=m_j=n_j$. Every $D_j$ is square and invertible.

Right multiplication by a suitable unitary is a unitary change of this identification commuting with the left algebra. The polar decomposition of each $D_j$ therefore lets us assume $D_j>0$. In one such block, for an arbitrary matrix $X=aD$, the Tomita operator isIts polar factors areIndeed, $J$ is an antiunitary involution and $J{\Delta }^{1/2}(X)=D^{-1}{X}^{\ast }D$. The fact that ${\Delta }^{1/2}$ is a positive operator is transparent in a basis diagonalizing $D$: it multiplies the matrix unit $E_{kl}$ by the positive ratio $d_k/d_l$. These are therefore the unique polar factors of $S$.

Finally, if $L_a$ denotes left multiplication,As $a$ ranges over the full matrix block, these are exactly its right multiplications. Applying this on every block and transporting back by the unitary identifications provesThis also establishes the required modular definitions intrinsically, independent of the chosen matrix coordinates.

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.