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The absolute value of a linear operator $L$ is the positive square root of an operator $L^{†}L$. Its eigenvalues are the singular values of $L$, its trace is the trace norm of $L$, and it appears in the polar decomposition of a bounded operator $L=U|L|$.

An operator is trace class exactly when it factors as $T=AB$ with $A$ and $B$ Hilbert--Schmidt. The factors can be chosen so that $\Vert T{\Vert }_1=\Vert A{\Vert }_{\mathrm{HS}}\Vert B{\Vert }_{\mathrm{HS}}$ by using the polar decomposition of a bounded operator.

Here a Hilbert space module means a Hilbert space with a unital star-representation of a Von Neumann algebra, rather than a Hilbert C-star module with an algebra-valued inner product. A closed invariant subspace is reducing because the algebra is closed under adjoints. Module equivalence means an intertwining unitary. The polar decomposition of a bounded operator converts an intertwiner into a unitary between its initial and final support subspaces.

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.

The polar decomposition of a bounded operator states that every bounded $T:H\to H$ is uniquelywhere $V$ is a partial isometry with initial space $\overline{\mathrm{Ran}|T|}=(\ker T{)}^{\perp }$ and final space $\overline{\mathrm{Ran}T}$. Its initial and final orthogonal projections are $V^{\ast }V$ and $V{V}^{\ast }$.

Existence of $|T|$ was proved in (a). For every $\xi$,Thus the map $|T|\xi \mapsto T\xi$ is well defined and isometric on $\mathrm{Ran}|T|$. It extends uniquely to an isometry between its closed range and $\overline{\mathrm{Ran}T}$. Extend it by zero on $\ker |T|=\ker T$ to obtain $V$. The equation and the projection assertions follow immediately. Any other normalized partial isometry with the same product must agree on the dense range of $|T|$ and vanish on its orthogonal complement, giving uniqueness. The kernel normalization is essential; without it an action on the kernel could be added.

A useful consequence for a Von Neumann algebra $M$ is that the polar factors of $T\in M$ also belong to $M$. The polynomial construction in (a) puts $|T|$ in the norm-closed algebra $M$. For $\epsilon >0$, the inverse of $|T|+\epsilon I$ belongs to $M$ by a convergent Neumann series, andThese operators are contractions, since $\Vert |T|\xi \Vert \le \Vert (|T|+\epsilon I)\xi \Vert$. On vectors $|T|\xi$ the difference from $V$ has norm at most $\epsilon \Vert \xi \Vert$, and on the kernel it is zero. Density and the uniform bound prove the strong limit. Strong closedness of $M$ then gives $V\in M$. This justifies the use of algebra-valued polar factors in Question 7(c).

Apply part (a) to the compact positive operator $T^{\ast }T$. Its positive square root $B=|T|$ is defined by multiplying each eigenvector of eigenvalue $\lambda$ by $\sqrt{\lambda }$ and vanishing on the kernel. The eigenvalues tend to zero, so $B$ is compact. For every $x$, $\Vert Tx{\Vert }^2=⟨T^{\ast }Tx,x⟩=\Vert Bx{\Vert }^2$.

Consequently $U(Bx)=Tx$ is a well-defined isometry on $\mathrm{ran}B$. Extend it to the closure of that range and set it to zero on its orthogonal complement. This gives a partial isometry with initial space $(\ker T{)}^{\perp }$, final space $\overline{\mathrm{ran}T}$, andFor uniqueness, the positive factor in any such polar decomposition of a bounded operator satisfies $B^2=T^{\ast }T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\mathrm{ran}B$ of its initial space and the kernel convention fixes it elsewhere. This proves uniqueness with the canonical initial-space condition. Without that condition the word “unique” would be false: $T=0$ can be written $U\cdot 0$ with many different unitary $U$.

The absolute value of an operator, trace norm, and operator norm are respectivelyThe positive square root of an operator makes $|L|$ well defined. Its eigenvalues $s_j\ge 0$ are the singular values of $L$, so $\Vert L{\Vert }_1={\sum }_j{s}_j$ and $\Vert L{\Vert }_{\mathrm{op}}={\max }_j{s}_j$.

Use the polar decomposition of a bounded operator $L=U|L|$, with the given unitary operator $U$. In an orthonormal eigenbasis $e_j$ for $|L|$,The operator norm assumption and unitarity imply $|⟨e_j|ZU|e_j⟩|\le \Vert Z{\Vert }_{\mathrm{op}}\le 1$. The triangle inequality therefore yields the required trace estimate:Choosing $Z=U^{†}$ attains equality, which also gives the finite-dimensional trace duality formula $\Vert L{\Vert }_1={\max }_{\Vert Z{\Vert }_{\mathrm{op}}\le 1}|\mathrm{Tr}(ZL)|$. A singular $L$ causes no difficulty: in a finite-dimensional square space the polar factor can be extended to a unitary operator on the complementary subspaces.

The Procrustes distance between covariance operators isFor unitary $R$,The polar decomposition of a bounded operator impliesTherefore

For an orthogonal $R$, unitary invariance of the Hilbert-Schmidt norm givesThe product $L_2^{\ast }{L}_1$ is trace-class. Its polar decomposition of a bounded operator and trace duality givewhere $({\sigma }_k)$ are its singular values. Taking the infimum proves

For factorizations $C_i=L_i{L}_i^{\ast }$ and $C_j=L_j{L}_j^{\ast }$, the Procrustes distance between covariance operators is the following infimum over unitary operators $R$:Unitary invariance of the Hilbert-Schmidt norm givesThe polar decomposition of a bounded operator and trace duality implywhere the last equality expresses the trace norm as the sum of the singular values. Taking the infimum proves

An element $x$ of a unital C-star algebra is positive when $x=y^{\ast }y$ for some $y$, equivalently when $x=x^{\ast }$ and $\sigma (x)\subseteq [0,\infty )$. The continuous functional calculus for the nonnegative function $t\mapsto \sqrt{t}$ defines a positive element $x^{1/2}$ satisfying $(x^{1/2}{)}^2=x$. If a positive $z$ also satisfies $z^2=x$, functional calculus for $z$ gives $z=\sqrt{z^2}=\sqrt{x}$, proving uniqueness. For a positive operator $T=S^{\ast }S$ on ${\ell }^2$,

For arbitrary $T$, put $|T|=(T^{\ast }T{)}^{1/2}$. Thenso $\ker |T|=\ker T$. Defineon $\mathrm{im}|T|$. The kernel identity makes this well-defined, and the norm identity makes it an isometry. Extend it continuously toand set it equal to zero on $\ker T$. The resulting $U$ is a partial isometry, has $\ker U=\ker T$, and satisfies the polar decomposition of a bounded operator $T=U|T|$.

Finally $U^{\ast }U$ is the orthogonal projection onto $(\ker T{)}^{\perp }$, which contains the range of $|T|$. Therefore