Original text: CC0 1.0. Prerequisite proofs and component terms.

Comparing weights through their finite-energy vectors

OA-MOD-DW-01 — Conventions and the exact comparison problem

Inner products are linear in the first variable. Fix a weight $\varphi:M_+\to[0,\infty]$, and write

The construction in OA-MOD-WG supplies a positive linear extension of $\varphi$ to $\mathfrak m_\varphi$, a Hilbert space $H_\varphi$, a dense-range map $\Lambda_\varphi$, and a unital representation $\pi_\varphi$ with

In particular $\mathfrak m_\varphi^+=\mathfrak m_\varphi\cap M_+=\{a\in M_+:\varphi(a)<\infty\}$, and every element of $\mathfrak m_\varphi$ is a complex linear combination of this cone. These are algebraic domains, not norm closures.

Let $\mathcal D_\varphi$ be the cone of positive linear forms $\ell:\mathfrak m_\varphi\to\mathbb C$ such that, for some finite $c\geq0$,

Positivity refers to the inherited cone above. It does not assert that $\ell$ is bounded for the operator norm, or that it has already been extended to a normal weight on all of $M_+$. The order on these forms is positivity of their difference on this cone.

OA-MOD-DW-02 — A factorization inside the algebra

Lemma. If $x,y\in M$ and $y^*y\leq x^*x$, there is a unique contraction $v\in M$ which vanishes on $\overline{xH}^{\perp}$ and satisfies $y=vx$, in any faithful concrete realization $M\subseteq B(H)$.

Proof. On $xH$ define $x\xi\mapsto y\xi$. The inequality says both that this is well defined and that its norm is at most one. Extend it continuously to $\overline{xH}$, and set it equal to zero on the orthogonal complement. This proves existence and uniqueness in $B(H)$. Every unitary $u\in M'$ preserves $\overline{xH}$; because $x,y$ commute with $u$, both $v$ and $uvu^*$ have the prescribed properties. Thus $v$ commutes with every unitary of $M'$. Every element of a unital C*-algebra is a linear combination of unitaries: a self-adjoint contraction $b$ is the real part of $b+i(1-b^2)^{1/2}$. Consequently $v\in M''=M$. This uses bounded continuous functional calculus and the bicommutant theorem, recorded foundation contracts. $\square$

For $a,b\geq0$ put $d=a+b$ and $p=s(d)$, the projection onto $\overline{dH}$. Applying the lemma to $a^{1/2},d^{1/2}$ and $b^{1/2},d^{1/2}$ gives contractions $v,w\in M$ with

To verify the last identity, its quadratic form agrees with that of $p$ on $d^{1/2}H$, by $a+b=d$; continuity extends the equality to $pH$. Both sides vanish on $(1-p)H$. This argument needs neither an inverse of $d$ nor a positive lower bound for it.

OA-MOD-DW-03 — The commutant correspondence

Theorem. There is an additive, positively homogeneous order isomorphism

uniquely characterized by

Moreover $T\leq cI$ if and only if $\ell_T\leq c\varphi$ on $\mathfrak m_\varphi^+$. Thus the least admissible comparison constant is $\|T\|$. The assertion includes the zero Hilbert space, where both cones consist only of zero.

Proof, from forms to operators. For $\ell\in\mathcal D_\varphi$, positivity applied to $(x+zy)^*(x+zy)$, for every $z\in\mathbb C$, gives

For completeness, when $\ell(y^*y)>0$, minimize this quadratic polynomial in $z$; when that diagonal term is zero, varying the magnitude and phase of $z$ forces the cross term to vanish. The inequality shows that the expression is independent of null representatives and defines a bounded sesquilinear form on $\Lambda_\varphi(\mathfrak n_\varphi)$. It extends uniquely to $H_\varphi$. The Hilbert-space representation theorem for bounded sesquilinear forms gives a unique positive operator $T$ with $0\leq T\leq cI$.

For $a\in M$ and $x,y\in\mathfrak n_\varphi$, associativity gives

The test vectors form a dense subspace, so $T\pi_\varphi(a)=\pi_\varphi(a)T$. This proves membership in the commutant without any appeal to a modular group.

Proof, from operators to forms. Fix $T\in\pi_\varphi(M)'_+$. For finite-weight positive $a$, define

We must prove additivity; merely writing a formula on products would not prove independence of their decompositions. Let $a,b\in\mathfrak m_\varphi^+$, and use $d,p,v,w$ from OA-MOD-DW-02. Here $d^{1/2}\in\mathfrak n_\varphi$. Put $\zeta=\Lambda_\varphi(d^{1/2})$. Then $\pi_\varphi(p)\zeta=\zeta$ and

Homogeneity follows from the square root of a scalar. Thus $f_T$ extends to a real linear form on $\mathfrak m_{\varphi,\mathrm{sa}}$: assign $f_T(a)-f_T(b)$ to $a-b$. If $a-b=a'-b'$, the equality $a+b'=a'+b$ and additivity prove independence. Complexification gives a positive linear form $\ell_T$ on $\mathfrak m_\varphi$. Also

If $x\in\mathfrak n_\varphi$, take its polar decomposition $x=u|x|$ in $M$. Then $\Lambda_\varphi(x)=\pi_\varphi(u)\Lambda_\varphi(|x|)$, and $u^*u$ fixes $|x|$. Commutation with $T$ yields

Polarization now proves the formula for $y^*x$. Uniqueness holds because these products span $\mathfrak m_\varphi$, and their GNS vectors are dense. The two constructions are inverse. Addition and scalar multiplication follow from the defining pairings. Finally positivity of $\ell_{T_2}-\ell_{T_1}$ is equivalent to nonnegativity of the quadratic form of $T_2-T_1$ on a dense subspace, hence on all of $H_\varphi$. Apply this to $cI-T$ to obtain the bound and the optimal constant. $\square$

OA-MOD-DW-04 — Comparing two weights and the precise target space

Theorem. Let $\psi$ be another weight on $M$, with $\psi\leq c\varphi$ on $M_+$ for a finite $c>0$. There is a unique bounded map

with norm at most $\sqrt c$. It intertwines the representations, and

The range closure is exactly $K=\overline{\Lambda_\psi(\mathfrak n_\varphi)}\subseteq H_\psi$. If $C=UT^{1/2}$ is its polar decomposition, then $U$ is a unitary from $s(T)H_\varphi$ onto $K$, and is zero on $\ker T$. These two subspaces reduce the corresponding representations, and $U$ intertwines their restrictions.

Proof. Domination gives $\mathfrak n_\varphi\subseteq\mathfrak n_\psi$ and $\|\Lambda_\psi(x)\|^2\leq c\|\Lambda_\varphi(x)\|^2$. Consequently the displayed assignment is well defined even for nonfaithful weights and extends from a dense domain. Its range closure is $K$ by construction. On the dense GNS domain,

Boundedness extends this identity to all vectors. Taking adjoints and using the identity for $a^*$ shows that $C^*$ intertwines in the reverse direction, so $C^*C$ commutes with $\pi_\varphi(M)$. The pairing formula is immediate. It agrees with OA-MOD-DW-03 applied to $\psi|_{\mathfrak m_\varphi}$.

The closed subspaces $\ker C$ and $K$ are invariant under the representations and their adjoints. Therefore their orthogonal projections commute with these representations. Since $T^{1/2}$ also commutes with $\pi_\varphi(M)$, the equality $U\pi_\varphi(a)T^{1/2}\xi=\pi_\psi(a)UT^{1/2}\xi$ holds first on $\operatorname{ran}T^{1/2}$, and then by continuity on its closure $s(T)H_\varphi$. Both sides vanish on its orthogonal complement. The initial and final subspaces of a polar decomposition give the claimed unitary. $\square$

Do not replace $K$ by $H_\psi$ in this level of generality. If $\varphi$ is zero at zero and infinite at every nonzero positive element, then $\mathfrak n_\varphi=\{0\}$, while every weight $\psi$ satisfies $\psi\leq\varphi$. The comparison map is zero even when $H_\psi\neq0$. This also shows why restriction to a finite domain need not determine the whole weight.

Editable source · Proof dependencies and component terms