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

General weights: finite domains, GNS spaces, and normal representations

OA-MOD-WG-001. Conventions and precise prerequisite contract

Throughout, $M$ is a unital von Neumann algebra, represented faithfully and normally on some Hilbert space $K$. Neither $K$ nor the predual of $M$ is assumed separable. All unqualified inner products are linear in their first variable. A bounded increasing net of positive operators has a supremum in $M_+$; the notation $a_i\uparrow a$ includes the assertion that $a$ is that supremum. Its convergence is strong and ultraweak. A sum over an arbitrary index set of nonnegative numbers means the supremum of its finite partial sums.

The bounded operator-algebra prerequisites used here are:

1. Positivity and continuous functional calculus in a unital C*-algebra, including $x^*x\leq\|x\|^2 1$, positive square roots, and order reversal under inversion of strictly positive invertible operators.
2. The spectral theorem for a bounded positive operator $b$, including its support projection $s(b)$, and monotone convergence for bounded increasing nets of positive operators.
3. On a norm-bounded set of operators, strong convergence implies ultraweak convergence. Multiplication by a fixed bounded operator is separately ultraweakly continuous.
4. A positive map between von Neumann algebras is normal precisely when it preserves the suprema of bounded increasing positive nets. Equivalently, for such a map, normality is ultraweak continuity. This equivalence is a bounded-map theorem; it is not being asserted here for extended-valued weights.

These prerequisites are stated exactly and are used here without proof. None is a theorem about closability of an unbounded operator or lower semicontinuity of a weight.

OA-MOD-WG-002. The finite part of an extended-valued weight

A weight is a map $\varphi:M_+\to[0,\infty]$ satisfying $$\varphi(0)=0,\qquad \varphi(a+b)=\varphi(a)+\varphi(b),\qquad \varphi(t a)=t\varphi(a)\quad(t\geq0).$$ We use $0\cdot\infty=0$. It follows immediately that $0\leq a\leq b$ implies $\varphi(a)\leq\varphi(b)$, since $b=a+(b-a)$.

Define three subsets with different purposes: $$F_\varphi=\{a\in M_+:\varphi(a)<\infty\},\qquad \mathfrak n_\varphi=\{x\in M:\varphi(x^*x)<\infty\},$$ $$\mathfrak m_\varphi =\operatorname{span}_{\mathbb C} \{y^*x:x,y\in\mathfrak n_\varphi\}.$$ The first is a positive cone, the second will be the domain of the GNS map, and the third will be the domain of a finite-valued linear extension. In particular, the original weight is not a complex-linear function on all of $M$.

We call $\varphi$ faithful if $a\in M_+$ and $\varphi(a)=0$ imply $a=0$. We call it normal if $$\varphi(a)=\sup_i\varphi(a_i) \quad\text{whenever }0\leq a_i\uparrow a\text{ is bounded in }M.$$ We call it semifinite if $\mathfrak m_\varphi$ is ultraweakly dense in $M$. This definition is essential: for general weights it is not replaced by a requirement that every nonzero positive element dominate a nonzero element of finite weight. Item OA-MOD-WG-013 gives a counterexample to that replacement.

OA-MOD-WG-003. Domain algebra and its positive cone

Proposition. For every weight, $F_\varphi$ is an additive hereditary cone, $\mathfrak n_\varphi$ is a left ideal, and $\mathfrak m_\varphi$ is a possibly nonunital *-subalgebra. More precisely, $$\mathfrak m_\varphi=\operatorname{span}_{\mathbb C}F_\varphi, \qquad \mathfrak m_\varphi\cap M_+=F_\varphi, \qquad \mathfrak m_\varphi\subseteq \mathfrak n_\varphi\cap\mathfrak n_\varphi^*.$$ Every element of $\mathfrak m_\varphi$ has the form $(a-b)+i(c-d)$, where $a,b,c,d\in F_\varphi$.

Proof. Additivity and homogeneity preserve finite values, and monotonicity makes $F_\varphi$ hereditary. For $x,y\in\mathfrak n_\varphi$, $$(x+y)^*(x+y)\leq2x^*x+2y^*y;$$ this follows by expanding $(x-y)^*(x-y)\geq0$. Consequently $x+y\in\mathfrak n_\varphi$. Scalar multiplication is immediate. For $a\in M$, $$(ax)^*(ax)=x^*a^*ax\leq\|a\|^2x^*x,$$ so $ax\in\mathfrak n_\varphi$. This proves the left ideal assertion.

Taking adjoints preserves the span defining $\mathfrak m_\varphi$. Products of its spanning elements remain in that span because $$(y^*x)(v^*u)=y^*(xv^*u),$$ and $xv^*u\in\mathfrak n_\varphi$. Each $y^*x$ itself belongs to $\mathfrak n_\varphi$, by the left ideal property, and so does its adjoint. This proves the subalgebra and inclusion statements.

The polarization identity $$4y^*x=\sum_{k=0}^{3}i^k(x+i^k y)^*(x+i^k y)$$ puts every spanning element in $\operatorname{span}_{\mathbb C}F_\varphi$. Conversely, for $a\in F_\varphi$, the square root $a^{1/2}$ lies in $\mathfrak n_\varphi$, and $a=(a^{1/2})^*a^{1/2}$. The two spans therefore coincide.

Grouping the positive and negative real and imaginary coefficients of a finite linear combination of members of $F_\varphi$ gives the four-term decomposition. If an element $h$ of that span is self-adjoint, averaging a decomposition with its adjoint removes the imaginary part and expresses $h=a-b$, with $a,b\in F_\varphi$. If also $h\geq0$, then $0\leq h\leq a$, so heredity gives $h\in F_\varphi$. The reverse inclusion was already established. ∎

Here “hereditary” for the subalgebra refers to its positive cone: if $0\leq b\leq a\in(\mathfrak m_\varphi)_+$, then $b\in\mathfrak m_\varphi$. No norm-closedness is asserted.

Corollary for arbitrary hereditary cones. Let $P\subseteq M_+$ be a nonempty cone closed under addition and multiplication by nonnegative scalars, and suppose $0\leq a\leq b\in P$ implies $a\in P$. Set $$\mathfrak n_P=\{x:x^*x\in P\},\qquad \mathfrak m_P=\operatorname{span}_{\mathbb C} \{y^*x:x,y\in\mathfrak n_P\}.$$ Then $\mathfrak n_P$ is a left ideal, $\mathfrak m_P$ is a hereditary *-subalgebra, $$\mathfrak m_P=\operatorname{span}_{\mathbb C}P,\qquad \mathfrak m_P\cap M_+=P,\qquad \mathfrak m_P\subseteq\mathfrak n_P\cap\mathfrak n_P^*,$$ and every element of $\mathfrak m_P$ is a complex linear combination of four members of $P$.

Proof. Define $\varphi_P(a)=0$ for $a\in P$ and $\varphi_P(a)=+\infty$ otherwise. Nonemptiness and the cone property give $0\in P$. For positive $a,b$, heredity gives $$a+b\in P\quad\Longleftrightarrow\quad a\in P\text{ and }b\in P.$$ It follows that $\varphi_P(a+b)=\varphi_P(a)+\varphi_P(b)$, with the extended-value conventions of WG-002. For $t>0$, $ta\in P$ is equivalent to $a\in P$, by applying the cone property also to $t^{-1}$. This proves homogeneity for $t>0$; for $t=0$ it follows from $0\cdot\infty=0$. Thus $\varphi_P$ is a weight with $F_{\varphi_P}=P$. The proposition applied to this weight gives all the stated conclusions. Neither normality nor semifiniteness of this auxiliary weight is claimed. ∎

OA-MOD-WG-004. Linear extension without infinite subtraction

Proposition. There is a unique positive complex-linear functional $$\widetilde\varphi:\mathfrak m_\varphi\longrightarrow\mathbb C$$ whose restriction to $F_\varphi$ equals $\varphi$. It satisfies $\widetilde\varphi(z^*)=\overline{\widetilde\varphi(z)}$.

Proof. For a self-adjoint element $h=a-b$, with $a,b\in F_\varphi$, set $\widetilde\varphi(h)=\varphi(a)-\varphi(b)$. If also $h=c-d$, then $a+d=c+b$, and additivity gives $$\varphi(a)+\varphi(d)=\varphi(c)+\varphi(b).$$ Every number here is finite. Subtracting proves independence of the decomposition. Addition of decompositions proves additivity on the self-adjoint part; positive scalar multiplication follows from homogeneity, and negative scalar multiplication follows by swapping the two terms. Thus this is a real-linear functional.

Every $z\in\mathfrak m_\varphi$ has the unique self-adjoint decomposition $$z=h+ik,\qquad h=(z+z^*)/2,\quad k=(z-z^*)/(2i).$$ Define $\widetilde\varphi(z)=\widetilde\varphi(h)+i\widetilde\varphi(k)$. Real linearity and the transformation $(h,k)\mapsto(-k,h)$ under multiplication by $i$ prove complex linearity. Positivity follows from OA-MOD-WG-003, because the positive elements of the domain are exactly $F_\varphi$. The formula for adjoints and uniqueness follow from the same decomposition. ∎

We retain the tilde when a complex argument is present, making the domain visible. A formula such as $\widetilde\varphi(y^*a x)$ is legitimate for $x,y\in\mathfrak n_\varphi$, $a\in M$, because $ax\in\mathfrak n_\varphi$. It is not legitimate for arbitrary $x,y\in M$ merely because the product exists.

OA-MOD-WG-005. Cauchy-Schwarz and the null ideal

Lemma. On $\mathfrak n_\varphi$, the formula $$B_\varphi(x,y)=\widetilde\varphi(y^*x)$$ defines a positive semidefinite sesquilinear form, linear in $x$, and $$|B_\varphi(x,y)|^2 \leq\varphi(x^*x)\varphi(y^*y).$$ Its null space is the left ideal $$N_\varphi=\{x\in M:\varphi(x^*x)=0\}\subseteq\mathfrak n_\varphi.$$

Proof. Sesquilinearity and conjugate symmetry follow from OA-MOD-WG-004; positivity follows from the definition. Put $A=B_\varphi(x,x)$, $C=B_\varphi(y,y)$, and $b=B_\varphi(x,y)$. For every $t\in\mathbb C$, $$0\leq B_\varphi(x+ty,x+ty) =A+t\overline b+\overline t b+|t|^2C.$$ If $C>0$, choose $t=-b/C$ to obtain $|b|^2\leq AC$. If $C=0$ and $b\neq0$, choosing $t=-r b$, with $r>0$, would make the right side $A-2r|b|^2$, which is eventually negative. Thus $b=0$ when $C=0$, proving the inequality in every case.

The inequality shows that a null vector is orthogonal to every element of $\mathfrak n_\varphi$. Hence sums and scalar multiples of null vectors are null. The estimate from OA-MOD-WG-003 gives $$\varphi((ax)^*(ax))\leq\|a\|^2\varphi(x^*x)=0 \quad(x\in N_\varphi),$$ so $N_\varphi$ is a left ideal. ∎

OA-MOD-WG-006. The GNS construction for an arbitrary weight

Let $H_\varphi$ be the Hilbert completion of $\mathfrak n_\varphi/N_\varphi$. Write $$\Lambda_\varphi:\mathfrak n_\varphi\to H_\varphi, \qquad x\mapsto[x].$$ The preceding lemma proves that the quotient inner product $$\langle\Lambda_\varphi(x),\Lambda_\varphi(y)\rangle =\widetilde\varphi(y^*x)$$ is well-defined and positive definite. By construction, the range of $\Lambda_\varphi$ is dense. For $a\in M$, define initially $$\pi_\varphi(a)\Lambda_\varphi(x)=\Lambda_\varphi(ax).$$ This is well-defined because both domain ideals are left ideals, and $$\|\Lambda_\varphi(ax)\|^2 \leq\|a\|^2\|\Lambda_\varphi(x)\|^2.$$ It therefore extends uniquely to a bounded operator on $H_\varphi$, with norm at most $\|a\|$.

Theorem. This construction gives a unital *-representation $\pi_\varphi:M\to B(H_\varphi)$, for every weight. Neither normality, faithfulness, nor semifiniteness is needed for this assertion. The triple $(H_\varphi,\pi_\varphi,\Lambda_\varphi)$ is the GNS semicyclic triple.

Proof. Linearity and multiplication hold on the dense range of $\Lambda_\varphi$, because the left action of $M$ has these properties. Boundedness extends these identities to all of $H_\varphi$. For $x,y\in\mathfrak n_\varphi$, $$\begin{aligned} \langle\pi_\varphi(a)\Lambda_\varphi(x),\Lambda_\varphi(y)\rangle &=\widetilde\varphi(y^*a x)\\ &=\langle\Lambda_\varphi(x),\pi_\varphi(a^*)\Lambda_\varphi(y)\rangle. \end{aligned}$$ Density yields $\pi_\varphi(a)^*=\pi_\varphi(a^*)$. Finally, $\pi_\varphi(1)\Lambda_\varphi(x)=\Lambda_\varphi(x)$, so $\pi_\varphi(1)=I$. In particular the representation is nondegenerate: $\overline{\pi_\varphi(M)H_\varphi}=H_\varphi$. When $H_\varphi=\{0\}$, its identity operator is the zero operator and these statements retain their usual meaning. ∎

Uniqueness. Suppose $(H',\pi',\Lambda')$ has a linear map $\Lambda':\mathfrak n_\varphi\to H'$ with dense range, the same inner-product formula, and $\pi'(a)\Lambda'(x)=\Lambda'(ax)$, with $\pi'$ a bounded-operator representation. Then $U\Lambda_\varphi(x)=\Lambda'(x)$ defines an isometry on a dense subspace: its well-definedness and norm preservation follow from the shared quadratic form. Its range is dense, so it extends to a unitary. The action identity gives $U\pi_\varphi(a)=\pi'(a)U$ first on GNS vectors and then everywhere. Density also proves uniqueness of $U$.

Here “semicyclic” means this representation together with its dense-range module map. Some treatments impose additional closedness conditions when defining an abstract semicyclic object. No such closedness is included silently in the present terminology.

OA-MOD-WG-007. Normality is a net argument

Theorem. If $\varphi$ is normal, then $\pi_\varphi$ is normal. Semifiniteness is not required.

Proof. Let $0\leq a_i\uparrow a$ in $M$. Strong convergence gives $x^*a_i x\uparrow x^*a x$ for each fixed $x\in\mathfrak n_\varphi$. All their weights are finite, since $$0\leq x^*a_i x\leq x^*a x\leq\|a\|x^*x.$$ By normality, $$\varphi(x^*a_i x)\uparrow\varphi(x^*a x).$$ Set $D_i=\pi_\varphi(a-a_i)$. This is positive, $0\leq D_i\leq\|a\|I$, and additivity with finite values gives $$\langle D_i\Lambda_\varphi(x),\Lambda_\varphi(x)\rangle =\varphi(x^*a x)-\varphi(x^*a_i x)\longrightarrow0.$$ Functional calculus gives $D_i^2\leq\|a\|D_i$. Thus $$\|D_i\Lambda_\varphi(x)\|^2 \leq\|a\|\langle D_i\Lambda_\varphi(x),\Lambda_\varphi(x)\rangle \longrightarrow0.$$ For an arbitrary $\xi\in H_\varphi$, approximate $\xi$ by a GNS vector $\eta$. The bound $$\|D_i\xi\|\leq\|a\|\|\xi-\eta\|+\|D_i\eta\|$$ proves $D_i\xi\to0$. Therefore $\pi_\varphi(a_i)\uparrow\pi_\varphi(a)$ strongly. The bounded-map normality criterion in OA-MOD-WG-001 proves normality. ∎

No sequence was extracted from the given net. The proof does not require a faithful normal state, a countable approximate unit, or a dominated-convergence theorem for the unbounded function $\varphi$.

OA-MOD-WG-008. Finite positive cutoffs characterize semifiniteness

Theorem. For an arbitrary weight, semifiniteness is equivalent to the existence of an increasing net $(e_i)$ of positive contractions in $F_\varphi$ with $e_i\uparrow1$. No normality assumption is needed for this equivalence.

Proof. Suppose $\mathfrak m_\varphi$ is ultraweakly dense. Order $F_\varphi$ by the operator order; it is directed because $a+b$ is a common upper bound. For $a\in F_\varphi$, put $$e_a=a(1+a)^{-1}=1-(1+a)^{-1}.$$ The inverse-order inequality shows that $a\leq b$ implies $e_a\leq e_b$. Functional calculus gives $0\leq e_a\leq1$ and $e_a\leq a$, so $e_a\in F_\varphi$. The increasing net has a strong supremum $e\leq1$.

Fix $b\in F_\varphi$. Every $t b$, $t>0$, is also an index. Spectral calculus gives $$e_{tb}=tb(1+tb)^{-1}\uparrow s(b) \quad(t\to\infty),$$ because the scalar functions increase to the indicator of $(0,\infty)$. Hence $e\geq s(b)$. For a positive contraction dominating a projection $p$, one has $ep=p$: indeed $p(1-e)p=0$, so $(1-e)^{1/2}p=0$. Applying this with $p=s(b)$ gives $eb=b$. Linearity gives $ez=z$ for every $z\in\mathfrak m_\varphi$. Separate ultraweak continuity of multiplication and density give this identity for every $z\in M$. Setting $z=1$ proves $e=1$.

Conversely, suppose finite positive contractions $e_i\uparrow1$ are given. Since $e_i^2\leq e_i$, each $e_i\in\mathfrak n_\varphi$. For $a\in M$, the left ideal property gives $a e_i\in\mathfrak n_\varphi$, so $$e_i a e_i=e_i^*(a e_i)\in\mathfrak m_\varphi.$$ These operators converge strongly to $a$: for $\xi\in K$, $$\|(e_i a e_i-a)\xi\| \leq\|a\|\|(e_i-1)\xi\|+\|(e_i-1)a\xi\|.$$ Their norms are bounded by $\|a\|$, so the convergence is also ultraweak. Thus $\mathfrak m_\varphi$ is ultraweakly dense. ∎

These cutoffs are generally positive contractions, not projections. The set of all finite-weight projections must not be assumed directed under inclusion.

OA-MOD-WG-009. Density of the two-sided finite domain in the GNS space

Theorem. If $\varphi$ is normal and semifinite, then $$\overline{\Lambda_\varphi(\mathfrak m_\varphi)} =\overline{\Lambda_\varphi(\mathfrak n_\varphi\cap\mathfrak n_\varphi^*)} =H_\varphi.$$ More explicitly, for the cutoffs in OA-MOD-WG-008 and every $x\in\mathfrak n_\varphi$, $$e_a x\in\mathfrak m_\varphi,\qquad \Lambda_\varphi(e_a x)\longrightarrow\Lambda_\varphi(x) \quad\text{in Hilbert norm}.$$

Proof. Both $e_a$ and $x$ belong to $\mathfrak n_\varphi$, and $e_a x=e_a^*x$, proving membership in $\mathfrak m_\varphi$. Since $0\leq1-e_a\leq1$, $$\begin{aligned} \|\Lambda_\varphi(x-e_a x)\|^2 &=\varphi\bigl(x^*(1-e_a)^2x\bigr)\\ &\leq\varphi\bigl(x^*(1-e_a)x\bigr)\\ &=\varphi(x^*x)-\varphi(x^*e_a x). \end{aligned}$$ The last equality subtracts finite numbers. The operators $x^*e_a x$ increase strongly to $x^*x$, so normality makes the final difference tend to zero. Density of $\Lambda_\varphi(\mathfrak n_\varphi)$ now proves the first assertion, and the inclusion in OA-MOD-WG-003 proves the second. ∎

This proves a dense domain for a prospective involution when the weight is faithful. It does not prove that the involution is closable.

OA-MOD-WG-010. Faithfulness and the finite-weight specialization

Proposition. If $\varphi$ is faithful and semifinite, then $\pi_\varphi$ is faithful. Normality is not needed for this particular conclusion.

Proof. If $\pi_\varphi(a)=0$, then for each finite cutoff $e_i$, $$0=\|\pi_\varphi(a)\Lambda_\varphi(e_i)\|^2 =\varphi((a e_i)^*(a e_i)).$$ Faithfulness gives $a e_i=0$. Since $e_i\to1$ strongly in the faithful concrete realization of $M$, we get $a=0$. ∎

If $\varphi(1)<\infty$, then $\varphi(a)\leq\|a\|\varphi(1)$ for all $a\in M_+$. Thus $\mathfrak n_\varphi=\mathfrak m_\varphi=M$, and $\Omega_\varphi=\Lambda_\varphi(1)$ exists. Its orbit is dense because $\pi_\varphi(a)\Omega_\varphi=\Lambda_\varphi(a)$. If the finite weight is also faithful, $\pi_\varphi$ is faithful and $\Omega_\varphi$ is separating for $\pi_\varphi(M)$, since $$\pi_\varphi(a)\Omega_\varphi=0 \Longrightarrow\varphi(a^*a)=0 \Longrightarrow a=0.$$ For an infinite weight, the symbol $\Lambda_\varphi(1)$ need not be defined. The dense family $\Lambda_\varphi(\mathfrak n_\varphi)$ is the replacement, and a cyclic vector may not exist at all.

Editable source · Proof dependencies and component terms