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Nonscalar commutant contractions and convex decompositions

Statement

Assume AC. Let $G$ be a topological group, let $φ \in P_{1} (G)$ , and let $(π_{φ}, H_{φ}, ξ_{φ})$ be its cyclic GNS triple, with $∥ ξ_{φ} ∥ = 1$ . A positive contraction means a bounded self-adjoint operator $T \in π_{φ} (G)^{'}$ for which $T$ and $I - T$ are positive. Every nonscalar positive contraction $T$ yields $s \in (0, 1)$ and distinct $φ_{1}, φ_{2} \in P_{1} (G)$ with $φ = s φ_{1} + (1 - s) φ_{2}$ . Conversely, for every such decomposition there is a nonscalar positive contraction $T \in π_{φ} (G)^{'}$ such that $⟨ π_{φ} (g) T ξ_{φ}, ξ_{φ} ⟩ = s φ_{1} (g)$ for all $g \in G$ .

Facts & Assumptions

Given: AC; a topological group $G$ ; a normalized continuous positive-type function $φ$ ; its canonical GNS triple; and the first-variable-linear complex Hilbert inner product. “Scalar operator” means $λ I$ for some $λ \in C$ .

[F1]

$P (G)$ is the cone of continuous functions of positive type and $P_{1} (G) = {ψ \in P (G) : ψ (e) = 1}$ ; positive real scalar multiples preserve positive type (Continuous positive-type functions and normalization).

[F2]

Under AC the GNS triple is cyclic, has diagonal coefficient $φ$ , and satisfies $∥ ξ_{φ} ∥^{2} = φ (e)$ (GNS construction for a continuous positive-type function).

[F3]

Under AC, each $0 \leq ψ \leq φ$ corresponds to a unique bounded self-adjoint operator in $π_{φ} (G)^{'}$ with both $T$ and $I - T$ positive and coefficient $ψ (g) = ⟨ π_{φ} (g) T ξ_{φ}, ξ_{φ} ⟩$ ; conversely every such operator gives a continuous positive-type function dominated by $φ$ (Dominated positive type and positive commutant contractions).

[F4]

Cyclicity means the complex-linear span of ${π_{φ} (g) ξ_{φ} : g \in G}$ is dense in $H_{φ}$ (Cyclic vector and cyclic unitary representation).

[F5]

For a bounded operator $A$ , self-adjoint means $A^{*} = A$ , and positive means $⟨ A x, x ⟩$ is real and nonnegative for every $x$ (Self-adjoint, positive, unitary and normal operators).

[F6]

The Hilbert adjoint satisfies $⟨ A x, y ⟩ = ⟨ x, A^{*} y ⟩$ (The Hilbert-space adjoint of a bounded operator).

[F7]

The inner product is linear in its first variable, conjugate-linear in its second, conjugate-symmetric, and positive definite; its induced norm is the nonnegative square root of $⟨ x, x ⟩$ (Real and complex inner-product spaces and their induced length).

[F8]

A bounded linear operator has a bound $C \geq 0$ with $∥ A x ∥ \leq C ∥ x ∥$ , and $B (H)$ consists of bounded linear operators (A bounded linear operator between normed spaces, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

[F9]

A unitary representation is a group homomorphism into the unitary operators, and its commutant is $π_{φ} (G)^{'} = {A \in B (H_{φ}) : A π_{φ} (g) = π_{φ} (g) A}$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners).

[F10]

AC implies DC and Countable Choice; the adjoint and positive-operator definitions assume Countable Choice (The Axiom of Choice, AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Hilbert-space adjoint of a bounded operator, Self-adjoint, positive, unitary and normal operators).

Proof

Bekka–de la Harpe–Valette's Proposition C.5.1 shows that if the GNS representation is irreducible, a positive-type summand is a scalar multiple of the original function; its proof constructs an intertwiner and applies Schur's lemma. Neeb's Proposition 5.1.11 and Theorem 5.1.12 give the analogous dominated-operator and extremal-ray correspondence for invariant reproducing kernels. The argument below proves the precise positive-contraction and convex-decomposition correspondence for every normalized cyclic GNS triple, using the local dominated-operator theorem.

Proof technique: direct.

1.1F5F6F7algebra

Let $A$ be a bounded self-adjoint positive operator and set $q_{A} (x, y) = ⟨ A x, y ⟩$ . It is sesquilinear, and self-adjointness with the adjoint identity gives $q_{A} (x, y) = ⟨ x, A y ⟩ = \overline{⟨ A y, x ⟩} = \overline{q_{A} (y, x)}$ ; positivity gives $q_{A} (x, x) \geq 0$ . If $q_{A} (x, x) = 0$ , fix any $y$ , put $b = q_{A} (x, y)$ and $c = q_{A} (y, y) \geq 0$ . For each real $r > 0$ , expansion gives $0 \leq q_{A} (x - r b y, x - r b y) = - r \overline{b} b - r b \overline{b} + r^{2} ∣ b ∣^{2} c = - 2 r ∣ b ∣^{2} + r^{2} ∣ b ∣^{2} c$ . Taking $r = (c + 1)^{- 1}$ makes the displayed value $- (c + 2) ∣ b ∣^{2} / (c + 1)^{2}$ , which is negative unless $b = 0$ ; positivity therefore gives $b = 0$ . This holds for every $y$ ; setting $y = A x$ gives $∥ A x ∥^{2} = q_{A} (x, A x) = 0$ , so $A x = 0$ . Thus zero quadratic value forces annihilation, including for a degenerate form.

1.2F2F7

By [F2], the GNS vector satisfies $∥ ξ_{φ} ∥^{2} = φ (e) = 1$ . Nonnegativity of the induced norm [F7] therefore gives $∥ ξ_{φ} ∥ = 1$ .

1.3F1F3

Conversely, suppose $φ = s φ_{1} + (1 - s) φ_{2}$ with $0 < s < 1$ and distinct $φ_{1}, φ_{2} \in P_{1} (G)$ . Positive scaling and [F1] give $s φ_{1} \in P (G)$ and $φ - s φ_{1} = (1 - s) φ_{2} \in P (G)$ ; thus $0 \leq s φ_{1} \leq φ$ . By [F3] there is a unique positive contraction $S \in π_{φ} (G)^{'}$ whose coefficient is $s φ_{1}$ .

2.1F3F5F7step 1.2algebra

Let $T$ be a positive contraction in the commutant and put $t = ⟨ T ξ_{φ}, ξ_{φ} ⟩$ . Positivity of $T$ and $I - T$ and [F7] give $t \in R$ and $1 - t = ⟨ (I - T) ξ_{φ}, ξ_{φ} ⟩ \geq 0$ . Hence $0 \leq t \leq 1$ .

2.2F1F2F3F7F9step 1.3algebra

If this $S$ were scalar, say $S = λ I$ , then evaluating its coefficient at $e$ and using [F1], [F2] gives $s = s φ_{1} (e) = ⟨ S ξ_{φ}, ξ_{φ} ⟩ = λ$ . For every $g$ , the coefficient would then be $λ ⟨ π_{φ} (g) ξ_{φ}, ξ_{φ} ⟩ = s φ (g)$ . Since the same coefficient is $s φ_{1} (g)$ and $s > 0$ , we get $φ_{1} = φ$ , and the decomposition with $s < 1$ then forces $φ_{2} = φ$ , a contradiction. Therefore $S$ is nonscalar.

3.1F3F4F8F9step 1.1step 2.1algebra

If $t = 0$ , step 1.1 applied to $T$ gives $T ξ_{φ} = 0$ ; if $t = 1$ , it applied to $I - T$ gives $(I - T) ξ_{φ} = 0$ . In either case let $A = T$ or $A = I - T$ , respectively. Commutation gives $A π_{φ} (g) ξ_{φ} = π_{φ} (g) A ξ_{φ} = 0$ for every $g$ , so $A$ vanishes on the cyclic orbit span. If $C$ is a bound for $A$ , density [F4] implies $A = 0$ : for any $v \in H_{φ}$ and $ε > 0$ choose one orbit-span vector $w$ with $∥ v - w ∥ < ε$ , giving $∥ A v ∥ \leq C ε$ . Thus $t = 0$ forces $T = 0$ , and $t = 1$ forces $T = I$ . A nonscalar $T$ therefore has $0 < t < 1$ .

4.1F1F3F7step 1.2step 2.1step 3.1algebra

Let $ψ_{T} (g) = ⟨ π_{φ} (g) T ξ_{φ}, ξ_{φ} ⟩$ . Since $I - T$ is also a positive contraction in the commutant, [F3] makes both $ψ_{T}$ and $ψ_{I - T}$ continuous and of positive type. By linearity and $T + (I - T) = I$ , their sum is $φ$ , and their identity values are $t$ and $1 - t$ . Positive scalar multiples preserve positive type by [F1], so $φ_{1} = ψ_{T} / t$ and $φ_{2} = ψ_{I - T} / (1 - t)$ belong to $P_{1} (G)$ . They satisfy $φ = t φ_{1} + (1 - t) φ_{2}$ .

5.1F1F3F5step 4.1algebra

If $φ_{1} = φ_{2}$ , their convex combination is $φ$ , so $ψ_{T} = t φ$ . The scalar operator $t I$ is a positive contraction in the commutant and has coefficient $t φ$ . Since $0 \leq t φ \leq φ$ , uniqueness in [F3] gives $T = t I$ , contradicting that $T$ is nonscalar. Hence the two normalized summands are distinct.

6.1F2F3F10step 1.1step 3.1algebra∎

AC is used through [F2] for the canonical GNS triple and [F3] for the dominated-positive-operator theorem. That theorem uses AC $\Rightarrow$ DC $\Rightarrow$ Countable Choice for Riesz representation and the adjoint and positive-operator interfaces [F10]. The positive-form expansion in step 1.1, the finite coefficient calculations, and the one-at-a-time density argument in step 3.1 use no further choice.

Depends on

Used by

Extreme normalized positive type is equivalent to irreducible GNS Theorem

Dependency tree · two levels

40 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Bekka, de la Harpe and Valette, Kazhdan's Property (T), Proposition C.5.1 and complete proof (standard reference, not scraped)
Karl-Hermann Neeb, An Introduction to Unitary Representations of Lie Groups, Proposition 5.1.11 and Theorem 5.1.12 with complete proof (standard reference, not scraped)