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Operator-Algebraic States and GNS Representations: Examples and Counterexamples

1 · Prerequisites

2 · Summary

The worked example computes two state quotient models on $M_{n} (C)$ : the first-coordinate vector state gives the defining representation on $C^{n}$ , while normalized trace gives left multiplication on the normalized Hilbert-Schmidt space $M_{n} (C)$ . For $n \geq 2$ , the trace state is non-extreme and its GNS representation has a proper invariant subspace; $n = 1$ is recorded as the boundary case.

Vector and normalized-trace states on matrix algebras proves the state bounds, quotient identifications, cyclicity, and reducibility directly from the matrix formulas using the first-variable-linear convention fixed on the A page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Vector and normalized-trace states on matrix algebras

Statement

Let $n \geq 1$ , $A = M_{n} (C)$ with its matrix C*-algebra structure, and $e_{1}, \dots, e_{n}$ the standard basis of $C^{n}$ . For a state $ρ$ on the unital algebra $A$ , its quotient GNS model here is the following construction when the displayed formulas are well-defined: set $N_{ρ} = {a \in A : ρ (a^{*} a) = 0},$ give $A / N_{ρ}$ the inner product $⟨ [a], [b] ⟩_{ρ} = ρ (b^{*} a),$ complete it, let $π_{ρ} (c) [b] = [c b]$ , and take $ξ_{ρ} = [I_{n}]$ .

For $a = (a_{i j})$ , define $Tr (a) = \sum_{j = 1}^{n} a_{j j}$ . Write $∥ b ∥_{τ} = \sqrt{⟨ b, b ⟩_{τ}}$ for the norm induced by the normalized-trace pairing below. In this item a representation is a bounded linear map $π : A \to B (H)$ preserving products and involution; the models below are unital.

For each $j$ , $ω_{j} (a) = ⟨ a e_{j}, e_{j} ⟩$ is a state and its quotient GNS model is unitarily equivalent to the defining representation of $M_{n} (C)$ on $C^{n}$ . The normalized trace $τ (a) = \frac{1}{n} Tr (a)$ has quotient GNS model $M_{n} (C)$ with inner product $⟨ b, c ⟩_{τ} = \frac{1}{n} Tr (c^{*} b)$ and left multiplication. If $n \geq 2$ , $τ$ is not an extreme state and its GNS representation is reducible. When $n = 1$ , $τ = ω_{1}$ and the one-dimensional representation is irreducible.

Facts & Assumptions

Given: An integer $n \geq 1$ , the C*-algebra $A = M_{n} (C)$ , its standard basis vectors $e_{1}, \dots, e_{n}$ , and the inner product linear in its first variable.

[F1]

Finite matrix algebras with the operator norm are C star algebras: $M_{n} (C)$ with conjugate transpose and the operator norm on $C^{n}$ is a unital complex C*-algebra for every $n \geq 1$ .

[F2]

States and positive functionals on a C star algebra: a state is a bounded positive linear functional of norm one, where positivity means $ρ (a^{*} a) \geq 0$ for every $a$ .

[F3]

Real and complex inner-product spaces and their induced length: complex inner products here are linear in the first variable, conjugate-linear in the second, and have induced norm $∥ x ∥ = \sqrt{⟨ x, x ⟩}$ .

[F4]

The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators: $B (H)$ is the space of bounded linear maps $H \to H$ .

[F5]

The operator norm as the least bound and as the unit-sphere or unit-ball supremum: the operator norm is the supremum of $∥ T x ∥$ over $∥ x ∥ \leq 1$ ; rescaling gives $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ .

[F6]

Hilbert space: a complex Hilbert space is a complete complex inner-product space for its induced norm.

[F7]

Every finite-dimensional normed space is Banach: a complex normed vector space with a finite basis is complete.

[F8]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ .

[F9]

The induced length is a norm: the length induced by any real or complex inner product satisfies absolute homogeneity and the triangle inequality, and is a norm.

An extreme state is a state that is not a convex combination $t α + (1 - t) β$ of distinct states with $0 < t < 1$ . A representation is reducible if it has a nonzero proper closed invariant subspace. A vector is cyclic if its representation orbit has dense linear span.

[L1]

In these finite-dimensional Hilbert-space models, an adjoint of a bounded map $T$ means a bounded map $S$ satisfying $⟨ T x, y ⟩ = ⟨ x, S y ⟩$ for all $x, y$ . Such an $S$ is unique: if $S_{1}, S_{2}$ both satisfy the identity, then $⟨ x, (S_{1} - S_{2}) y ⟩ = 0$ for every $x$ , and taking $x = (S_{1} - S_{2}) y$ gives $(S_{1} - S_{2}) y = 0$ . On $C^{n}$ , coordinate expansion shows that the adjoint of the matrix $a$ is its conjugate transpose $a^{*}$ .

Verification

Given: The data above. For each state $ρ$ considered, form $N_{ρ}$ , $A / N_{ρ}$ , and the quotient inner product displayed in the Statement.

1.1F1F2F3F5F8given

For each $j$ , $ω_{j} (a) = ⟨ a e_{j}, e_{j} ⟩$ is linear and satisfies $∣ ω_{j} (a) ∣ \leq ∥ a e_{j} ∥_{2} \leq ∥ a ∥$ , by [F5] and [F8]. The $j$ th diagonal entry of $a^{*} a$ is $\sum_{i} ∣ a_{i j} ∣^{2} = ∥ a e_{j} ∥_{2}^{2}$ , so $ω_{j} (a^{*} a) = ∥ a e_{j} ∥_{2}^{2} \geq 0$ ; also $ω_{j} (I_{n}) = 1$ . Since $n \geq 1$ , a unit vector exists and $∥ I_{n} ∥ = \sup_{∥ x ∥_{2} \leq 1} ∥ I_{n} x ∥_{2} = 1$ . Thus $∥ ω_{j} ∥ \leq 1$ and $∥ ω_{j} ∥ \geq ∣ ω_{j} (I_{n}) ∣ = 1$ , so each $ω_{j}$ is a state by [F2].

2.1F1F2F3F5F8givenstep 1.1

The functional $τ (a) = n^{- 1} Tr (a)$ satisfies $τ (a^{*} a) = n^{- 1} \sum_{i, j} ∣ a_{i j} ∣^{2} \geq 0$ and $τ (I_{n}) = 1$ . Moreover, $∣ τ (a) ∣ = n^{- 1} ∣ \sum_{j} ⟨ a e_{j}, e_{j} ⟩ ∣ \leq n^{- 1} \sum_{j} ∥ a e_{j} ∥_{2} \leq ∥ a ∥$ by [F5] and [F8]. Since $∥ I_{n} ∥ = 1$ by step 1.1, this gives $∥ τ ∥ \geq ∣ τ (I_{n}) ∣ = 1$ , hence $∥ τ ∥ = 1$ and $τ$ is a state by [F2].

2.2F1F3F6F7F9step 1.1given

For each $j$ , $N_{ω_{j}} = {a : a e_{j} = 0}$ by step 1.1, and $U_{j} ([a]) = a e_{j}$ is well-defined and injective on $A / N_{ω_{j}}$ . For matrices $a, b$ , $ω_{j} (b^{*} a) = (b^{*} a)_{j j} = \sum_{i} a_{i j} \overline{b_{i j}} = ⟨ a e_{j}, b e_{j} ⟩$ , so the quotient pairing is well-defined and $U_{j}$ is an isometry. It is onto because an arbitrary vector $v \in C^{n}$ is the $j$ th column of a matrix $a$ , giving $a e_{j} = v$ ; the classes $[E_{i j}]$ for $1 \leq i \leq n$ map to the standard basis, so the quotient has a finite basis. The norm is the inner-product norm and satisfies the norm axioms by [F9]; [F7] makes this finite-dimensional space complete, hence each quotient is already the Hilbert space $C^{n}$ by [F6].

3.1F1F3F4F5F6F7F9L1step 2.1algebra

The null space $N_{τ}$ is ${0}$ , since $τ (a^{*} a) = 0$ forces every matrix entry of $a$ to vanish. The quotient pairing is $⟨ b, c ⟩_{τ} = n^{- 1} Tr (c^{*} b) = n^{- 1} \sum_{i, j} b_{i j} \overline{c_{i j}}$ ; this coordinate formula is linear in $b$ , conjugate-linear in $c$ , conjugate symmetric, and positive-definite. Its induced length is a norm by [F9], and the matrix units form a finite basis, so [F7] makes it complete; it is a Hilbert space by [F6]. Left multiplication is bounded: $∥ π_{τ} (a) b ∥_{τ}^{2} = n^{- 1} \sum_{j} ∥ a b e_{j} ∥_{2}^{2} \leq ∥ a ∥^{2} n^{- 1} \sum_{j} ∥ b e_{j} ∥_{2}^{2} = ∥ a ∥^{2} ∥ b ∥_{τ}^{2}$ by [F5]. Bilinearity of matrix multiplication gives $π_{τ} (a + c) = π_{τ} (a) + π_{τ} (c)$ and $π_{τ} (z a) = z π_{τ} (a)$ , associativity gives $π_{τ} (a) π_{τ} (c) = π_{τ} (a c)$ , and $π_{τ} (I_{n}) b = [I_{n} b] = [b]$ makes the representation unital. Also $⟨ a b, c ⟩_{τ} = τ (c^{*} a b) = τ ((a^{*} c)^{*} b) = ⟨ b, a^{*} c ⟩_{τ}$ . Therefore $π_{τ} (a^{*})$ satisfies the adjoint identity for $π_{τ} (a)$ and equals its adjoint by [L1]. The vector $I_{n}$ is unit and cyclic because $π_{τ} (a) I_{n} = a$ , and its coefficient is $⟨ a I_{n}, I_{n} ⟩_{τ} = τ (a)$ . This verifies the trace quotient GNS model.

3.2F1F3F4L1step 2.2algebra

If $b \in N_{ω_{j}}$ then $(c b) e_{j} = c (b e_{j}) = 0$ , so $π_{ω_{j}} (c) [b] = [c b]$ is well-defined. For $a, c \in A$ , $z \in C$ , and every $[b]$ , $π_{ω_{j}} (a + c) [b] = [(a + c) b] = [a b] + [c b]$ and $π_{ω_{j}} (z a) [b] = [(z a) b] = z [a b]$ ; associativity also gives $π_{ω_{j}} (a) π_{ω_{j}} (c) [b] = [a c b] = π_{ω_{j}} (a c) [b]$ , and $π_{ω_{j}} (I_{n}) [b] = [I_{n} b] = [b]$ makes the representation unital. Under $U_{j}$ it is exactly the matrix $a$ , so it is bounded and its adjoint is the action of $a^{*}$ by [L1]. The vector $[I_{n}]$ is unit and cyclic because $π_{ω_{j}} (a) [I_{n}] = [a]$ for every $a \in A$ ; moreover, its coefficient is $⟨ π_{ω_{j}} (a) [I_{n}], [I_{n}] ⟩_{ω_{j}} = ⟨ a e_{j}, e_{j} ⟩ = ω_{j} (a)$ . Thus $U_{j}$ identifies this quotient GNS model with the defining representation for every $j$ .

4.1F1step 3.1givenalgebra

For $n \geq 2$ , let $W$ be the subspace of matrices whose columns after the first are zero. It contains $E_{11}$ and omits $E_{12}$ , so it is nonzero and proper. It is closed because it is the intersection of the kernels of the finitely many entry functionals $b \mapsto b_{i j}$ for $j > 1$ ; each functional is bounded for the normalized Hilbert-Schmidt norm because step 3.1 gives $∥ b ∥_{τ}^{2} = n^{- 1} \sum_{k, l} ∣ b_{k l} ∣^{2}$ , so $∣ b_{i j} ∣^{2} \leq \sum_{k, l} ∣ b_{k l} ∣^{2} = n ∥ b ∥_{τ}^{2}$ and hence $∣ b_{i j} ∣ \leq \sqrt{n} ∥ b ∥_{τ}$ . For $a \in M_{n} (C)$ and $b \in W$ , every column of $a b$ after the first is still zero, so left multiplication preserves $W$ . Thus the trace GNS representation is reducible for $n \geq 2$ .

4.2F2F5step 1.1step 2.2step 3.1step 3.2algebra

If $n = 1$ , then $τ = ω_{1}$ by their formulas, and steps 2.2 and 3.2 identify its GNS space with $C$ ; the only subspaces of a one-dimensional space are ${0}$ and the whole space, so the representation is irreducible. If $n \geq 2$ , define $ψ = (n - 1)^{- 1} \sum_{j = 2}^{n} ω_{j}$ . Step1.1 gives $ψ (a^{*} a) \geq 0$ and $∣ ψ (a) ∣ \leq (n - 1)^{- 1} \sum_{j = 2}^{n} ∣ ω_{j} (a) ∣ \leq ∥ a ∥$ , while $ψ (I_{n}) = 1$ . Since $∥ I_{n} ∥ = 1$ by step 1.1, these bounds give $∥ ψ ∥ = 1$ , so $ψ$ is a state by [F2]. Since $Tr (a) = \sum_{j} ω_{j} (a)$ , $τ = n^{- 1} ω_{1} + ((n - 1) / n) ψ$ . For $p = E_{11}$ , $ω_{1} (p) = 1$ and $ψ (p) = 0$ , so the two states are distinct and this is a nontrivial convex decomposition. Therefore $τ$ is not an extreme state for $n \geq 2$ .

5.1step 2.1step 2.2step 3.1step 3.2step 4.1step 4.2∎

Steps2.2 and 3.2 verify the quotient GNS model and defining representation for every vector state, while steps 2.1 and 3.1 verify the normalized-trace state and its left-multiplication quotient model. Steps4.1 and 4.2 give the $n = 1$ and $n \geq 2$ boundaries for the trace state. These are the claims in the Statement.

Remarks

The quotient inner products are linear in the first variable. The general C*-state GNS route and the distinct topological-group positive-type route are summarized on the A page.