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State-preserving symmetries implemented unitarily

Statement

Assume Countable Choice ${A C}_{ω}$ , the hypothesis of GNS Hilbert reconstruction with invariant operator domain. Let $G$ be a topological group with identity $e$ . Let $A$ be a unital complex star algebra with positive normalized state $ω$ , and let $α : G \to {Aut}^{*} (A)$ be a group action: each $α_{g}$ is a bijective complex-linear map preserving product, unit and involution, with $α_{e} = i d$ and $α_{g} α_{h} = α_{g h}$ . Suppose $ω (α_{g} (a)) = ω (a)$ for all $g \in G$ and $a \in A$ . Also suppose that, for every $a, b \in A$ , the coefficient function $g \mapsto ω (b^{*} α_{g} (a))$ is continuous at $e$ .

Use the GNS representation from GNS Hilbert reconstruction with invariant operator domain, with common dense domain $D_{ω}$ and cyclic vector $Ω_{ω}$ . Then $U_{g} [a] : = [α_{g} (a)]$ extends to a strongly continuous unitary representation $U : G \to U (H_{ω})$ , fixes $Ω_{ω}$ , and satisfies

$U_{g} π_{ω} (a) U_{g}^{- 1} ∣_{D_{ω}} = π_{ω} (α_{g} (a)) (a \in A, g \in G) .$

Here strongly continuous means that $g \mapsto U_{g} ξ$ is norm-continuous for every $ξ \in H_{ω}$ . No differentiability, generator domain or spectral theorem follows from these hypotheses alone.

Facts & Assumptions

Given: Countable Choice ${A C}_{ω}$ , a topological group $G$ , a unital complex star algebra $A$ , a positive normalized state $ω$ , a state-preserving action $α$ by unital star automorphisms, and continuity at $e$ of every $g \mapsto ω (b^{*} α_{g} (a))$ .

[F1]

GNS Hilbert reconstruction with invariant operator domain: the state has a Hilbert-space GNS representation on a common invariant dense domain $D_{ω} = A / N_{ω}$ , with $π_{ω} (a) [b] = [a b]$ , $Ω_{ω} = [1_{A}]$ , and inner product $([a], [b])_{ω} = ω (b^{*} a)$ .

[A1]

By the action assumptions, each $α_{g}$ is a unital star automorphism, $α_{g h} = α_{g} \circ α_{h}$ , and $ω \circ α_{g} = ω$ ; multiplication and inversion in the topological group are continuous.

[A2]

For every $a, b \in A$ , the scalar function $g \mapsto ω (b^{*} α_{g} (a))$ is continuous at $e$ .

Proof

technique · direct

1.1F1A1given

If $n \in N_{ω}$ , then $ω ((α_{g} n)^{*} α_{g} n) = ω (α_{g} (n^{*} n)) = ω (n^{*} n) = 0$ , so $α_{g} (N_{ω}) \subseteq N_{ω}$ . Applying the same argument to $g^{- 1}$ gives equality. For $a, b \in A$ , $([α_{g} (a)], [α_{g} (b)])_{ω} = ω (α_{g} (b^{*} a)) = ω (b^{*} a)$ , so $[a] \mapsto [α_{g} (a)]$ is an isometry on $D_{ω}$ with inverse induced by $α_{g^{- 1}}$ ; it extends uniquely to a unitary $U_{g}$ on $H_{ω}$ . The action law gives $U_{g} U_{h} = U_{g h}$ , and $U_{g} Ω_{ω} = Ω_{ω}$ .

2.1F1A1step 1.1

For $[b] \in D_{ω}$ , $U_{g} π_{ω} (a) U_{g}^{- 1} [b] = U_{g} [a α_{g^{- 1}} (b)] = [α_{g} (a) b] = π_{ω} (α_{g} (a)) [b]$ , using the representation rule and multiplicativity of $α_{g}$ . This is the claimed covariance identity on the common domain.

2.2F1A1A2step 1.1algebra

For $a \in A$ , state invariance and the quotient inner product give $∥ U_{g} [a] - [a] ∥^{2} = 2 ω (a^{*} a) - ω (a^{*} α_{g} (a)) - ω ((α_{g} (a))^{*} a)$ . The first cross term tends to $ω (a^{*} a)$ as $g \to e$ by [A2]. For the second, invariance gives $ω ((α_{g} (a))^{*} a) = ω (α_{g^{- 1}} ((α_{g} (a))^{*} a)) = ω (a^{*} α_{g^{- 1}} (a))$ , which also tends to $ω (a^{*} a)$ by [A2] and continuity of inversion. Thus $U_{g} [a] \to [a]$ for every $[a] \in D_{ω}$ .

3.1step 1.1step 2.2givenalgebra∎

Let $ξ \in H_{ω}$ and $ε > 0$ . Choose $y \in D_{ω}$ with $∥ ξ - y ∥ < ε / 3$ . Since each $U_{g}$ is unitary, $∥ U_{g} ξ - ξ ∥ \leq 2 ∥ ξ - y ∥ + ∥ U_{g} y - y ∥$ ; step 2.2 makes the last term less than $ε / 3$ for $g$ sufficiently near $e$ . Hence every orbit is continuous at $e$ . For any $g_{0} \in G$ , $∥ U_{g} ξ - U_{g_{0}} ξ ∥ = ∥ U_{g_{0}^{- 1} g} ξ - ξ ∥$ , so continuity of multiplication transfers continuity to $g_{0}$ . The unitary representation is strongly continuous on all of $G$ .

Depends on

GNS Hilbert reconstruction with invariant operator domain

State-preserving symmetries implemented unitarily

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources