Unital star algebra and positive normalized state

A unital complex star algebra is a complex vector space $A$ with a complex-bilinear associative product $(a,b)\mapsto ab$, a specified unit $1_A$, and a conjugate-linear map $a\mapsto a^{\ast }$ such that, for all $a,b\in A$,

$(ab{)}^{\ast }=b^{\ast }{a}^{\ast },(a^{\ast }{)}^{\ast }=a,1_A^{\ast }=1_A\mathrm{.}$

A positive normalized state on $A$ is a complex-linear functional $\omega :A\to \mathbb{C}$ such that $\omega (1_A)=1$ and $\omega (a^{\ast }a)$ is a real nonnegative number for every $a\in A$. This is an algebraic definition: no norm, topology, boundedness condition, or completion is part of the data.

A sesquilinear form $B:A\times A\to \mathbb{C}$ is taken to be linear in its first variable and conjugate-linear in its second. It is Hermitian when $B(a,b)=\overline{B(b,a)}$ for all $a,b\in A$, positive semidefinite when $B(a,a)\in [0,\mathrm{\infty })$ for every $a\in A$, and has radical

$\mathrm{rad}B:=\{a\in A:B(a,b)=0\text{ for every }b\in A\}\mathrm{.}$

For a Hermitian form this left radical also equals $\{a\in A:B(b,a)=0\text{ for every }b\in A\}$.

Let $A$ be a unital complex star algebra and let $\omega$ be a positive normalized state on $A$ (Unital star algebra and positive normalized state). Define $q(a):=\omega (a^{\ast }a)$ and $N_{\omega }:=\{a\in A:q(a)=0\}$. Then

$\mathrm{\mid }\omega (b^{\ast }a){\mathrm{\mid }}^2\le q(a)q(b)(a,b\in A)\mathrm{.}$

The form $(a,b{)}_{\omega }:=\omega (b^{\ast }a)$ is Hermitian and positive semidefinite, $N_{\omega }$ is its radical and a complex linear subspace, and $N_{\omega }$ is a left ideal: $ca\in N_{\omega }$ whenever $a\in N_{\omega }$ and $c\in A$.

Let $H$ be a complex Hilbert space with inner product linear in its first variable. An operator on $H$ is a linear map $T:D(T)\to H$ whose domain $D(T)$ is a linear subspace of $H$; it is densely defined when $D(T)$ is dense in $H$. It is bounded when some $C\ge 0$ satisfies $\mathrm{\parallel }Tx\mathrm{\parallel }\le C\mathrm{\parallel }x\mathrm{\parallel }$ for all $x\in D(T)$, and unbounded when no such $C$ exists. No boundedness of the operators ${\pi }_{\omega }(a)$ is assumed here.

For a complex vector space $E$, $\mathrm{End}(E)$ denotes the set of complex-linear maps from $E$ to itself. The product norm on $H\times H$ is $\mathrm{\parallel }(x,y){\mathrm{\parallel }}_{\times }:=(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$.

For a densely defined $T$, define its Hilbert-space adjoint by

$D(T^{\ast }):=\{y\in H:\text{there is }z\in H\text{ such that }(Tx,y)=(x,z)\text{ for every }x\in D(T)\},T^{\ast }y:=z\mathrm{.}$

The vector $z$ is unique because $D(T)$ is dense. The graph of $T$ is $G(T):=\{(x,Tx):x\in D(T)\}\subseteq H\times H$. The operator $T$ is symmetric when $D(T)\subseteq D(T^{\ast })$ and $T^{\ast }x=Tx$ for every $x\in D(T)$; it is self-adjoint when $D(T)=D(T^{\ast })$ and the same equality holds there. It is closable when the closure of $G(T)$ in the product norm on $H\times H$ is the graph of an operator, denoted $\bar{T}$. A symmetric operator is essentially self-adjoint when it is closable and $\bar{T}$ is self-adjoint.

Let $A$ be a complex vector space with a complex-bilinear associative product, unit $1_A$, and conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and $1_A^{\ast }=1_A$. Let $\omega :A\to \mathbb{C}$ be complex-linear with $\omega (1_A)=1$ and real nonnegative $\omega (a^{\ast }a)$ for every $a\in A$. Assume the Axiom of Countable Choice ${\mathrm{A}\mathrm{C}}_{\omega }$: every countable family of nonempty sets has a choice function. Put $N_{\omega }:=\{a\in A:\omega (a^{\ast }a)=0\}$ and ${\mathcal{D}}_{\omega }:=A\mathrm{/}{N}_{\omega }$, with

$([a],[b]{)}_{\omega }:=\omega (b^{\ast }a),$

using the first-variable-linear convention. Let $H_{\omega }$ be the Hilbert completion of this inner-product space, and identify ${\mathcal{D}}_{\omega }$ with its canonical dense image in $H_{\omega }$. For each $a\in A$ define left multiplication on the common domain ${\mathcal{D}}_{\omega }$ by ${\pi }_{\omega }(a)[b]:=[ab]$. Then ${\pi }_{\omega }$ is a unital algebra representation of $A$ on the invariant dense domain ${\mathcal{D}}_{\omega }$, and

${\mathcal{D}}_{\omega }\subseteq D({\pi }_{\omega }(a{)}^{\ast }),{\pi }_{\omega }(a{)}^{\ast }{\mathrm{\mid }}_{{\mathcal{D}}_{\omega }}={\pi }_{\omega }(a^{\ast })\mathrm{.}$

Each ${\pi }_{\omega }(a)$ is closable, meaning that the closure of its graph in $H_{\omega }\times H_{\omega }$ is the graph of an operator. The vector ${\mathrm{\Omega }}_{\omega }:=[1_A]$ is a unit cyclic vector, ${\pi }_{\omega }(A){\mathrm{\Omega }}_{\omega }={\mathcal{D}}_{\omega }$, and $\omega (a)=({\pi }_{\omega }(a){\mathrm{\Omega }}_{\omega },{\mathrm{\Omega }}_{\omega })$. For uniqueness, let $K$ be a Hilbert space, ${\mathcal{D}}^{\mathrm{\prime }}\subseteq K$ a dense linear subspace invariant under every operator $\rho (a)$, and $\rho :A\to \mathrm{End}({\mathcal{D}}^{\mathrm{\prime }})$ a unital algebra representation satisfying $(\rho (a)x,y)=(x,\rho (a^{\ast })y)$ for all $a\in A$ and $x,y\in {\mathcal{D}}^{\mathrm{\prime }}$. If ${\mathrm{\Omega }}^{\mathrm{\prime }}\in {\mathcal{D}}^{\mathrm{\prime }}$, ${\mathcal{D}}^{\mathrm{\prime }}=\rho (A){\mathrm{\Omega }}^{\mathrm{\prime }}$, and $\omega (a)=(\rho (a){\mathrm{\Omega }}^{\mathrm{\prime }},{\mathrm{\Omega }}^{\mathrm{\prime }})$, then there is a unique unitary $V:H_{\omega }\to K$ with $V{\mathrm{\Omega }}_{\omega }={\mathrm{\Omega }}^{\mathrm{\prime }}$ and $V{\pi }_{\omega }(a)x=\rho (a)Vx$ for all $a\in A$ and $x\in {\mathcal{D}}_{\omega }$. This theorem supplies neither a C*-norm nor essential self-adjointness of ${\pi }_{\omega }(a)$ when $a=a^{\ast }$.

Assume Countable Choice ${\mathrm{A}\mathrm{C}}_{\omega }$, 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 $\omega$, and let $\alpha :G\to {\mathrm{Aut}}^{\ast }(A)$ be a group action: each ${\alpha }_g$ is a bijective complex-linear map preserving product, unit and involution, with ${\alpha }_e=\mathrm{i}\mathrm{d}$ and ${\alpha }_g{\alpha }_h={\alpha }_{gh}$. Suppose $\omega ({\alpha }_g(a))=\omega (a)$ for all $g\in G$ and $a\in A$. Also suppose that, for every $a,b\in A$, the coefficient function $g\mapsto \omega (b^{\ast }{\alpha }_g(a))$ is continuous at $e$.

Use the GNS representation from GNS Hilbert reconstruction with invariant operator domain, with common dense domain ${\mathcal{D}}_{\omega }$ and cyclic vector ${\mathrm{\Omega }}_{\omega }$. Then $U_g[a]:=[{\alpha }_g(a)]$ extends to a strongly continuous unitary representation $U:G\to \mathcal{U}(H_{\omega })$, fixes ${\mathrm{\Omega }}_{\omega }$, and satisfies

$U_g{\pi }_{\omega }(a)U_g^{-1}{\mathrm{\mid }}_{{\mathcal{D}}_{\omega }}={\pi }_{\omega }({\alpha }_g(a))(a\in A,g\in G)\mathrm{.}$

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