Cyclic representation of a quasi-local state

For $a,b\in {\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$, positivity makes $Q(z):=\omega ((a+zb{)}^{\ast }(a+zb))\ge 0$ for every $z\in \mathbb{C}$. Expanding gives $Q(z)=\omega (a^{\ast }a)+zu+\bar{z}v+\mathrm{\mid }z{\mathrm{\mid }}^{2}c$, where $u=\omega (a^{\ast }b)$, $v=\omega (b^{\ast }a)$, and $c=\omega (b^{\ast }b)\ge 0$. Taking real $z=t$ and purely imaginary $z=it$ shows respectively that $u+v$ is real and $i(u-v)$ is real, so $v=\bar{u}$. If $c>0$, take $z=-v\mathrm{/}c$ to obtain $Q(z)=\omega (a^{\ast }a)-\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}c\ge 0$. If $c=0$ and $v\mathrm{\ne }0$, taking $z=-tv$ for sufficiently large real $t>0$ gives $Q(z)=\omega (a^{\ast }a)-2t\mathrm{\mid }v{\mathrm{\mid }}^2<0$, a contradiction; hence $v=0$. Therefore $\mathrm{\mid }\omega (b^{\ast }a){\mathrm{\mid }}^2\le \omega (a^{\ast }a)\omega (b^{\ast }b)$.

Define $N:=\{a\in {\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}:\omega (a^{\ast }a)=0\}$. If $a,b\in N$, step 1.1 makes both mixed terms $\omega (a^{\ast }b)$ and $\omega (b^{\ast }a)$ zero, so $\omega ((a+b{)}^{\ast }(a+b))=0$; scalar multiples stay in $N$ by homogeneity. Thus $N$ is a vector subspace. The same inequality makes $⟨[a],[b]⟩:=\omega (b^{\ast }a)$ independent of representatives in ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}\mathrm{/}N$. This form is linear in its first argument, conjugate-linear in the second, conjugate symmetric, and positive definite on the quotient: $⟨[a],[a]⟩=0$ implies $a\in N$ and hence $[a]=0$.

Let $a,b\in {\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and choose one finite region containing both. In its matrix representation, $c:=\mathrm{\parallel }a{\mathrm{\parallel }}^{2}I-a^{\ast }a$ is Hermitian and positive, since $⟨c\xi ,\xi ⟩=\mathrm{\parallel }a{\mathrm{\parallel }}^2\mathrm{\parallel }\xi {\mathrm{\parallel }}^2-\mathrm{\parallel }a\xi {\mathrm{\parallel }}^2\ge 0$ for every vector $\xi$. By [F2], $c^{1\mathrm{/}2}$ exists in that same finite matrix algebra. Therefore $b^{\ast }cb=(c^{1\mathrm{/}2}b{)}^{\ast }(c^{1\mathrm{/}2}b)$ and positivity of $\omega$ gives $\omega (b^{\ast }{a}^{\ast }ab)\le \mathrm{\parallel }a{\mathrm{\parallel }}^2\omega (b^{\ast }b)$. In particular, the null space $N$ of step 2.1 is a left ideal under multiplication by local $a$: if $b\in N$, the inequality gives $\omega ((ab{)}^{\ast }(ab))=0$.

Define ${\pi }_0(a)[b]:=[ab]$ for $a,b\in {\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$. Step 3.1 shows this is well defined on the quotient. The bound there gives $\mathrm{\parallel }{\pi }_0(a)[b]{\mathrm{\parallel }}^2\le \mathrm{\parallel }a{\mathrm{\parallel }}^2\mathrm{\parallel }[b]{\mathrm{\parallel }}^2$, so ${\pi }_0(a)$ extends uniquely to a bounded operator on the Hilbert completion ${\mathcal{H}}_{\omega }$, with $\mathrm{\parallel }{\pi }_0(a)\mathrm{\parallel }\le \mathrm{\parallel }a\mathrm{\parallel }$. On quotient vectors, multiplication gives ${\pi }_0(a){\pi }_0(d)[b]={\pi }_0(ad)[b]$, and $⟨{\pi }_0(a)[b],[c]⟩=\omega (c^{\ast }ab)=⟨[b],{\pi }_0(a^{\ast })[c]⟩$; continuity extends these identities to ${\mathcal{H}}_{\omega }$.

For $A\in \mathcal{A}$, represent it by a Cauchy sequence $(a_n)$ in ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$, as in the norm-completion construction. The estimate $\mathrm{\parallel }{\pi }_0(a_n)-{\pi }_0(a_m)\mathrm{\parallel }\le \mathrm{\parallel }{a}_n-a_m\mathrm{\parallel }$ makes the operators operator-norm Cauchy. For each $\xi \in {\mathcal{H}}_{\omega }$, completeness gives $T\xi :={\lim }_n{\pi }_0(a_n)\xi$; this is linear and bounded because $(a_n)$ is norm bounded and $\mathrm{\parallel }{\pi }_0(a_n)\xi \mathrm{\parallel }\le \mathrm{\parallel }{a}_n\mathrm{\parallel }\mathrm{\parallel }\xi \mathrm{\parallel }$. Given $\epsilon >0$, for all sufficiently large $n,m$ the operator difference has norm at most $\epsilon$; fixing such $n$ and passing $m\to \mathrm{\infty }$ gives $\mathrm{\parallel }({\pi }_0(a_n)-T)\xi \mathrm{\parallel }\le \epsilon \mathrm{\parallel }\xi \mathrm{\parallel }$, hence $\mathrm{\parallel }{\pi }_0(a_n)-T\mathrm{\parallel }\le \epsilon$. Thus $T\in \mathcal{B}({\mathcal{H}}_{\omega })$ is the operator-norm limit. Equivalent representatives give the same limit by the same estimate, defining ${\pi }_{\omega }(A):=T$. Linearity follows because local approximants to $A,B$ and $\lambda \in \mathbb{C}$ give approximants $a_n+\lambda b_n$ to $A+\lambda B$, and each ${\pi }_0$ is linear. Passing $\mathrm{\parallel }{\pi }_0(a_n)\mathrm{\parallel }\le \mathrm{\parallel }{a}_n\mathrm{\parallel }$ to the limit also gives $\mathrm{\parallel }{\pi }_{\omega }(A)\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }$, so ${\pi }_{\omega }$ is bounded and contractive. If $(b_n)$ represents $B$, then $(a_n{b}_n)$ represents $AB$ by continuity of multiplication in $\mathcal{A}$, and ${\pi }_0(a_n{b}_n)={\pi }_0(a_n){\pi }_0(b_n)$; taking operator-norm limits gives multiplicativity. For any $x$ in a C*-algebra, $\mathrm{\parallel }{x}^{\ast }{\mathrm{\parallel }}^2=\mathrm{\parallel }x{x}^{\ast }\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }{x}^{\ast }\mathrm{\parallel }$ by the C*-identity and submultiplicativity; this gives $\mathrm{\parallel }{x}^{\ast }\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }$, and applying it to $x^{\ast }$ yields equality. Hence $(a_n^{\ast })$ represents $A^{\ast }$; the adjoint identity from step 4.1 and norm continuity of Hilbert adjoints [F6] give ${\pi }_{\omega }(A^{\ast })={\pi }_{\omega }(A{)}^{\ast }$. The constant sequence at $I$ gives ${\pi }_{\omega }(I)=I$.

Put ${\mathrm{\Omega }}_{\omega }:=[I]$. Its squared norm is $\omega (I)=1$. For every local $a$, ${\pi }_{\omega }(a){\mathrm{\Omega }}_{\omega }=[a]$; these classes are dense by the definition of ${\mathcal{H}}_{\omega }$, so ${\mathrm{\Omega }}_{\omega }$ is cyclic. For local $a$, $⟨{\pi }_{\omega }(a){\mathrm{\Omega }}_{\omega },{\mathrm{\Omega }}_{\omega }⟩=\omega (a)$ by the quotient inner product. Both sides are continuous in $a$ and ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ is dense in $\mathcal{A}$, so the identity holds for every $A\in \mathcal{A}$.