Local spin matrix algebras and isometric inclusions

The identity map shows that $\mathbb{N}$ is at most countable by [F1], so [F3] gives that $\mathbb{N}\times \mathbb{N}$ is at most countable. This product is nonempty, so [F2] supplies a surjection $s:\mathbb{N}\to \mathbb{N}\times \mathbb{N}$. By [F4], the quotient map $\pi :\mathbb{N}\times \mathbb{N}\to \mathbb{Z}$ is surjective; hence $\pi \circ s:\mathbb{N}\to \mathbb{Z}$ is a surjection.

The surjection in step 1.1 makes $\mathbb{Z}$ nonempty, and [F2] then shows that $\mathbb{Z}$ is at most countable in the sense of [F1].

Since $d$ is finite and positive, [F3] with $C=\mathbb{Z}$ and $m=d$ gives that $\mathrm{\Gamma }={\mathbb{Z}}^d$ is at most countable.

Under [F1], $T\mapsto T\otimes I$ preserves products and adjoints because $(T\otimes I)(S\otimes I)=TS\otimes I$ and $(T\otimes I{)}^{\ast }=T^{\ast }\otimes I$; it sends $I_{\mathrm{\Lambda }}$ to $I_{{\mathrm{\Lambda }}^{\mathrm{\prime }}}$. The configuration-basis factorization is associative, so for nested regions the two successive inclusions and the direct inclusion act identically on every configuration basis vector and hence are equal operators.

Let $K={\mathcal{H}}_{{\mathrm{\Lambda }}^{\mathrm{\prime }}\setminus \mathrm{\Lambda }}$, and choose its finite orthonormal configuration basis $e_1,\dots ,e_s$. Every $\xi \in {\mathcal{H}}_{\mathrm{\Lambda }}\otimes K$ has a unique expansion $\xi ={\sum }_{j=1}^s{\xi }_j\otimes e_j$, with $\mathrm{\parallel }\xi {\mathrm{\parallel }}^2={\sum }_j\mathrm{\parallel }{\xi }_j{\mathrm{\parallel }}^2$. Therefore $\mathrm{\parallel }(T\otimes I)\xi {\mathrm{\parallel }}^2={\sum }_j\mathrm{\parallel }T{\xi }_j{\mathrm{\parallel }}^2\le \mathrm{\parallel }T{\mathrm{\parallel }}^2\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$, so $\mathrm{\parallel }T\otimes I\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$.

If $T=0$, both norms are zero. Otherwise, for every $\epsilon >0$ the supremum definition in [F2] gives a unit vector $u\in {\mathcal{H}}_{\mathrm{\Lambda }}$ with $\mathrm{\parallel }Tu\mathrm{\parallel }>\mathrm{\parallel }T\mathrm{\parallel }-\epsilon$. Since $q\ge 1$, $K$ is nonzero; choose a unit configuration vector $v\in K$. Then $\mathrm{\parallel }(T\otimes I)(u\otimes v)\mathrm{\parallel }=\mathrm{\parallel }Tu\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }>\mathrm{\parallel }T\mathrm{\parallel }-\epsilon$, so $\mathrm{\parallel }T\otimes I\mathrm{\parallel }\ge \mathrm{\parallel }T\mathrm{\parallel }-\epsilon$ for every $\epsilon >0$.

Letting $\epsilon \downarrow 0$ in step 1.3 and combining with step 1.2 gives $\mathrm{\parallel }{\iota }_{\mathrm{\Lambda },{\mathrm{\Lambda }}^{\mathrm{\prime }}}(T)\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$. Step 1.1 supplies the unital $\ast$-homomorphism and composition identities, completing the claim.

For $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$ and $B\in {\mathcal{A}}_{\mathrm{\Delta }}$, both embed into ${\mathcal{A}}_{\mathrm{\Lambda }\cup \mathrm{\Delta }}$ by [F1]. Thus their sum, product and adjoint are defined in a common finite matrix algebra, so ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ is a unital $\ast$-algebra. If $A$ has two localization regions, embedding into their union and isometry [F1] show that both assigned norms equal its norm in that common algebra; hence the local norm is well defined.

By [F3] and [F4], the norm completion is a Banach space and sums, scalar multiples and the limit norm are well defined on Cauchy-sequence classes; the constant sequence at the local unit is its candidate unit, and the local algebra is dense by the definition of completion.

In any common finite matrix algebra, the operator norm is submultiplicative, the adjoint preserves norm, and [F2] supplies the C*-identity $\mathrm{\parallel }{A}^{\ast }A\mathrm{\parallel }=\mathrm{\parallel }A{\mathrm{\parallel }}^2$. Isometric inclusions [F1] transfer these identities to ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$, so it is a normed $\ast$-algebra with the C*-identity.

Let $(A_n)$ and $(B_n)$ be Cauchy sequences in ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$. Each is bounded: after some index its terms lie within one unit of a fixed term, and the finitely many earlier terms have a finite maximum norm. Then $\mathrm{\parallel }{A}_n{B}_n-A_m{B}_m\mathrm{\parallel }\le \mathrm{\parallel }{A}_n\mathrm{\parallel }\mathrm{\parallel }{B}_n-B_m\mathrm{\parallel }+\mathrm{\parallel }{A}_n-A_m\mathrm{\parallel }\mathrm{\parallel }{B}_m\mathrm{\parallel }$ shows that $(A_n{B}_n)$ is Cauchy. If either sequence is replaced by an equivalent one, the same estimate makes the product sequences equivalent. Thus multiplication $[A_n][B_n]:=[A_n{B}_n]$ is well defined on the completion and obeys $\mathrm{\parallel }xy\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$.

Since $\mathrm{\parallel }{A}_n^{\ast }-A_m^{\ast }\mathrm{\parallel }=\mathrm{\parallel }{A}_n-A_m\mathrm{\parallel }$ in the local algebra by step 2.1, $[A_n{]}^{\ast }:=[A_n^{\ast }]$ is a well-defined involution on the completion. The Cauchy-sequence identities for product and adjoint show $(xy{)}^{\ast }=y^{\ast }{x}^{\ast }$ and associativity; continuity of multiplication and the limit norm pass $\mathrm{\parallel }{x}^{\ast }x\mathrm{\parallel }=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ from [F2] to every $x$ in the completion. The unit class $[I_{\mathrm{l}\mathrm{o}\mathrm{c}}]$ acts as an identity by termwise multiplication and has norm one by isometry, so $\mathcal{A}$ satisfies the C*-algebra definition in [F5].

If $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$ and $B\in {\mathcal{A}}_{\mathrm{\Delta }}$ with $\mathrm{\Lambda }\cap \mathrm{\Delta }=\mathrm{\varnothing }$, in ${\mathcal{A}}_{\mathrm{\Lambda }\cup \mathrm{\Delta }}$ they have forms $A_{\mathrm{\Lambda }}\otimes I_{\mathrm{\Delta }}$ and $I_{\mathrm{\Lambda }}\otimes B_{\mathrm{\Delta }}$. Both products equal $A_{\mathrm{\Lambda }}\otimes B_{\mathrm{\Delta }}$, so $AB=BA$. This proves locality for disjoint finite local factors.

For $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$, define ${\omega }_0(A):={\mathrm{Tr}}_{\mathrm{\Lambda }}({\rho }_{\mathrm{\Lambda }}A)$. This is independent of the chosen localization region: if $A$ is also viewed in ${\mathcal{A}}_{\mathrm{\Delta }}$, embed both representatives into ${\mathcal{A}}_{\mathrm{\Lambda }\cup \mathrm{\Delta }}$; the two partial-trace consistency equations make both pairings equal to the one computed using ${\rho }_{\mathrm{\Lambda }\cup \mathrm{\Delta }}$. Thus ${\omega }_0$ is a well-defined functional on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$.

Linearity follows by placing any two local arguments and their sum in one common finite region. The identity has value ${\mathrm{Tr}}_{\mathrm{\Lambda }}({\rho }_{\mathrm{\Lambda }}I)=1$, and for local $A$, [F2] gives ${\omega }_0(A^{\ast }A)={\mathrm{Tr}}_{\mathrm{\Lambda }}({\rho }_{\mathrm{\Lambda }}{A}^{\ast }A)\ge 0$. Therefore ${\omega }_0$ is normalized and positive on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$.

For local $A$, its restriction is the normalized finite-matrix state from [F2], whose functional norm is one. Hence $\mathrm{\mid }{\omega }_0(A)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }$.

The bound of step 3.1 makes ${\omega }_0$ continuous and uniquely extendible from the dense subalgebra ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ to a bounded linear functional $\omega$ on $\mathcal{A}$, with $\mathrm{\parallel }\omega \mathrm{\parallel }\le 1$. It remains normalized because $I\in {\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and ${\omega }_0(I)=1$; since $\mathrm{\parallel }I\mathrm{\parallel }=1$, this also gives $\mathrm{\parallel }\omega \mathrm{\parallel }\ge 1$, hence $\mathrm{\parallel }\omega \mathrm{\parallel }=1$. For any $X\in \mathcal{A}$, choose local $X_n\to X$; then $X_n^{\ast }{X}_n\to X^{\ast }X$ by continuity of multiplication and adjoint, and $\omega (X^{\ast }X)={\lim }_n{\omega }_0(X_n^{\ast }{X}_n)\ge 0$. Thus $\omega$ is an infinite algebraic state.

Any state with the prescribed local restrictions agrees with ${\omega }_0$ on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and hence agrees with $\omega$ on its norm closure, since both are continuous and ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ is dense. The extension is unique.

For each $m\in \mathbb{N}$, let $R_m$ be the set of ranges of all $m$-tuples in ${\mathrm{\Gamma }}^m$, including tuples with repeated entries. Since ${\mathrm{\Gamma }}^m$ is at most countable by [F1], a surjection $\mathbb{N}\to {\mathrm{\Gamma }}^m$ composed with the range map shows $R_m$ is at most countable by [F2]. Every finite region is the range of a tuple of its cardinality, with the empty region the range of the unique $0$-tuple. Thus all finite regions form the countable union ${\bigcup }_m{R}_m$, which is at most countable by [F1] and [A1]. For each region $\mathrm{\Lambda }$, rational real and imaginary matrix entries give a countable subset of ${\mathcal{A}}_{\mathrm{\Lambda }}$: it is a finite Cartesian power of $\mathbb{Q}$, so [F1]–[F2] apply. These matrices are norm dense: for a $D_{\mathrm{\Lambda }}\times D_{\mathrm{\Lambda }}$ error matrix $E$ whose entries have modulus at most $\delta$, Cauchy–Schwarz gives $\mathrm{\parallel }Ev{\mathrm{\parallel }}_2^2\le {\sum }_{i,j}\mathrm{\mid }{E}_{ij}{\mathrm{\mid }}^2\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\le D_{\mathrm{\Lambda }}^2{\delta }^2\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$, so $\mathrm{\parallel }E{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le D_{\mathrm{\Lambda }}\delta$. A second countable union, using [A1], therefore gives a countable norm-dense set $D$ in ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$, hence in its completion $\mathcal{A}$.

By [F2], choose a surjection $s:\mathbb{N}\to D$ and enumerate its values as $D_1,D_2,\dots$. For each fixed $j\ge 1$, the scalar sequence $z_n^{(j)}:={\omega }_n(D_j)$ is bounded: $D_j$ is local, and the finite-dimensional state/density interface [F3] gives $\mathrm{\mid }{\omega }_n(D_j)\mathrm{\mid }\le \mathrm{\parallel }{D}_j\mathrm{\parallel }$. The same estimate extends each ${\omega }_n$ from the dense local algebra to all of $\mathcal{A}$, so $\mathrm{\parallel }{\omega }_n\mathrm{\parallel }=1$ by normalization.

We specify a deterministic convergent-subsequence rule for any bounded complex sequence $z_n$. Choose $M>0$ with $z_n\in [-M,M{]}^2$ under $\mathbb{C}\cong {\mathbb{R}}^2$. Subdivide the current closed square into its four closed equal subsquares in fixed lexicographic order; at least one contains infinitely many terms, and the rule takes the first such square. At the first stage choose the least index whose term lies in that square; at every later stage choose the least index exceeding the previous one whose term lies in the selected square. Writing the nested squares as $I_j\times J_j$, their closed coordinate intervals are nested and have lengths tending to zero. [F5] gives unique $x\in {\bigcap }_j{I}_j$ and $y\in {\bigcap }_j{J}_j$, so $(x,y)$ is the unique common point of the squares. The subsequence terms after stage $j$ lie in the $j$th square, whose diameter tends to zero, hence converge to $(x,y)$. Every choice in this rule is determined by a finite ordering or a least natural index; it uses no additional choice axiom.

Apply the deterministic rule of step 1.3 recursively to the coordinate sequences $z_n^{(j)}$ along nested subsequences for $j\ge 1$. More precisely, start with $i_0(n)=n$; at stage $j$, write the resulting subsequence as $i_j(n)=i_{j-1}(s_j(n))$ with $s_j$ strictly increasing, so $s_j(j)\ge j$. By [F5] and recursion, these maps are defined for every positive $j$. Put $r_j=i_j(j)$ for $j\ge 1$. For $j\ge 2$, $r_j=i_{j-1}(s_j(j))\ge i_{j-1}(j)>i_{j-1}(j-1)=r_{j-1}$, so $(r_j{)}_{j\ge 1}$ is increasing. For fixed $m\ge 1$ and $j\ge m$, nestedness gives $r_j=i_m(t_{m,j})$ where $t_{m,j}\ge j$; hence these are tails of the convergent coordinate subsequence selected at stage $m$, and ${\omega }_{r_j}(D_m)$ converges.

For any $A\in {\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and $\epsilon >0$, choose $D_m$ with $\mathrm{\parallel }A-D_m\mathrm{\parallel }<\epsilon \mathrm{/}4$ by density. Since each ${\omega }_{r_j}$ has norm one, $\mathrm{\mid }{\omega }_{r_j}(A)-{\omega }_{r_k}(A)\mathrm{\mid }\le 2\mathrm{\parallel }A-D_m\mathrm{\parallel }+\mathrm{\mid }{\omega }_{r_j}(D_m)-{\omega }_{r_k}(D_m)\mathrm{\mid }\mathrm{.}$ The last term tends to zero as $j,k\to \mathrm{\infty }$ by step 2.1. Thus ${\omega }_{r_j}(A)$ is Cauchy and converges for every local $A$. Define ${\omega }_0(A):={\lim }_j{\omega }_{r_j}(A)$ on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$. For local $A,B$ and $\lambda \in \mathbb{C}$, the equalities ${\omega }_{r_j}(A+\lambda B)={\omega }_{r_j}(A)+\lambda {\omega }_{r_j}(B)$ pass to the limit, and ${\omega }_0(I)=1$. For every local $X$, ${\omega }_0(X^{\ast }X)={\lim }_j{\omega }_{r_j}(X^{\ast }X)\ge 0$. Finally $\mathrm{\mid }{\omega }_0(A)\mathrm{\mid }={\lim }_j\mathrm{\mid }{\omega }_{r_j}(A)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }$. Thus ${\omega }_0$ is a bounded normalized positive functional on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$.

On each finite region $\mathrm{\Lambda }$, write ${\rho }_{\mathrm{\Lambda }}^{(j)}$ for the density of ${\omega }_{r_j}{\mathrm{\mid }}_{{\mathcal{A}}_{\mathrm{\Lambda }}}$. For each matrix unit $E_{ab}$, its entries are rational, so $E_{ba}\in D$; the trace-pairing formula gives $({\rho }_{\mathrm{\Lambda }}^{(j)}{)}_{ab}={\omega }_{r_j}(E_{ba})$. Thus every matrix coefficient converges by step 2.1. The entrywise limit ${\rho }_{\mathrm{\Lambda }}$ is Hermitian and trace one because these are finite linear equalities; for every vector $v$, $⟨{\rho }_{\mathrm{\Lambda }}v,v⟩={\lim }_j⟨{\rho }_{\mathrm{\Lambda }}^{(j)}v,v⟩\ge 0$, so it is positive. Every partial-trace consistency equation is a finite linear equation in these entries and therefore passes to the limit. Hence $\{{\rho }_{\mathrm{\Lambda }}\}$ is a consistent family.

By [F6], the consistent family of step 4.1 extends to a unique state $\omega$ on $\mathcal{A}$, and $\omega {\mathrm{\mid }}_{{\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}}={\omega }_0$. For any $A\in \mathcal{A}$ and $\epsilon >0$, choose local $B$ with $\mathrm{\parallel }A-B\mathrm{\parallel }<\epsilon \mathrm{/}4$. Then $\mathrm{\mid }{\omega }_{r_j}(A)-\omega (A)\mathrm{\mid }\le 2\mathrm{\parallel }A-B\mathrm{\parallel }+\mathrm{\mid }{\omega }_{r_j}(B)-\omega (B)\mathrm{\mid },$ whose right side is eventually less than $\epsilon$ by step 3.1. Therefore ${\omega }_{r_j}(A)\to \omega (A)$ for every $A\in \mathcal{A}$, proving sequential compactness in the pointwise topology.

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 $A\in \mathcal{A}$.