Quasi-local algebra completion and locality

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.