Local spin matrix algebras and isometric inclusions — Examples

Direct matrix multiplication gives $X^{\ast }=X$, $Z^{\ast }=Z$, $X^2=Z^2=I$, and $\mathrm{\parallel }X\mathrm{\parallel }=\mathrm{\parallel }Z\mathrm{\parallel }=1$ because each preserves the Euclidean norm and is nonzero. By [F2], the embedded $X_x$ and $Z_y$ have the same adjoint, square and norm properties.

If $x<y$, lexicographic factor order gives $X_x=X\otimes I$ and $Z_y=I\otimes Z$, so $(X\otimes I)(I\otimes Z)=X\otimes Z=(I\otimes Z)(X\otimes I)$. If $y<x$, the factors are interchanged and both products are $Z\otimes X$. Thus in either ordering $X_x{Z}_y=Z_y{X}_x$.

The supports $\{x\}$ and $\{y\}$ are disjoint; the computed commutation is therefore the stated instance of the general disjoint-locality clause.

Fix any finite region $\mathrm{\Lambda }\subseteq \mathbb{Z}$. Enumerate the finite set $F:=\mathrm{\Lambda }\cap \{n\in \mathbb{Z}:0<n\}$ as $n_1,\dots ,n_r$, and put $S:={\sum }_{j=1}^r{n}_j$, taking $S=0$ when $r=0$. By induction on the number of terms, every finite sum of positive integers is nonnegative: the empty sum is $0$, and if a partial sum $R\ge 0$ then adding a positive integer $n$ gives $R+n>R\ge 0$, since $0<n$ translates to $R+0\le R+n$ by [F1], and equality would cancel $R$ in the additive group $\mathbb{Z}$ to give $0=n$. If $k\in F$, write $S=k+R$, where $R$ is the sum of the remaining positive terms; the same induction gives $R\ge 0$, and order-compatibility yields $S\ge k$. Set $m:=S+1$. Since $1>0$ by [F2] and $S\ge 0$, order-compatibility gives $m>S\ge 0$, so $m>0$. If $m\in \mathrm{\Lambda }$, then $m\in F$ and the preceding bound gives $S\ge m$, contradicting $m>S$. Thus every finite region omits a positive integer site.

If $M>N$, the triangle inequality and isometric inclusions give $\mathrm{\parallel }{H}_M-H_N\mathrm{\parallel }\le {\sum }_{n=N+1}^M{2}^{-n}<2^{-N}$. Thus $(H_N)$ is Cauchy in ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and converges to some $H\in \mathcal{A}$ by completion. Since each $H_N$ is self-adjoint and the adjoint extends continuously, $H^{\ast }=H$.

Fix $k\ge 1$ and define $X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ on ${\mathbb{C}}^2$, and let $X_k$ be its one-site embedding at site $k$. For $N\ge k$, every $Z_n$ with $n\mathrm{\ne }k$ commutes with $X_k$ by disjoint locality, while direct multiplication gives $[Z,X]=ZX-XZ=2iY$, where $Y=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)$ and $Y^{\ast }Y=I$. Hence $[H_N,X_k]=2^{-k}(2iY{)}_k$ for all $N\ge k$. By continuity of multiplication and norm convergence, $[H,X_k]=2^{1-k}i{Y}_k$, whose norm is $2^{1-k}>0$.

If $H$ had a finite localization region $\mathrm{\Lambda }$, step 1.1 supplies $k\ge 1$ outside $\mathrm{\Lambda }$. Then [F4] would give $[H,X_k]=0$, contradicting step 2.1. Therefore $H$ has no finite localization region.

Each ${\rho }_{\mathrm{\Lambda }}$ is positive because it is a positive scalar multiple of the identity. Its trace is $q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{\mathrm{Tr}}_{\mathrm{\Lambda }}(I_{\mathrm{\Lambda }})=1$ by [F1], including $\mathrm{\Lambda }=\mathrm{\varnothing }$ where $q^0=1$.

Put $\mathrm{\Delta }={\mathrm{\Lambda }}^{\mathrm{\prime }}\setminus \mathrm{\Lambda }$ and $r=\mathrm{\mid }\mathrm{\Delta }\mathrm{\mid }$. For every $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$, the defining partial-trace pairing gives ${\mathrm{Tr}}_{\mathrm{\Lambda }}\text{\hspace{0.17em}⁣}(A{\mathrm{Tr}}_{\mathrm{\Delta }}(I_{\mathrm{\Lambda }}\otimes I_{\mathrm{\Delta }}))={\mathrm{Tr}}_{{\mathrm{\Lambda }}^{\mathrm{\prime }}}((A\otimes I_{\mathrm{\Delta }})(I_{\mathrm{\Lambda }}\otimes I_{\mathrm{\Delta }}))=q^r{\mathrm{Tr}}_{\mathrm{\Lambda }}(A),$ where the last equality follows by summing the diagonal matrix coefficients over a configuration basis of ${\mathcal{H}}_{\mathrm{\Delta }}$, which has $q^r$ elements. The trace pairing is nondegenerate: if $\mathrm{Tr}(AT)=0$ for every matrix $A$, taking $A=E_{ji}$ makes each entry $T_{ij}$ zero. Hence ${\mathrm{Tr}}_{\mathrm{\Delta }}(I_{\mathrm{\Lambda }}\otimes I_{\mathrm{\Delta }})=q^r{I}_{\mathrm{\Lambda }}$. Since $\mathrm{\mid }{\mathrm{\Lambda }}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }+r$, ${\mathrm{Tr}}_{\mathrm{\Delta }}({\rho }_{{\mathrm{\Lambda }}^{\mathrm{\prime }}})=q^{-\mathrm{\mid }{\mathrm{\Lambda }}^{\mathrm{\prime }}\mathrm{\mid }}{q}^r{I}_{\mathrm{\Lambda }}=q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{I}_{\mathrm{\Lambda }}={\rho }_{\mathrm{\Lambda }}\mathrm{.}$ Thus the family is consistent for every inclusion, including equality of regions ($r=0$).

By [F3], this family extends to a unique algebraic state $\omega$ on $\mathcal{A}$, and its local restriction is $\omega (A)={\mathrm{Tr}}_{\mathrm{\Lambda }}({\rho }_{\mathrm{\Lambda }}A)=q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{\mathrm{Tr}}_{\mathrm{\Lambda }}(A)$ for $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$.

If $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$ and $B\in {\mathcal{A}}_{\mathrm{\Delta }}$ for disjoint finite regions, choose orthonormal configuration bases $(e_i)$ and $(f_j)$ for their local Hilbert spaces. The tensor basis gives ${\mathrm{Tr}}_{\mathrm{\Lambda }\cup \mathrm{\Delta }}(A\otimes B)={\sum }_{i,j}⟨A{e}_i,e_i⟩⟨B{f}_j,f_j⟩={\mathrm{Tr}}_{\mathrm{\Lambda }}(A){\mathrm{Tr}}_{\mathrm{\Delta }}(B)\mathrm{.}$ Substitution into step 2.1 gives $\omega (AB)=\omega (A)\omega (B)$ on these disjoint local factors; this product formula follows from the displayed density family and is not an assumption about arbitrary states.