Finite-excitation vacuum spin representation and normality

Let $s\mathrm{\ne }t$ in $S_{\mathrm{f}\mathrm{i}\mathrm{n}}$. They differ at some site $x$; set $a=s_x$ and $Q:=\pi (E_x^{aa})$. Then $Q{e}_s=e_s$ and $Q{e}_t=0$. Since $TQ=QT$, self-adjointness of $Q$ gives $⟨T{e}_s,e_t⟩=⟨QT{e}_s,e_t⟩=⟨T{e}_s,Q{e}_t⟩=0.$ Thus every off-diagonal coefficient of $T{e}_s$ vanishes, so completeness of the specified orthonormal basis implies $T{e}_s=c_s{e}_s$ for some scalar $c_s$. The representation and completion are under the exact Countable Choice hypothesis A1.

If two configurations $s,t$ differ at only one site $x$, choose the matrix unit $E_x^{t_x{s}_x}$. By [F2], $\pi (E_x^{t_x{s}_x})e_s=e_t$. Commuting this operator through $T$ gives $c_t{e}_t=T{e}_t=T\pi (E_x^{t_x{s}_x})e_s=\pi (E_x^{t_x{s}_x})T{e}_s=c_s{e}_t$, hence $c_t=c_s$. Any $s,t\in S_{\mathrm{f}\mathrm{i}\mathrm{n}}$ differ on a finite set, and changing those finitely many labels one site at a time connects them; therefore all $c_s$ equal a single scalar $c$.

Step 2.1 proves every $T$ in the commutant is $cI$, while scalars commute with all operators, so $\pi (\mathcal{A}{)}^{\mathrm{\prime }}=\mathbb{C}I$. The definition of bicommutant then gives $\pi (\mathcal{A}{)}^{\mathrm{\prime }\mathrm{\prime }}=(\mathbb{C}I{)}^{\mathrm{\prime }}=\mathcal{B}({\mathcal{H}}_{\mathrm{v}\mathrm{a}\mathrm{c}})$. Faithfulness follows from the isometry established in Finite-excitation vacuum spin representation and normality.

The finite Hermitian spectral calculus [F11] diagonalizes $h$ with real eigenvalues; exponentiating them gives $Z>0$ and a positive trace-one matrix $r$. For nested finite regions $\mathrm{\Lambda }\subseteq {\mathrm{\Lambda }}^{\mathrm{\prime }}$, the product-basis trace factorization and $\mathrm{Tr}r=1$ give ${\mathrm{Tr}}_{{\mathrm{\Lambda }}^{\mathrm{\prime }}\setminus \mathrm{\Lambda }}(r^{\otimes {\mathrm{\Lambda }}^{\mathrm{\prime }}})=r^{\otimes \mathrm{\Lambda }}$. Thus the family is consistent, and [F5] defines a unique state ${\omega }_r$ with the displayed local restrictions.

Let ${\mathrm{\Lambda }}_n=[-n,n{]}^d\cap \mathrm{\Gamma }$. In a finite-volume product eigenbasis, $H_{{\mathrm{\Lambda }}_n}={\sum }_{x\in {\mathrm{\Lambda }}_n}{h}_x$ has eigenvalues equal to sums of the corresponding one-site eigenvalues by [F11]. Therefore $e^{-\beta H_{{\mathrm{\Lambda }}_n}}=(e^{-\beta h}{)}^{\otimes {\mathrm{\Lambda }}_n}$ and its partition function is $Z^{\mathrm{\mid }{\mathrm{\Lambda }}_n\mathrm{\mid }}$, so its normalized Gibbs density is $r^{\otimes {\mathrm{\Lambda }}_n}$. For every finite $X\subseteq \mathrm{\Gamma }$, define ${\rho }_X^{(n)}:={\mathrm{Tr}}_{(X\cup {\mathrm{\Lambda }}_n)\setminus X}\text{\hspace{0.17em}⁣}(r^{\otimes {\mathrm{\Lambda }}_n}\otimes (q^{-1}I{)}^{\otimes (X\setminus {\mathrm{\Lambda }}_n)}),$ placing the tensor factors on ${\mathrm{\Lambda }}_n$ and $X\setminus {\mathrm{\Lambda }}_n$ respectively. These densities are consistent: removing a site outside ${\mathrm{\Lambda }}_n$ traces a factor $q^{-1}I$ to $1$, while tracing sites in ${\mathrm{\Lambda }}_n$ commutes with the displayed partial trace because both are finite tensor-basis sums. Therefore [F5] gives an algebraic state ${\omega }_n$ on $\mathcal{A}$. If $A$ is local in ${\mathcal{A}}_X$, then for all sufficiently large $n$, $X\subseteq {\mathrm{\Lambda }}_n$ and ${\omega }_n(A)={\omega }_r(A)$. For arbitrary $A\in \mathcal{A}$, choose local $C$ with $\mathrm{\parallel }A-C\mathrm{\parallel }<\epsilon \mathrm{/}4$ using [F2]; since both states have norm one, $\mathrm{\mid }{\omega }_n(A)-{\omega }_r(A)\mathrm{\mid }\le 2\mathrm{\parallel }A-C\mathrm{\parallel }+\mathrm{\mid }{\omega }_n(C)-{\omega }_r(C)\mathrm{\mid }<\epsilon$ for all sufficiently large $n$. Hence ${\omega }_n\to {\omega }_r$ pointwise on $\mathcal{A}$.

Applying the local-pair conclusion of [F7] to this particular pointwise-convergent finite Gibbs sequence gives, for every local $C,D$, a bounded continuous strip function $F_{C,D}$, holomorphic inside, with lower and upper edges ${\omega }_r(C{\alpha }_t(D))$ and ${\omega }_r({\alpha }_t(D)C)$. This is precisely the local conclusion of [F7]; no whole-algebra KMS conclusion is attributed to that supplier.

Let $A,B\in \mathcal{A}$ and choose local sequences $A_n\to A$, $B_n\to B$. On either boundary, the correlation difference for local pairs is bounded by $\mathrm{\parallel }{A}_n-A_m\mathrm{\parallel }\mathrm{\parallel }{B}_n\mathrm{\parallel }+\mathrm{\parallel }{A}_m\mathrm{\parallel }\mathrm{\parallel }{B}_n-B_m\mathrm{\parallel },$ because ${\omega }_r$ has norm one and ${\alpha }_t$ is isometric by [F4]. Decomposing the difference into the two local-pair differences $(A_n-A_m,B_n)$ and $(A_m,B_n-B_m)$, [F8] gives the same bound throughout the strip; its zero-edge uniqueness makes this decomposition equal to $F_{A_n,B_n}-F_{A_m,B_m}$. The right side tends to zero, since the convergent sequences have bounded norms. Thus $F_{A_n,B_n}$ is uniformly Cauchy on the closed strip. Its limit $F_{A,B}$ is bounded and continuous there and is holomorphic inside by [F10]. Norm continuity of ${\omega }_r$, ${\alpha }_t$, and multiplication gives the required lower and upper boundary values. Therefore ${\omega }_r$ is KMS for all $A,B\in \mathcal{A}$.

Let $\nu$ be any $\beta$-KMS state on $\mathcal{A}$. For a finite region $\mathrm{\Lambda }$, every larger finite-volume Hamiltonian has the form $H_{{\mathrm{\Lambda }}^{\mathrm{\prime }}}=H_{\mathrm{\Lambda }}+K_{{\mathrm{\Lambda }}^{\mathrm{\prime }}\setminus \mathrm{\Lambda }}$, where $K_{{\mathrm{\Lambda }}^{\mathrm{\prime }}\setminus \mathrm{\Lambda }}$ is a sum of onsite terms supported outside $\mathrm{\Lambda }$ and therefore commutes with every $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$. Hence ${\alpha }_t^{{\mathrm{\Lambda }}^{\mathrm{\prime }}}(A)=e^{it{H}_{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}A{e}^{-it{H}_{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}$ whenever ${\mathrm{\Lambda }}^{\mathrm{\prime }}\supseteq \mathrm{\Lambda }$. Taking the thermodynamic limit in [F4] shows that ${\mathcal{A}}_{\mathrm{\Lambda }}$ is invariant under ${\alpha }_t$ and its restriction is exactly this finite-dimensional conjugation group. Restricting the KMS functions to $A,B\in {\mathcal{A}}_{\mathrm{\Lambda }}$ shows that $\nu {\mathrm{\mid }}_{{\mathcal{A}}_{\mathrm{\Lambda }}}$ is a finite-dimensional $\beta$-KMS state. By [F9], its density is $e^{-\beta H_{\mathrm{\Lambda }}}\mathrm{/}{Z}^{\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}=r^{\otimes \mathrm{\Lambda }}$. Thus $\nu$ and ${\omega }_r$ agree on every finite local algebra; continuity and density of ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ imply $\nu ={\omega }_r$. This proves uniqueness.

For $s\in S_{\mathrm{f}\mathrm{i}\mathrm{n}}$ and finite $F\subseteq \mathrm{\Gamma }$, let $P_F^s\in {\mathcal{A}}_F$ be the projection onto the one-dimensional local basis vector with configuration $s{\mathrm{\mid }}_F$. If $t\in S_{\mathrm{f}\mathrm{i}\mathrm{n}}$, then $\pi (P_F^s)e_t=e_t$ exactly when $t{\mathrm{\mid }}_F=s{\mathrm{\mid }}_F$, and otherwise it is zero. Thus, as $F$ increases over finite subsets ordered by inclusion, $\pi (P_F^s)$ decreases strongly to the rank-one projection $E_s:=\mathrm{\mid }{e}_s⟩⟨e_s\mathrm{\mid }$: each basis vector $e_t$ with $t\mathrm{\ne }s$ eventually fails at a site where it differs from $s$, and $e_s$ remains fixed. Since all these projections have norm at most one and the finite span of the basis is dense, convergence on basis vectors extends to strong convergence on every vector.

Let $k_s:=\mathrm{\mid }\{x:s_x\mathrm{\ne }0\}\mathrm{\mid }$. Once $F$ contains this finite support, the product-state expectation is ${\omega }_r(P_F^s)={\prod }_{x\in F}⟨s_x,r{s}_x⟩=({\prod }_{x:s_x\mathrm{\ne }0}⟨s_x,r{s}_x⟩)p_0^{\mathrm{\mid }F\mathrm{\mid }-k_s}\mathrm{.}$ Because $0<p_0<1$ and $\mathrm{\Gamma }$ is infinite, this tends to zero along the directed set of finite regions: for every $N$ all sufficiently large $F$ have $\mathrm{\mid }F\mathrm{\mid }-k_s\ge N$.

Suppose a normal state $\tilde{\omega }$ on $\mathcal{B}({\mathcal{H}}_{\mathrm{v}\mathrm{a}\mathrm{c}})$ extended ${\omega }_r$. The complements $Q_F:=I-\pi (P_F^s)$ are increasing projections. The projection onto the closure of the union of their increasing ranges is their least upper bound: it dominates each $Q_F$, and any upper bound dominates that closed span. Strong convergence in step 1.1 identifies this supremum as $I-E_s$. Normality and step 1.2 give $\tilde{\omega }(I-E_s)={\sup }_F\tilde{\omega }(I-\pi (P_F^s))={\sup }_F(1-{\omega }_r(P_F^s))=1,$ so $\tilde{\omega }(E_s)=0$ for every $s\in S_{\mathrm{f}\mathrm{i}\mathrm{n}}$. For finite $J\subseteq S_{\mathrm{f}\mathrm{i}\mathrm{n}}$, the projections $R_J:={\sum }_{s\in J}{E}_s$ increase to $I$ as $J$ ranges over finite subsets, because $(e_s)$ is a complete orthonormal basis. Applying normality again gives $1=\tilde{\omega }(I)={\sup }_J\tilde{\omega }(R_J)=0,$ a contradiction. Therefore no normal extension exists.

Suppose instead a positive trace-class density $\rho$ represented an extension. By [F4], for each $s$ and finite $F$, $\mathrm{Tr}(\rho \pi (P_F^s))={\sum }_j{\lambda }_j⟨\pi (P_F^s)u_j,u_j⟩\mathrm{.}$ For every fixed $N$, strong convergence in step 1.1 makes the first $N$ terms converge to ${\sum }_{j\le N}{\lambda }_j⟨E_s{u}_j,u_j⟩$. The tails of both the $\pi (P_F^s)$ sum and its proposed $E_s$ limit are bounded by ${\sum }_{j>N}{\lambda }_j$, because each operator is a projection. Hence their tail difference is at most $2{\sum }_{j>N}{\lambda }_j$, which tends to zero; therefore $\mathrm{Tr}(\rho \pi (P_F^s))\to \mathrm{Tr}(\rho E_s)$. Step 1.2 then gives $\mathrm{Tr}(\rho E_s)=0$ for each $s$. The same finite-tail argument applied to $R_J\uparrow I$ gives $\mathrm{Tr}(\rho I)={\sup }_J\mathrm{Tr}(\rho R_J)=0$, contradicting $\mathrm{Tr}\rho =1$. Thus no such trace-class density exists.