Thermodynamic spectral transport along a uniformly gapped path

First justify positivity in [F5] on the completion. For $C\in \mathcal{A}$, take local $C_m\to C$. Continuity of multiplication and contractivity give $E_{{\mathrm{\Lambda }}_n}(C^{\ast }C)={\lim }_m{E}_{{\mathrm{\Lambda }}_n}(C_m^{\ast }{C}_m)$. Each matrix on the right is positive by the local twirl; its norm limit is positive because its quadratic form on every vector is the limit of nonnegative quadratic forms. Thus $A\mapsto \mathrm{Tr}(\rho E_{{\mathrm{\Lambda }}_n}(A))$ is positive for every finite density $\rho$, and is bounded and normalized by contractivity and unitality. For each $n$, [F2] gives a norm-continuous self-adjoint $D_n(s)$. The finite-dimensional matrix ODE $U_n^{\mathrm{\prime }}=i{D}_n{U}_n$, $U_n(0)=I$, has a solution by its uniformly convergent Picard series; differentiating $U_n^{\ast }{U}_n$ gives $U_n^{\ast }{U}_n=I$. Differentiate $Q_n(s):=U_n(s{)}^{\ast }{P}_n(s)U_n(s)$. Since $U_n^{\mathrm{\prime }}=i{D}_n{U}_n$, $U_n^{{\ast }^{\mathrm{\prime }}}=-i{U}_n^{\ast }{D}_n$, and $P_n^{\mathrm{\prime }}=i[D_n,P_n]$, the three terms in $Q_n^{\mathrm{\prime }}$ cancel. Hence $Q_n(s)=P_n(0)$, or $P_n(s)=U_n(s)P_n(0)U_n(s{)}^{\ast }$. If ${\rho }_0$ is a finite-volume band density at $s=0$, then $U_n{\rho }_0{U}_n^{\ast }$ is a band density at $s$. The normalized-trace expectation satisfies $E_{{\mathrm{\Lambda }}_n}(U_n^{\ast }A{U}_n)=U_n^{\ast }{E}_{{\mathrm{\Lambda }}_n}(A)U_n$, first on finite tensor products and then by continuity. Cyclicity [F9] therefore gives ${\omega }_{{\rho }_0}^{\mathrm{\Lambda }}\circ {\alpha }_{n,s}={\omega }_{U_n{\rho }_0{U}_n^{\ast }}^{\mathrm{\Lambda }}$. The density/state correspondence [F7, F10] makes both extensions states on $\mathcal{A}$. Reversing $U_n$ proves equality of the finite-volume state sets in both directions.

By [F3], the generator interactions ${\mathrm{\Psi }}_{{\mathrm{\Lambda }}_n}$ have a common $F_{\rho }$-norm bound $K_{\rho }$ for every $\rho >d$. The path $U_n^{\mathrm{\prime }}=i{D}_n{U}_n$ is the Heisenberg flow of the interaction $-{\mathrm{\Psi }}_{{\mathrm{\Lambda }}_n}$ with $\mathrm{\hslash }=1$; its inverse is the reversed path with the same norm bound.

Fix a local $A\in {\mathcal{A}}_X$. For a finite ball $B_R\supset X$, let ${\alpha }_{n,s}^{(R)}$ be the flow generated by the restriction of $-{\mathrm{\Psi }}_{{\mathrm{\Lambda }}_n}$ to supports $Z\subseteq B_R$. For two finite-volume propagators $U_J,U_K$, set $F(u)={\alpha }_{u,0}^J({\alpha }_{s,u}^K(A))$. The composition rule gives $F(0)={\alpha }_{s,0}^K(A)$ and $F(s)={\alpha }_{s,0}^J(A)$. The endpoint evolution equations are ${\mathrm{\partial }}_u{\alpha }_{u,0}^J(B)=i{\alpha }_{u,0}^J([J(u),B])$ and ${\mathrm{\partial }}_u{\alpha }_{s,u}^K(A)=-i[K(u),{\alpha }_{s,u}^K(A)]$. Product differentiation and integration therefore give ${\alpha }_{s,0}^J(A)-{\alpha }_{s,0}^K(A)=i{\int }_0^s{\alpha }_{u,0}^J\text{\hspace{0.17em}⁣}([J(u)-K(u),{\alpha }_{s,u}^K(A)])du\mathrm{.}$ Apply this with $J=-{\mathrm{\Psi }}_{{\mathrm{\Lambda }}_n}$ and $K=J{\mathrm{\mid }}_{Z\subseteq B_R}$. In the Duhamel integrand, the removed terms are exactly those ${\mathrm{\Psi }}_{{\mathrm{\Lambda }}_n}(Z,u)$ with $Z\subseteq ̸B_R$. For terms with $Z\cap X=\mathrm{\varnothing }$, the initial supports are disjoint, so [F4] gives the commutator estimate. Their total contribution is at most $L_{\rho }\mathrm{\parallel }A\mathrm{\parallel }{K}_{\rho }{\sum }_{x\in X}{\sum }_{y\in {\mathbb{Z}}^d}{F}_{\rho }(d_{\mathrm{\infty }}(x,y)){\sum }_{z\mathrm{\notin }{B}_R}{F}_{\rho }(d_{\mathrm{\infty }}(y,z)),$ where $L_{\rho }=2C_{F_{\rho }}^{-1}(e^{2C_{F_{\rho }}{K}_{\rho }}-1)$ is the uniform commutator coefficient from [F4]. For each removed support choose an outside site $z\in Z\setminus B_R$, sum over $y\in Z$, and use ${\sum }_{Z\ni y,z}\mathrm{\parallel }{\mathrm{\Psi }}_{{\mathrm{\Lambda }}_n}(Z,u)\mathrm{\parallel }\le K_{\rho }{F}_{\rho }(d_{\mathrm{\infty }}(y,z))$. The displayed tail tends to zero by dominated convergence: for fixed $y$, the inner tail tends to zero and is bounded by the full finite lattice sum of $F_{\rho }$, while the outer sum is finite by [F3]. If $Z\cap X\mathrm{\ne }\mathrm{\varnothing }$, the disjoint-support hypothesis of [F4] is unavailable, so use $\mathrm{\parallel }[{\alpha }_{s,u}^K(A),{\mathrm{\Psi }}_{{\mathrm{\Lambda }}_n}(Z,u)]\mathrm{\parallel }\le 2\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }{\mathrm{\Psi }}_{{\mathrm{\Lambda }}_n}(Z,u)\mathrm{\parallel }$. Summing over $x\in Z\cap X$ and $z\in Z\setminus B_R$ bounds this contribution by $2\mathrm{\parallel }A\mathrm{\parallel }{K}_{\rho }{\sum }_{x\in X}{\sum }_{z\mathrm{\notin }{B}_R}{F}_{\rho }(d_{\mathrm{\infty }}(x,z)),$ which also tends to zero by summability. Thus the full-flow versus restricted-flow error tends to zero uniformly in $n,s$. For fixed $R$, only finitely many $Z\subseteq B_R$ occur. By [F3], their generator coefficients converge uniformly in $s$; the finite-dimensional Duhamel bound makes ${\alpha }_{n,s}^{(R)}(A)$ uniformly Cauchy. Combining this local convergence with the uniform tail shows ${\alpha }_{n,s}(A)$ is uniformly Cauchy as $n\to \mathrm{\infty }$. Applying the same argument to the inverse paths gives a uniform pointwise limit for ${\alpha }_{n,s}^{-1}$.

Each ${\alpha }_{n,s}$ and inverse preserves the norm and the $C^{\ast }$-operations. Their uniform limits on the dense local algebra extend by isometry to $\mathcal{A}$; continuity of multiplication and adjoint passes the $C^{\ast }$-operations to the limits. Taking limits in the two composition identities on local observables shows that the two limits are inverses. Thus ${\alpha }_s$ is an automorphism, norm-continuous in $s$ on each local observable. Taking limits in [F4] gives the stated $F_{\rho }$ commutator bound, with $K_{\rho }$ from [F3].

For every $n$, the normalized projection $P_n(s)\mathrm{/}\mathrm{Tr}{P}_n(s)$ is a band density, since the band is nonempty. [F5, F7, F10] extend its trace pairing to a state on $\mathcal{A}$, and [F8] gives at least one subsequential limit; hence each $\mathcal{S}(s)$ is nonempty. Let ${\omega }_0\in \mathcal{S}(0)$, witnessed by band states ${\omega }_k^0$ along a subsequence $n_k$. Step 1.1 gives ${\omega }_k^s={\omega }_k^0\circ {\alpha }_{n_k,s}\in {\mathcal{S}}_{{\mathrm{\Lambda }}_{n_k}}(s)$. For local $A$, $\mathrm{\mid }{\omega }_k^s(A)-({\omega }_0\circ {\alpha }_s)(A)\mathrm{\mid }\le \mathrm{\parallel }{\alpha }_{n_k,s}(A)-{\alpha }_s(A)\mathrm{\parallel }+\mathrm{\mid }{\omega }_k^0({\alpha }_s(A))-{\omega }_0({\alpha }_s(A))\mathrm{\mid },$ which tends to zero by pointwise convergence of the states and uniform norm convergence of the automorphisms. Density and the state norm-one bounds extend convergence to all $A\in \mathcal{A}$. Thus ${\omega }_0\circ {\alpha }_s\in \mathcal{S}(s)$. Conversely, for any ${\omega }_s\in \mathcal{S}(s)$, use the inverse finite-volume transport in step 1.1 and the convergent inverse flows from step 2.1 to obtain ${\omega }_s\circ {\alpha }_s^{-1}\in \mathcal{S}(0)$. Both inclusions prove the asserted set equality.