Strongly continuous thermodynamic spin automorphism group

Fix local $A\in {\mathcal{A}}_X$; if its displayed localization is empty, embed the scalar $A\in \mathbb{C}I$ into any one-site algebra, so take $X\mathrm{\ne }\mathrm{\varnothing }$. Let $C_n=\{x\in {\mathbb{Z}}^d:d_{\mathrm{\infty }}(x,0)\le n\}$. For all sufficiently large $n$, $C_n\supseteq X$, and for $m\ge n$ the boundary estimate [F3] bounds $\mathrm{\parallel }{\alpha }_t^{C_m}(A)-{\alpha }_t^{C_n}(A)\mathrm{\parallel }$ by its right side with $\mathrm{\Lambda }=C_n$. That exterior exponential tail tends to zero as $n\to \mathrm{\infty }$, uniformly for $\mathrm{\mid }t\mathrm{\mid }\le T$ because its time factor is bounded there. Hence ${\alpha }_t^{C_n}(A)$ is Cauchy uniformly on compact time intervals; completeness in [A1], [A2], [F1], and [F2] gives a limit ${\alpha }_t(A)\in \mathcal{A}$. Given any other increasing finite exhaustion ${\mathrm{\Lambda }}_j$, choose $n$ so large that both the canonical limit error and the [F3] bound for $C_n$ are below $\epsilon \mathrm{/}2$. Since $C_n$ is finite and ${\mathrm{\Lambda }}_j\uparrow {\mathbb{Z}}^d$, eventually $C_n\subseteq {\mathrm{\Lambda }}_j$; applying [F3] to this nested pair proves ${\alpha }_t^{{\mathrm{\Lambda }}_j}(A)\to {\alpha }_t(A)$ uniformly on $[-T,T]$. Thus the limit is independent of the exhaustion. The finite-matrix premise [A1] is supplied by Finite matrix C-star algebra and density-state duality; this proof uses that earlier result rather than reproving it.

Every finite-volume map is an isometric unital $\ast$-automorphism by conjugation with the unitary matrix exponential in [F4]. Norm limits of these complex-linear maps preserve addition and scalar multiplication, so the limit on local observables is complex-linear. The limits also preserve the unit, adjoints, products on local observables, and norm. For products, if local sequences $A_n\to A$ and $B_n\to B$ are used, submultiplicativity gives $\mathrm{\parallel }{A}_n{B}_n-AB\mathrm{\parallel }\le \mathrm{\parallel }{A}_n\mathrm{\parallel }\mathrm{\parallel }{B}_n-B\mathrm{\parallel }+\mathrm{\parallel }{A}_n-A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$, so multiplication passes continuously to the completion; the adjoint is norm-continuous as well. By density of ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and the isometry bound, each ${\alpha }_t$ extends uniquely to an isometric complex-linear unital $\ast$-homomorphism of $\mathcal{A}$.

For local $A$, ${\alpha }_0(A)={\lim }_{\mathrm{\Lambda }}{\alpha }_0^{\mathrm{\Lambda }}(A)=A$, and density extends ${\alpha }_0={\mathrm{id}}_{\mathcal{A}}$ to all $A\in \mathcal{A}$. For local $A$ and fixed $s,t$, choose local $C$ with $\mathrm{\parallel }C-{\alpha }_t(A)\mathrm{\parallel }<\epsilon$. For large $\mathrm{\Lambda }$, both ${\alpha }_t^{\mathrm{\Lambda }}(A)$ is within $\epsilon$ of ${\alpha }_t(A)$ and ${\alpha }_s^{\mathrm{\Lambda }}(C)$ is within $\epsilon$ of ${\alpha }_s(C)$. By isometry, replacing ${\alpha }_t^{\mathrm{\Lambda }}(A)$ with $C$ in ${\alpha }_s^{\mathrm{\Lambda }}{\alpha }_t^{\mathrm{\Lambda }}(A)$ costs at most $2\epsilon$, and replacing ${\alpha }_s^{\mathrm{\Lambda }}(C)$ by ${\alpha }_s(C)$ costs at most $\epsilon$. The finite-volume group law ${\alpha }_s^{\mathrm{\Lambda }}{\alpha }_t^{\mathrm{\Lambda }}(A)={\alpha }_{s+t}^{\mathrm{\Lambda }}(A)$ therefore gives $\mathrm{\parallel }{\alpha }_s{\alpha }_t(A)-{\alpha }_{s+t}(A)\mathrm{\parallel }\le 4\epsilon$ after taking limits. Letting $\epsilon \downarrow 0$ proves the group law on local $A$; density extends it to every $A\in \mathcal{A}$. In particular, ${\alpha }_{-t}$ is the inverse of ${\alpha }_t$, so each is bijective and hence an automorphism.

If $A\in {\mathcal{A}}_X$ is local, finite-volume differentiation gives $\mathrm{\parallel }{\alpha }_t^{\mathrm{\Lambda }}(A)-A\mathrm{\parallel }\le \frac{\mathrm{\mid }t\mathrm{\mid }}{\mathrm{\hslash }}\mathrm{\parallel }[H_{\mathrm{\Lambda }},A]\mathrm{\parallel }\le \frac{2\kappa \mathrm{\mid }X\mathrm{\mid }\mathrm{\parallel }A\mathrm{\parallel }}{\mathrm{\hslash }}\mathrm{\mid }t\mathrm{\mid },$ because only terms meeting $X$ contribute and ${\sum }_{Z\cap X\mathrm{\ne }\mathrm{\varnothing }}\mathrm{\parallel }\mathrm{\Phi }(Z)\mathrm{\parallel }\le \kappa \mathrm{\mid }X\mathrm{\mid }$. Passing to the limit gives continuity at $t=0$ for local $A$. For general $A\in \mathcal{A}$, choose local $C$ close to $A$ and use $\mathrm{\parallel }{\alpha }_t(A)-A\mathrm{\parallel }\le 2\mathrm{\parallel }A-C\mathrm{\parallel }+\mathrm{\parallel }{\alpha }_t(C)-C\mathrm{\parallel }$. The group law gives continuity at every time.