Invertible Subalgebras

Abstract

We introduce invertible subalgebras of local operator algebras on lattices. An invertible subalgebra is defined to be one such that every local operator can be locally expressed by elements of the inveritible subalgebra and those of the commutant. On a two-dimensional lattice, an invertible subalgebra hosts a chiral anyon theory by a commuting Hamiltonian, which is believed to be impossible on any full local operator algebra. We prove that the stable equivalence classes of ${\textsf{d}}$-dimensional invertible subalgebras form an abelian group under tensor product, isomorphic to the group of all ${\textsf{d}}+1$ dimensional quantum cellular automata (QCA) modulo blending equivalence and shifts. In an appendix, we consider a metric on the group of all QCA on infinite lattices and prove that the metric completion contains the time evolution by local Hamiltonians, which is only approximately locality-preserving. Our metric topology is strictly finer than the strong topology.

Appendices

Appendix A: Topology of Linear Transformations on Local Operator Algebras

We are going to show that time evolution by a local Hamiltonian for any given evolution time is a limit of a sequence of QCA, each of which is a finite depth quantum circuit. This formalizes the statement that quantum circuits approximate Hamiltonian time evolution, which is nontrivial since our operator space is infinite dimensional. The content of this statement may depend significantly on the topology of the space to which the time evolution and quantum circuits belong. By choosing a sufficiently fine topology on this space, we wish to settle on an interesting and fruitful class of approximately locality-preserving $*$-automorphisms. This perspective is perhaps complementary to the result of [24]. We will comment on it after A.17 below.

Our exposition will contain some standard notions and results such as trace, norm, local Hamiltonians, derivations, and time evolution. These can all be found in a textbook [5], but we remain as elementary as possible.

We simply write $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ instead of $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}}^{\textsf{d}},p)$ when the spatial dimension ${\textsf{d}}$ and local dimension assignment p are not important. Nowhere in our analysis will the local dimension appear. The operator space $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ and its completion $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$ are always endowed with the standard norm topology. Consider a real vector space

$$\begin{aligned} {\mathcal {L}}= \{ \varphi : \mathop {\mathrm {{\textsf{Mat}}}}\limits \rightarrow \mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits ~|~ \varphi \text { is } *\text {-linear.} \} \end{aligned}$$

(38)

No locality-preserving property is enforced, though we will primarily interested in those with a locality-preserving property. Every $*$-algebra automorphism of $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ is a member of ${\mathcal {L}}$. As we will see, a derivation $x \mapsto [{\textbf{i}}H,x]$ by a local Hamiltonian H is a member of ${\mathcal {L}}$.

Definition A.1

For any $\alpha ,\beta \in {\mathcal {L}}$, we define

$$\begin{aligned} {{\,\mathrm{{dist}}\,}}(\alpha ,\beta ) = \sup _{x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits \setminus {\mathbb {C}}{\textbf{1}}} \frac{\Vert \alpha (x) - \beta (x)\Vert }{\Vert x\Vert \cdot |\mathop {\mathrm {{Supp}}}\limits (x)|} \end{aligned}$$

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where $|\mathop {\mathrm {{Supp}}}\limits (x)|$ denotes the number of sites in $\mathop {\mathrm {{Supp}}}\limits (x)$. Note that the supremum is taken over all finitely supported elements of $\mathop {\mathrm {{\textsf{Mat}}}}\limits$.

The appearance of $\Vert x\Vert$ in the denominator is necessary, because, otherwise, the supremum will be simply infinite as seen by scaling x by a scalar. The choice of the other factor $|\mathop {\mathrm {{Supp}}}\limits (x)|$ is somewhat arbitrary, but there is some reason we would want such a factor that is a growing function in $|\mathop {\mathrm {{Supp}}}\limits (x)|$, as we will explain in A.12 below.

Our metric topology on ${\mathcal {L}}$ sits between the strong topology and the induced norm topology. The induced norm topology is obtained by removing $|\mathop {\mathrm {{Supp}}}\limits (x)|$ factor in the definition of ${{\,\mathrm{{dist}}\,}}$. It is equivalently defined as one in which $\alpha _i$ converges to $\beta$ if and only if $\lim _{i \rightarrow \infty } \sup _{x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}}^{\textsf{d}}): \Vert x\Vert = 1} \Vert \alpha _i(x) - \beta (x)\Vert = 0$. This topology is too fine to be useful; we will see in A.10 below that the time evolution operator of a noninteracting local Hamiltonian is not continuous in time. In the strong topology, a sequence of linear transformations $\alpha _i$ converges to another $\beta$ if and only if $\lim _{i \rightarrow \infty } \Vert \alpha _i(x) - \beta (x)\Vert = 0$ for any $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}}^{\textsf{d}})$. It is pointwise convergence. The strong topology is a conventional choice in a classic book by Bratteli and Robinson [5] when deriving time evolution by a lattice local Hamiltonian as a limit of those on finite subsystems of increasing volume. We will show in A.13 that our metric topology is strictly finer than the strong topology.

Proposition A.2

${\mathcal {L}}$ is a complete metric space under ${{\,\mathrm{{dist}}\,}}$.

Proof

First, we have to show that ${{\,\mathrm{{dist}}\,}}$ is a metric. It is obvious that ${{\,\mathrm{{dist}}\,}}(\alpha ,\beta ) = {{\,\mathrm{{dist}}\,}}(\beta ,\alpha )$. Suppose ${{\,\mathrm{{dist}}\,}}(\alpha ,\beta ) = 0$. Clearly, $\delta = \alpha - \beta = 0$ on $\mathop {\mathrm {{\textsf{Mat}}}}\limits \setminus {\mathbb {C}}{\textbf{1}}$. If $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits \setminus {\mathbb {C}}{\textbf{1}}$, we have $\Vert \delta ({\textbf{1}})\Vert = \Vert \delta ({\textbf{1}}- x) + \delta (x)\Vert \le \Vert \delta ({\textbf{1}}- x)\Vert + \Vert \delta (x)\Vert = 0$, so $\delta = 0$ on the entire $\mathop {\mathrm {{\textsf{Mat}}}}\limits$. Triangle inequality follows by

$$\begin{aligned} \frac{\Vert \alpha (x) - \gamma (x)\Vert }{\Vert x\Vert \cdot |\mathop {\mathrm {{Supp}}}\limits (x)| }&\le \frac{\Vert \alpha (x) - \beta (x)\Vert }{\Vert x\Vert \cdot |\mathop {\mathrm {{Supp}}}\limits (x)| } + \frac{\Vert \beta (x) - \gamma (x)\Vert }{\Vert x\Vert \cdot |\mathop {\mathrm {{Supp}}}\limits (x)| } \nonumber \\&\le \sup _y \frac{\Vert \alpha (y) - \beta (y)\Vert }{\Vert y\Vert \cdot |\mathop {\mathrm {{Supp}}}\limits (y)| } + \sup _z \frac{\Vert \beta (z) - \gamma (z)\Vert }{\Vert z\Vert \cdot |\mathop {\mathrm {{Supp}}}\limits (z)| }\nonumber \\&= {{\,\mathrm{{dist}}\,}}(\alpha ,\beta ) + {{\,\mathrm{{dist}}\,}}(\beta ,\gamma ) \end{aligned}$$

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and taking the supremum over x.

To show the completeness, let $\alpha _1,\alpha _2,\ldots \in {\mathcal {L}}$ be a Cauchy sequence under ${{\,\mathrm{{dist}}\,}}$. For any $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$, the sequence $\alpha _1(x),\alpha _2(x),\ldots \in \mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$ is Cauchy under $\Vert \cdot \Vert$, and hence converges in $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$. Define $\alpha (x) = \lim _{i \rightarrow \infty } \alpha _i(x)$. It is readily checked that $\alpha$ is $*$-linear and hence is a member of ${\mathcal {L}}$. $\square$

Lemma A.3

On the subset of ${\mathcal {L}}$, consisting of all norm-nonincreasing maps, the evaluation map $(\alpha ,x) \mapsto \alpha (x)$ is continuous.

Proof

Let $\epsilon > 0$. Let $(\alpha ,x)$ be in the domain of evaluation map. Let $z \in \mathop {\mathrm {{\textsf{Mat}}}}\limits \setminus {\mathbb {C}}{\textbf{1}}$ be a single-site operator such that $\Vert z\Vert \le 1$ and $x-z \notin {\mathbb {C}}{\textbf{1}}$. We take $\delta = \epsilon / (2 + (1+\Vert x\Vert )(1+|\mathop {\mathrm {{Supp}}}\limits (x)|))$ and an open neighborhood of $(\alpha ,x)$ consisting of $(\beta ,y)$ such that ${{\,\mathrm{{dist}}\,}}(\alpha ,\beta ) < \delta$ and $\Vert x- y\Vert < \delta$. Then,

$$\begin{aligned} \Vert \alpha (x) - \beta (y)\Vert\le & {} \Vert \alpha (x) - \beta (x)\Vert + \Vert \beta (x) - \beta (y)\Vert \nonumber \\\le & {} \Vert \alpha (x-z) - \beta (x-z)\Vert + \Vert \alpha (z) - \beta (z)\Vert + \Vert x-y\Vert \nonumber \\\le & {} {{\,\mathrm{{dist}}\,}}(\alpha , \beta )\Vert x-z\Vert (|\mathop {\mathrm {{Supp}}}\limits (x)| + 1) + {{\,\mathrm{{dist}}\,}}(\alpha ,\beta )\Vert z\Vert + \Vert x - y\Vert \nonumber \\< & {} \delta (1 + \Vert x\Vert )(1+|\mathop {\mathrm {{Supp}}}\limits (x)|) + \delta + \delta = \epsilon . \end{aligned}$$

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$\square$

Proposition A.4

Every $*$-algebra homomorphism from $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ into $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$, a member of ${\mathcal {L}}$, preserves trace and norm.

Proof

Let $\phi$ be a $*$-algebra homomorphism.

      The trace preserving property follows from the uniqueness of the trace and that $x \mapsto {{\,\mathrm{{{tr}}}\,}}(\phi (x))$ is a trace. Alternatively, we can prove it as follows. An arbitrary operator $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$ can be written as the sum of its hermitian part $\sum _i a_i \pi _i$ and antihermitian part ${\textbf{i}}\sum _i b_i \pi '_i$ where $a_i,b_i \in {\mathbb {R}}$ and $\pi _i = \pi _i^\dag = \pi _i^2$, $\pi '_i = (\pi '_i)^\dag = (\pi '_i)^2$ are projectors. Hence, we simply show that the trace of a finitely supported projector is preserved under $\phi$. Any finite supported projector is a sum of finitely many rank 1 mutually orthogonal projectors $\tau _j$ on its finite support, so we further reduce the proof to showing that ${{\,\mathrm{{{tr}}}\,}}(\phi (\tau _j)) = {{\,\mathrm{{{tr}}}\,}}(\tau _j)$. We can complete $\{\tau _i\}$ to a resolution of identity: $\sum _{j=1}^n \tau _j = {\textbf{1}}$. Note that ${{\,\mathrm{{{tr}}}\,}}(\tau _j) = {{\,\mathrm{{{tr}}}\,}}(\tau _{j+1}) = 1/n$. Observe that there is a unitary u such that $u \tau _j u^\dag = \tau _{j+1}$ for all j where $\tau _{n+1} = \tau _1$. Being a $*$-homomorphism, $\phi (u)$ is also a unitary: $\phi (u)^\dag \phi (u) = \phi (u^\dag u) = \phi ({\textbf{1}}) = {\textbf{1}}$. Therefore, ${{\,\mathrm{{{tr}}}\,}}(\phi (\tau _j)) = {{\,\mathrm{{{tr}}}\,}}(\phi ( u \tau _j u^\dag ) ) = {{\,\mathrm{{{tr}}}\,}}( \phi (\tau _{j+1}))$ and $\sum _j {{\,\mathrm{{{tr}}}\,}}(\phi (\tau _j)) = 1$ implies that ${{\,\mathrm{{{tr}}}\,}}(\phi (\tau _j)) = 1/n$. This complete the proof that $\phi$ preserves the trace.

      For norm, it suffices to show that $\phi$ preserves norm of positive semidefinite operators $y \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$, since $\Vert x\Vert ^2 = \Vert x^\dag x\Vert$ for all $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$. Let $y = \sum _{i=1}^n a_i \pi _i$ be the eigenvalue decomposition with orthogonal projectors $\pi _i$ which sum to ${\textbf{1}}$ and real numbers $a_1 \ge \cdots \ge a_n \ge 0$, so $a_1 = \Vert y\Vert$. Observe that $\phi$ maps mutually orthogonal projectors to mutually orthogonal projectors. So, $\phi (y) = \sum _i a_i \pi '_i$ where $\pi '_i$ are mutually orthogonal projectors. For any positive semidefinite $\rho \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$ with ${{\,\mathrm{{{tr}}}\,}}(\rho ) \le 1$, we have ${{\,\mathrm{{{tr}}}\,}}(\rho \pi '_i) = {{\,\mathrm{{{tr}}}\,}}(\pi '_i \rho \pi '_i) \ge 0$ that sum to ${{\,\mathrm{{{tr}}}\,}}(\rho )$. So, ${{\,\mathrm{{{tr}}}\,}}(\phi (y)\rho ) = \sum _i a_i {{\,\mathrm{{{tr}}}\,}}(\pi '_i \rho )$ is a convex combination of $a_i$. Therefore, $\Vert \phi (y)\Vert = \sup _\rho {{\,\mathrm{{{tr}}}\,}}(\phi (y) \rho ) \le a_1 = \Vert y \Vert$. To show the opposite inequality, let $\epsilon \in (0, 1)$ and choose $\tau \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$ such that $\Vert \pi '_1 - \tau \Vert < \epsilon$. Then, $\Vert \tau \Vert < 1 + \epsilon$, implying $\Vert \tau ^\dag \tau - \pi '_1\Vert \le \Vert \tau ^\dag \Vert \Vert \tau - \pi '_1\Vert + \Vert \tau ^\dag - \pi '_1\Vert \Vert \pi '_1\Vert< (1+\epsilon )\epsilon + \epsilon < 3 \epsilon$, implying ${{\,\mathrm{{{tr}}}\,}}(\tau ^\dag \tau ) < 1 + 3 \epsilon$. Put $\sigma = \tau ^\dag \tau / (1+3\epsilon ) \succeq 0$, so ${{\,\mathrm{{{tr}}}\,}}( \sigma ) < 1$. Then, ${{\,\mathrm{{{tr}}}\,}}(\phi (y) \sigma ) \ge a_1 - O(n \epsilon )$. Since $\epsilon$ was arbitrary, $\Vert \phi (y)\Vert = \sup _\rho {{\,\mathrm{{{tr}}}\,}}(\phi (y) \rho ) \ge a_1$. $\square$

1.1 A.1 Hamiltonian time evolution

By a bounded strength strictly local Hamiltonian H on $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ we mean a collection $\{h_X | X \subset {\mathbb {Z}}^{\textsf{d}}, \Vert h_X\Vert \le 1, \mathop {\mathrm {{Supp}}}\limits (h_X) \subseteq X \}$ of hermitian elements of $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ such that for some $R > 0$, called range, it holds that $h_X = 0$ whenever $\mathop {\mathrm {{diam}}}\limits X > R$. The collection may have infinitely many nonzero terms; otherwise, one would be studying finite systems. We write $H = \sum _X h_X$ where the sum is formal to denote the collection. The most important role of a bounded strength strictly local Hamiltonian is that it gives a locality-preserving ${\mathbb {C}}$-linear $*$-map (not an algebra homomorphism) from $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ to itself, denoted by $[{\textbf{i}}H, \cdot ]: y \mapsto [{\textbf{i}}H, y] = \sum _X {\textbf{i}}[h_X, y]$ where this sum is now meaningful since it is always finite for any given $y \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$. It belongs to ${\mathcal {L}}$ and satisfies the Leibniz rule:

$$\begin{aligned}{}[{\textbf{i}}H, x y] = [{\textbf{i}}H, x] y + x [{\textbf{i}}H, y]. \end{aligned}$$

(42)

A member $\beta \in {\mathcal {L}}$ is a derivation with spread $\ell$ if it obeys the Leibniz rule and $\mathop {\mathrm {{Supp}}}\limits (\beta (x)) \subseteq \mathop {\mathrm {{Supp}}}\limits (x)^{+\ell }$ for all $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$. Every bounded strength strictly local Hamiltonian gives a derivation.

      By abuse of notation, for any $t \in {\mathbb {R}}$ we will write $y \mapsto y(t)$, called time evolution by H where H is implicit in notation y(t), to mean

$$\begin{aligned} y(t)&= \lim _{n \rightarrow \infty } y(t)_n\nonumber \\ y(t)_n&= \sum _{k = 0}^n \frac{t^k}{k!} [{\textbf{i}}H, y]_k \qquad \text { where }[{\textbf{i}}H, y]_k = {\left\{ \begin{array}{ll} [{\textbf{i}}H,[{\textbf{i}}H, y]_{k-1}] &{} (k > 0)\\ y &{} (k=0) \end{array}\right. } . \end{aligned}$$

(43)

The power series may not converge in $\mathop {\mathrm {{\textsf{Mat}}}}\limits$, but we will show that it absolutely converges, at least for some nonzero t, in the norm completion $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$. Unless otherwise specified, $H = \sum _X h_X$ will denote a bounded strength strictly local Hamiltonian with range R on $\mathop {\mathrm {{\textsf{Mat}}}}\limits$.

Lemma A.5

For any integer $k > 0$ and any $y \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$,

$$\begin{aligned} \frac{\Vert [H, y]_k \Vert }{k!} \le C \zeta ^k |\mathop {\mathrm {{Supp}}}\limits (y)| \Vert y\Vert \end{aligned}$$

(44)

where $C,\zeta > 0$ depends only on ${\textsf{d}}$ and R.

Proof

If $y \in {\mathbb {C}}{\textbf{1}}$, there is nothing to prove. Let s denote a site in the support of $y \in \mathop {\mathrm {{\textsf{Mat}}}}\limits \setminus {\mathbb {C}}{\textbf{1}}$. Introduce a graph $G_s$ with nodes labeled by s and all nonzero terms $h_X$ of H where an edge is present iff two operators have overlapping supports or an operator’s support contains s. The degree of G is bounded because H is strictly local; an upper bound on the degree depends on ${\textsf{d}}$ and R. The k-th nested commutator is a k-fold finite sum

$$\begin{aligned} \sum _{X_1,\ldots ,X_k} [h_{X_1},\cdots ,[h_{X_k}, y]\cdots ], \end{aligned}$$

(45)

each term of which may not vanish only if $\{s,h_{X_1},h_{X_2},\ldots ,h_{X_k}\}$ defines a connected subgraph J of $G_s$ for some $s \in \mathop {\mathrm {{Supp}}}\limits (y)$. The norm of a term here is at most $2^k \Vert y\Vert$. Given a tuple $(s,X_1,\ldots ,X_k)$ we get a unique connected subgraph $J \subset G_s$ of $k+1$ nodes, but given a connected subgraph $J \ni s$ with $k+1$ nodes of $G_s$, there are at most k! different tuples. Therefore,

$$\begin{aligned} \frac{1}{k!} \Vert [H, y]_k \Vert \le \sum _{s \in \mathop {\mathrm {{Supp}}}\limits (y)} ~~\sum _{J: y \in J \subset G_s, \text { connected}, |J| = k+1 } 2^k \Vert y\Vert . \end{aligned}$$

(46)

Hence, it remains to show that the number of rooted, connected subgraphs of $G_s$ with $k+1$ nodes is at most exponential in k. This counting is well known [25, Lemma 2.1], and an upper bound is $(ed)^{k+1}$ where $e \approx 2.718$ and d is the maximum degree of $G_s$. $\square$

Using A.5, we set

$$\begin{aligned} t_0 = \frac{1}{128 \zeta }. \end{aligned}$$

(47)

By A.5, for $t \in (-t_0,t_0)$ the sequence $y(t)_0, y(t)_1, y(t)_2,\ldots$ is a Cauchy sequence, and hence converges in $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$ absolutely. Therefore, the short time evolution belongs to ${\mathcal {L}}$ and preserves ${\textbf{1}}$. The following inequalities will be handy to show that short time evolution is a $*$-automorphism of $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$.

Lemma A.6

There exist constants $C',C'' > 0$ such that for any $t,t' \in (-t_0, t_0)$, any $x,y\in \mathop {\mathrm {{\textsf{Mat}}}}\limits$, and any positive integers n, m such that $n \le m$, it holds that

$$\begin{aligned} \Vert y(t)_n(t')_m - y(t+t')_n \Vert&\le 2^{-n} C' \Vert y\Vert |\mathop {\mathrm {{Supp}}}\limits (y)| ,\nonumber \\ \Vert x(t)_n y(t)_n - (xy)(t)_n \Vert&\le 2^{-n} C'' \Vert x\Vert \Vert y\Vert |\mathop {\mathrm {{Supp}}}\limits (x)| |\mathop {\mathrm {{Supp}}}\limits (y)|. \end{aligned}$$

(48)

Proof

This is a straightforward calculation using A.5 and using the Leibniz rule, which implies

$$\begin{aligned} \sum _{j=0}^n \left( {\begin{array}{c}n\\ j\end{array}}\right) [{\textbf{i}}H, x]_j [{\textbf{i}}H, y]_{n-j} = [{\textbf{i}}H, xy]_n. \end{aligned}$$

(49)

$\square$

Corollary A.7

Any short time evolution by H for time t with $|t| < t_0$ is a $*$-algebra homomorphism from $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ to $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$ preserving trace and norm.

Proof

Use the second inequality of A.6. Apply A.4. $\square$

Being continuous, the map $y \mapsto y(t)$ can be extended to the entire $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$ uniquely: if ${\bar{y}} = \lim _{n \rightarrow \infty } y_n \in \mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$ where $y_n \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$, then we define

$$\begin{aligned} {\bar{y}}(t) = \lim _{n \rightarrow \infty } y_n(t). \end{aligned}$$

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This limit exists by the metric completeness because $\{y_n(t)\}_n$ is Cauchy whenever $\{y_n\}$ is, implied by A.7. It is obvious that ${\bar{y}}(t)$ does not depend on the sequence $y_n$ converging to ${\bar{y}}$.

Corollary A.8

For any bounded strength strictly local Hamiltonian H, the time evolution $U^H_t: x \mapsto x(t)$ for any $t \in (-t_0,t_0)$ gives a $*$-automorphism of $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$ preserving trace and norm.

Proof

An inverse exists: $x \mapsto x(t) \mapsto x(t)(-t)$ where $x(t)(-t) = x$ by A.6. Use A.7. $\square$

Although we have not shown that $U^H_t$ is well defined for arbitrary $t\in {\mathbb {R}}$ by the formula (43), since $U^H_{t'}$ for $|t'| < t_0$ is now an automorphism on $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$, we can define $U^H_t$ for arbitrary t by a composition of finitely many $(U^H_{t_0/2})^{\pm 1}$ and some $U^H_{t'}$ for $|t'| < t_0$. It follows from A.6 that $U^H_t U^H_{t'} = U^H_{t'} U^H_t$ for $t,t' \in (-t_0, t_0)$. Hence, the composition can have arbitrary order, and we have a group homomorphism from the additive group of ${\mathbb {R}}$ into the group of all $*$-automorphisms of $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits$, generated by H.¹

Lemma A.9

The path $t \mapsto U^H_t$ is continuous.

Proof

It suffices to show the continuity at $t=0$. Let $\epsilon > 0$. For any $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$, if $|t| < \min (\epsilon / 4 C \zeta , t_0)$ where the constants $C,\zeta$ are from A.5, then

$$\begin{aligned} \Vert U^H_t(x) - x\Vert= & {} \left\Vert {\sum _{k \ge 1} [{\textbf{i}}t H, x]_k / k!}\right\Vert \le C \frac{\zeta t}{ 1- \zeta |t| } \Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)|, \nonumber \\ {{\,\mathrm{{dist}}\,}}(U^H_t, U^H_0)= & {} \sup _{x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits \setminus {\mathbb {C}}{\textbf{1}}} \frac{\Vert U^H_t(x) - U^H_0(x)\Vert }{\Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)|} < \epsilon . \end{aligned}$$

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$\square$

Remark A.10

The seemingly trivial statement of A.9 would have been false if we used a different topology on ${\mathcal {L}}$. This will give some reason why we wanted the factor $|\mathop {\mathrm {{Supp}}}\limits (x)|$ in the denominator when we introduced ${{\,\mathrm{{dist}}\,}}$. Let $H = {\frac{1}{2}}\sum _j \sigma _j^Z$ be a Hamiltonian on $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}},p=2)$. This is a noninteracting Hamiltonian on a one-dimensional chain of qubits; being one-dimensional will be immaterial. For any positive integer n, let

$$\begin{aligned} g_n(\phi ) = {\frac{1}{2}}(|0^{\otimes n}\rangle + \phi |1^{\otimes n}\rangle )(\langle 0^{\otimes n}| + \phi ^{-1} \langle 1^{\otimes n}|) \end{aligned}$$

(52)

be a projector where $\phi$ is a complex phase factor of magnitude one. Here, $\sigma ^Z |0\rangle = |0\rangle$ and $\sigma ^Z |1\rangle = - |1\rangle$. The precise support of $g_n(\phi )$ does not matter. Calculation shows that $U_t^H(g_n(1)) = g_n(e^{{\textbf{i}}t n})$ and thus

$$\begin{aligned}&\sup _{x \ne 0} \frac{ \Vert U^H_t(x) - U^H_0(x)\Vert }{\Vert x\Vert } \ge \sup _n \Vert g_n(e^{{\textbf{i}}t n}) - g_n(1)\Vert \nonumber \\&\quad = {\left\{ \begin{array}{ll} 1 &{} \text { if }t\text { is an irrational multiple of }\pi \\ 0 &{} \text { if } t = 0 \end{array}\right. }. \end{aligned}$$

(53)

Therefore, in the induced norm topology of ${\mathcal {L}}$ the time evolution is not continuous in time. $\square$

Proposition A.11

The time evolution by any bounded strength strictly local Hamiltonian is differentiable with respect to time:

$$\begin{aligned} \lim _{t \rightarrow 0} {{\,\mathrm{{dist}}\,}}\left( \frac{1}{t} (U^H_{s+t} - U^H_s) , ~~U^H_s \circ [{\textbf{i}}H, \cdot ] \right) = 0. \end{aligned}$$

(54)

Proof

Since $U^H_s$ preserves ${{\,\mathrm{{dist}}\,}}$, it suffices to consider $s = 0$. From A.5, we see for $|t| < t_0$ and any $x\in \mathop {\mathrm {{\textsf{Mat}}}}\limits$,

$$\begin{aligned} \frac{1}{t} (U^H_t(x) - x) - [{\textbf{i}}H, x]= & {} \sum _{k = 2}^{\infty } \frac{t^{k-1}}{k!} [{\textbf{i}}H, x]_k, \nonumber \\ \left\Vert {\frac{1}{t} (U^H_t(x) - x) - [{\textbf{i}}H, x] }\right\Vert\le & {} \sum _{k \ge 2} |t| ^{k-1} C \zeta ^k \Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)| = \frac{ C \zeta ^2 |t|}{1 - \zeta |t|} \Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)|,n\nonumber \\ {{\,\mathrm{{dist}}\,}}\left( \frac{U^H_t - U^H_0}{t - 0}, [{\textbf{i}}H, \cdot ] \right)\le & {} \frac{C\zeta ^2|t|}{1-\zeta |t|} \xrightarrow {\quad t \rightarrow 0 \quad } 0. \end{aligned}$$

(55)

$\square$

Remark A.12

With the differentiability, we can give a stronger reason that we want the factor $|\mathop {\mathrm {{Supp}}}\limits (x)|$ in the definition of ${{\,\mathrm{{dist}}\,}}$. Consider the same Hamiltonian $H = \sum _j \sigma ^Z_j$ as in A.10. Let $x_n = \bigotimes _{j=1}^n \sigma ^X_j$ be a tensor product of Pauli matrices, which has $\Vert x_n\Vert = 1$ for all $n \ge 1$. The derivation $x \mapsto [{\textbf{i}}H,x]$ by H blows the norm up:

$$\begin{aligned}{}[{\textbf{i}}H, x_n]&= - 2 \sum _{j=1}^n \sigma ^Y_j \prod _{i \ne j} \sigma ^X_i,\nonumber \\ \Vert [{\textbf{i}}H, x_n] \Vert&= 2 n = 2 |\mathop {\mathrm {{Supp}}}\limits (x_n)|, \nonumber \\ \left\Vert {\frac{1}{t} (U^H_t(x_n) - x_n) - [{\textbf{i}}H, x_n] } \right\Vert&\ge 2 |\mathop {\mathrm {{Supp}}}\limits (x_n)| - 2/t \qquad (t > 0), \end{aligned}$$

(56)

where the norm calculation uses the fact that different summands in the first line commute. The third line shows that in order for $U^H_t$ to be differentiable in t with respect to a metric

$$\begin{aligned} {{\,\mathrm{{dist}}\,}}_\eta (\alpha ,\beta ) = \sup _{x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits \setminus {\mathbb {C}}{\textbf{1}}} \frac{\Vert \alpha (x) - \beta (x)\Vert }{\Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)|^\eta }, \end{aligned}$$

(57)

we must have $\eta \ge 1$. Our metric ${{\,\mathrm{{dist}}\,}}$ is ${{\,\mathrm{{dist}}\,}}_{\eta =1}$.

If we consider a sequence $x_n / \sqrt{n}$ that converges to zero as $n \rightarrow \infty$ in $\mathop {\mathrm {{\textsf{Mat}}}}\limits$ with the norm topology, we see that $\lim _{n \rightarrow \infty } \Vert [{\textbf{i}}H, x_n / \sqrt{n}]\Vert = \infty$. This shows that the derivation by H is not norm-continuous. This may be thought of as the origin that we need the factor $|\mathop {\mathrm {{Supp}}}\limits (x)|$ in ${{\,\mathrm{{dist}}\,}}$.  $\square$

Remark A.13

Our metric topology is strictly finer than the strong topology. To show this, we find a sequence in ${\mathcal {L}}$ which converges in the strong topology, but not in our ${{\,\mathrm{{dist}}\,}}$ topology. Consider a depth 1 quantum circuit P consisting of single-site unitaries $\sigma ^X_j$ (a Pauli operator) on every site j of $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}},p=2)$. For each integer $n > 0$, we define $P_n = \prod _{j=-n}^n \sigma ^X_j$, a unitary of $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}},2)$. For any $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}},2)$ we know x is supported on $[-n_x, n_x] \subset {\mathbb {Z}}$ for some $n_x > 0$. Hence, $P_n(x) \equiv P_n x P_n^\dag = P(x)$ if $n > n_x$. In particular, for all $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$

$$\begin{aligned} \lim _{n \rightarrow \infty } \Vert P_n(x) - P(x)\Vert = 0. \end{aligned}$$

(58)

That is, the sequence $\{ P_n \in {\mathcal {L}}\}$ converges to $P \in {\mathcal {L}}$ in the strong topology. However, $P_n(\sigma ^Z_{k}) = \sigma ^Z_{k}$ if $k > n$ whereas $P(\sigma ^Z_{k}) = - \sigma ^Z_{k}$. Therefore,

$$\begin{aligned} {{\,\mathrm{{dist}}\,}}(P_n, P) = \sup _{x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits \setminus {\mathbb {C}}{\textbf{1}}} \frac{\Vert P_n(x) - P(x)\Vert }{\Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)|} \ge \sup _k \Vert P_n(\sigma ^Z_k) - P(\sigma ^Z_k)\Vert = 2, \end{aligned}$$

(59)

which does not converge to zero as $n \rightarrow \infty$. $\square$

1.2 A.2 A limit of quantum circuits

Given any region $\Omega \subseteq {\mathbb {Z}}^{\textsf{d}}$, we define

$$\begin{aligned} H_\Omega = \sum _{X \subseteq \Omega } h_X, \end{aligned}$$

(60)

the collection of all terms of H supported on $\Omega$.

Lemma A.14

(A Lieb–Robinson bound [27]). For any $t \in (-t_0,t_0)$ and any $\Omega \subseteq {\mathbb {Z}}^{\textsf{d}}$,

$$\begin{aligned} \Vert U^{H}_t(x) - U^{H_\Omega }_t(x) \Vert \le C (|t| / 4 t_0)^{L/R} \Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)| \end{aligned}$$

(61)

where L is the distance between $\mathop {\mathrm {{Supp}}}\limits (x)$ and the complement of $\Omega$, and the constant C depends only on $R,{\textsf{d}}$.

Proof

The defining expansions (43) are the same up to k-th order, where $k = L/R$, beyond which the series converges geometrically by A.5. $\square$

Lemma A.15

(Lemma 6 of [27]). Let $A, B, C \subset {\mathbb {Z}}^{\textsf{d}}$ be pairwise disjoint finite subsets. Then,

$$\begin{aligned} \left\Vert { e^{{\textbf{i}}t H_{A\cup B \cup C}} - e^{{\textbf{i}}t H_{A \cup B}} e^{-{\textbf{i}}t H_B} e^{{\textbf{i}}t H_{B \cup C}} }\right\Vert \le C |t| e^{-L/R} | A^{+R} | \end{aligned}$$

(62)

for some $C > 0$ depending only on ${\textsf{d}},R$. Here, L is the $\ell _\infty$-distance (that is our convention for ${\mathbb {Z}}^{\textsf{d}}$) between $A^{+R}$ and C.

      This is slightly weaker than [27, Lemma 6] as the upper bound depends on the volume of A. We will use this lemma with small A.

Proof

Since A, B, C are finite, those exponentials are unitaries of $\mathop {\mathrm {{\textsf{Mat}}}}\limits$. Without loss of generality, assume $t > 0$. Suppress the union symbol $\cup$. Put $W(s) = e^{{\textbf{i}}s H_{A B C}} e^{-{\textbf{i}}s H_{B C}}$ and $V(s) = e^{{\textbf{i}}s H_{A B}} e^{-{\textbf{i}}s H_{B}}$ for $s \in [0,t]$. They are unique solutions to first order differential equations $\partial _s W(s) = {\textbf{i}}W'(s) W(s)$ and $\partial _s V(s) = {\textbf{i}}V'(s) V(s)$ with initial condition $W(0) = V(s) = {\textbf{1}}$, where

$$\begin{aligned} {\textbf{i}}W'(s)&= U^{H_{A B C}}_s (H_{A B C} - H_{B C}) = U^{H_{A B C}}_s (H_{A^{+R}}),\nonumber \\ {\textbf{i}}V'(s)&= U^{H_{A B}}_s (H_{A B} - H_{B}) = U^{H_{A B}}_s (H_{A^{+R}}),\nonumber \\ \Vert W'(s) - V'(s)\Vert&\le \sum _{X \subseteq A^{+R}} \Vert U_s^{H_{ABC}}(h_X) - U_s^{H_{AB}}(h_X) \Vert \nonumber \\&\le O(1) |{A^{+R}}| R^{\textsf{d}}e^{-L/R}&\text {by }\hbox {A}.14. \end{aligned}$$

(63)

      On the other hand, a differential equation $\partial _s Y(s) = {\textbf{i}}V(s)^\dag (-V'(s) + W'(s) ) V(s) Y(s)$ with initial condition $Y(0) = {\textbf{1}}$, gives a unique unitary solution $Y(s) = V(s)^\dag W(s)$ for $s \in [0,t]$. Jensen’s inequality $\Vert Y(t) - {\textbf{1}}\Vert \le \int _0^t {\textrm{d}}s \Vert \partial _s Y(s)\Vert$ gives the result:

$$\begin{aligned} \left\Vert { e^{{\textbf{i}}t H_{A B C}} - e^{{\textbf{i}}t H_{A B}} e^{-{\textbf{i}}t H_B} e^{{\textbf{i}}t H_{B C}} }\right\Vert= & {} \left\Vert { e^{{\textbf{i}}t H_{A B C}} e^{-{\textbf{i}}t H_{B C}} - e^{{\textbf{i}}t H_{A B}} e^{-{\textbf{i}}t H_B} }\right\Vert \nonumber \\= & {} \Vert W(t) - V(t) \Vert \nonumber \\= & {} \Vert Y(t) - {\textbf{1}}\Vert \le O(1) t e^{-L/R} |A^{+R}|. \end{aligned}$$

(64)

$\square$

Lemma A.16

Given any $L > 10R$ and $t \in {\mathbb {R}}$ with $|t| < t_0$, there exists a quantum circuit $Q_L$ of spread $10(2\textsf{d}+1)L$ such that

$$\begin{aligned} {{\,\mathrm{{dist}}\,}}(U^H_t, Q_L) \le C |t| L^{2D} e^{-L/R} \end{aligned}$$

(65)

for some constant C depending on ${\textsf{d}},R$.

Proof

Assume $t > 0$ for brevity.

      We use the approximating circuit constructed in [27]. It is built out of local unitaries, each of which is supported on a ball of diameter at most L. Divide ${\mathbb {Z}}^{\textsf{d}}$ into disjoint “cells” each of which contains a radius 7L ball and is contained in the concentric radius 10L ball such that the cells are colored with ${\textsf{d}}+1$ colors $\{0,1,\ldots ,{\textsf{d}}\}$, and any two cells of a given color are separated by distance 5L. Such a division exists as seen for example by considering a regular triangulation of ${\mathbb {R}}^{\textsf{d}}$, and fattening all cells. For ${\textsf{d}}=2$, one can consider a honeycomb tiling. Let ce(k) denote the union of all cells colored $k = 0,1,2,\ldots ,{\textsf{d}}$, and let $ce(k)^+$ denote the L-neighborhood of ce(k). The $(2\textsf{d}+1)$st layer, the last, of the circuit is $e^{{\textbf{i}}t H_{ce({\textsf{d}})^+}}$. Here, we abused the notation; the Hamiltonian $H_{ce(k)^+}$ consists of terms that act on disjoint cells of linear size $\sim L$, so the exponential is an infinite product unitaries, which must be interpreted merely as a collection of those local unitaries. The layer beneath, the $2\textsf{d}$-th, is $e^{-{\textbf{i}}t H_{ce({\textsf{d}})^+ \cap ({\mathbb {Z}}^{\textsf{d}}\setminus ce({\textsf{d}}))}}$. Inductively, having defined 2k-th layer for $k > 1$, we define two more layers by declaring that the starting Hamiltonian is $H_{\bigcup _{j=0}^{k-1} ce(j)}$. This is a Hamiltonian on k-colored cells, and gives circuit elements down to the second layer. The first layer is simply $U_t^{H_{ce(0)}}$. This defines $Q_L$ of depth $2\textsf{d}+1$ and spread $10(2\textsf{d}+1)L$.

      Next, we have to show that this is a desired quantum circuit. Let $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits {\setminus } {\mathbb {C}}{\textbf{1}}$. Let $\Omega \subset {\mathbb {Z}}^{\textsf{d}}$ be a finite subset consisting of the cells that intersect $\mathop {\mathrm {{Supp}}}\limits (x)^{+123(2\textsf{d}+1)L}$. So, $|\Omega | \le O(1) L^{\textsf{d}}|\mathop {\mathrm {{Supp}}}\limits (x)|$. By A.14, we have

$$\begin{aligned} \Vert U^H_t(x) - U^{H_\Omega }_t(x)\Vert \le O(1) e^{-L/R} \Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)|. \end{aligned}$$

(66)

Apply the circuit construction above to $H_\Omega$, keeping the cells intact. The gates in this $\Omega$-dependent circuit $Q'$ are exactly the same as those of $Q_L$, except those near the boundary of $\Omega$. But $Q'$ contains the “geometric lightcone” of $\mathop {\mathrm {{Supp}}}\limits (x)$, and therefore $Q'(x) = Q_L(x)$.

      Now, by A.15 we can estimate $\Vert e^{{\textbf{i}}t H_\Omega } - Q'\Vert$. Consider one cell associated with the top layer, the $(2\textsf{d}+1)$st. Call this cell A and put $B = A^{+L} \setminus A$ and $C = \Omega \setminus (A\cup B)$. The distance between $A^{+R}$ and C is at least L. The gate $e^{{\textbf{i}}t H_{A\cup B}}$ is precisely a gate of $Q'$ at $(2\textsf{d}+1)$st layer, and $e^{-{\textbf{i}}t H_B}$ is a gate at the $2\textsf{d}$th layer of $Q'$. The full two top layers of $Q'$ are obtained by such decompositions, iterated at most $|\Omega |$ times, incurring accumulated “error” $O(1) t e^{-L/R} L^{\textsf{d}}|\Omega |$. Inductively, similar decompositions give the full $Q'$. Since $|\Omega | \le |\mathop {\mathrm {{Supp}}}\limits (x)|(2\,L)^{\textsf{d}}$, we have

$$\begin{aligned} \Vert e^{{\textbf{i}}t H_\Omega } - Q'\Vert \le O(1) {\textsf{d}}t e^{-L/R} L^{\textsf{d}}|\Omega | \le O(1) t e^{-L/R} L^{2\textsf{d}} |\mathop {\mathrm {{Supp}}}\limits (x)|. \end{aligned}$$

(67)

We conclude that

$$\begin{aligned} \Vert U^H_t(x) - Q_L(x)\Vert&= \Vert U^H_t(x) - Q'(x)\Vert \nonumber \\&\le \Vert U^H_t(x) - U^{H_\Omega }_t(x)\Vert + 2 \Vert e^{{\textbf{i}}t H_\Omega } - Q'\Vert \Vert x\Vert \nonumber \\&\le O(1) t e^{-L/R} L^{2\textsf{d}} \Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (x)| \end{aligned}$$

(68)

where we abused notations to write $Q'$ for the unitary and $Q'(x)$ for the conjugation of x by the unitary. $\square$

Theorem A.17

The time evolution $U^H_t$ by any bounded strength strictly local Hamiltonian H on $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}}^{\textsf{d}})$ for any time $t \in {\mathbb {R}}$ is a limit of a sequence of finite depth quantum circuits. Therefore, the ${{\,\mathrm{{dist}}\,}}$-closure of the group of all QCA contains the time evolution by any bounded strength strictly local Hamiltonian.

Proof

Take a positive integer n sufficiently large that $\tau = t/ n \in (-t_0, t_0)$. Let $Q_L$ be a quantum circuit corresponds to $U^H_\tau$ by A.16. Let $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits {\setminus } {\mathbb {C}}{\textbf{1}}$. Then, for any $k = 1,2,\ldots , n$,

$$\begin{aligned}&\left\Vert { \left( U^H_\tau \right) ^k \left( Q_L\right) ^{n-k}(x) - \left( U^H_\tau \right) ^{k-1} \left( Q_L\right) ^{n-k+1}(x) }\right\Vert \nonumber \\&= \left\Vert { U^H_\tau (Q_L^{n-k}(x) ) - Q_L( Q_L^{n-k}(x) ) }\right\Vert&\text {by}~\hbox {A}.8\nonumber \\&\le {{\,\mathrm{{dist}}\,}}(U^H_\tau , Q_L) \Vert x\Vert |\mathop {\mathrm {{Supp}}}\limits (Q_L^{n-k}(x))|&\text {by definition}~\hbox {A}.1\nonumber \\&\le C t_0 L^{2\textsf{d}} e^{-L/R} \Vert x\Vert (20(2\textsf{d}+1)nL)^{\textsf{d}}|\mathop {\mathrm {{Supp}}}\limits (x)|&\text {by}~\hbox {A}.16. \end{aligned}$$

(69)

Therefore,

$$\begin{aligned} {{\,\mathrm{{dist}}\,}}(U^H_t, Q_L^{n})&\le \sum _{k=1}^n {{\,\mathrm{{dist}}\,}}( \left( U^H_\tau \right) ^{k} \left( Q_L\right) ^{n-k}, \nonumber \\ \left( U^H_\tau \right) ^{k-1} \left( Q_L\right) ^{n-k+1} )&\le C' n^{{\textsf{d}}+1} L^{3\textsf{d}} e^{-L/R} \end{aligned}$$

(70)

for some constant $C'$ that depends only on ${\textsf{d}},R$. Since ${\mathcal {L}}$ is ${{\,\mathrm{{dist}}\,}}$-complete by A.2, taking $L \rightarrow \infty$ we complete the proof. $\square$

[24, Thm. 5.6] says that a certain class of approximately locality-preserving $*$-automorphisms $\alpha$ of $\mathop {\mathrm {{{\overline{\mathop {\mathrm {{\textsf{Mat}}}}\limits }}}}}\limits ({\mathbb {Z}}^{\textsf{d}})$ with ${\textsf{d}}=1$, is contained in the ${{\,\mathrm{{dist}}\,}}$-completion of the group of QCA. To state the studied class of $\alpha$, we denote by $[S] \subset {\mathbb {Z}}$ the smallest interval containing S for any $S \subset {\mathbb {Z}}$. The required condition for $\alpha$ is that for any given $\epsilon > 0$ there is a length scale $\ell$ (corresponding to the spread of a QCA) such that for every $x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits$ there exists $y_x \in \mathop {\mathrm {{\textsf{Mat}}}}\limits ([\mathop {\mathrm {{Supp}}}\limits (x)]^{+\ell })$ with $\Vert \alpha (x) - y_x\Vert < \epsilon \Vert x\Vert$. A specific relation between $\epsilon$ and $\ell$ encodes the decay rate of the tails of $\alpha$. It appears that an important condition is that $\ell$ should be uniform across the lattice, but not the specific decay rate. In course of proving [24, 5.6], they also show [24, 4.10] that two one-dimensional QCA on the same $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}},p)$ that are $\epsilon _0$-close in ${{\,\mathrm{{dist}}\,}}$ must blend where $\epsilon _0$ is a positive threshold depending only on the spread. This raises a natural question: in higher dimensions, if two QCA are $\epsilon$-close in ${{\,\mathrm{{dist}}\,}}$, do they give boundary algebras of the same Brauer class?

Appendix B: One-Dimensional Brauer Group is Trivial

In this appendix, we are going to show that the Brauer group of invertible subalgebras in one dimension is trivial. The Brauer triviality of an invertible subalgebra means that there is a locality-preserving isomorphism into the invertible subalgebra (up to stabilization) from a full local operator algebra. In fact, we will not need stabilization and will find a decomposition of the invertible subalgebra into mutually commuting central simple $*$-subalgebras on intervals of ${\mathbb {Z}}$. These interval subalgebras are images of single-site matrix algebras. It is not too difficult to find some central subalgebra around an interval. Then, we are led to think about central subalgebras on far separated intervals, and try to fill the “gaps” by taking commutants. We will do this, but there is one difficulty. Namely, it does not immediately follow that the commutants are again central since the existence argument (B.2) of a central algebra on an interval is too abstract to give any useful information about potentially central elements. So, we take a “dual” construction for interval central subalgebras, defined as commutants of some local operators such that any potential central elements have restricted support.

      For the rest of this section, we let ${\mathcal {A}}\subseteq \mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}})$ denote any one-dimensional invertible subalgebra of spread $\ell$. For any interval $[a,b] \cap {\mathbb {Z}}$, we introduce an abbreviation

$$\begin{aligned} {\mathcal {A}}_{[a,b]} = {\mathcal {A}}\cap \mathop {\mathrm {{\textsf{Mat}}}}\limits ( [a,b] \cap {\mathbb {Z}}). \end{aligned}$$

(71)

We also use a notation for any two sets ${\mathcal {X}}, {\mathcal {Z}}\subseteq {\mathcal {A}}$

$$\begin{aligned} {{\,\mathrm{{Comm}}\,}}( {\mathcal {X}}\,|\, {\mathcal {Z}}) = \{ z \in {\mathcal {Z}}~|~ \forall x \in {\mathcal {X}}:~[z,x] = 0 \}. \end{aligned}$$

(72)

In this notation, ${\mathcal {X}}$ need not be a subset of ${\mathcal {Z}}$. The center of an algebra ${\mathcal {Y}}$ will be denoted by

$$\begin{aligned} {{\,\mathrm{{Cent}}\,}}( {\mathcal {Y}}) = {{\,\mathrm{{Comm}}\,}}({\mathcal {Y}}\,|\, {\mathcal {Y}}). \end{aligned}$$

(73)

      First we will need a tool to examine operator components individually that are far apart in space.

Lemma B.1

Suppose an element x of an invertible subalgebra ${\mathcal {A}}\subseteq \mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}}^{\textsf{d}})$ of spread $\ell$ is decomposed as $x = \sum _i y_i z_i$ such that $\{y_i\}$ and $\{z_i\}$ are each linearly independent (obtained by e.g. the Schmidt decomposition with respect to the Hilbert–Schmidt inner product). If $Y = \bigcup _i \mathop {\mathrm {{Supp}}}\limits (y_i)$ and $Z = \bigcup _i \mathop {\mathrm {{Supp}}}\limits (z_i)$ are separated by distance $> \ell$, then $y_i, z_i \in {\mathcal {A}}$ for all i.

In [2] such a subalgebra ${\mathcal {A}}$ is called “locally factorizable.”

Proof

The commutant ${\mathcal {B}}$ of ${\mathcal {A}}$ within $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}}^{\textsf{d}})$ is $\ell$-locally generated (2.6, 2.7). Let b be any $\ell$-local generator of ${\mathcal {B}}$ whose support overlaps with Z. Since Y and Z are far apart, b does not overlap with any $y_i$. Hence, $0 = [b,x] = \sum _i y_i [b,z_i] = \sum _i y_i \otimes [b,z_i]$. The linear independence of $\{y_i\}$ implies that $[b,z_i] = 0$ for all i. This means that each $z_i$ commutes with every generator of ${\mathcal {B}}$. Therefore, $z_i$ belongs to the commutant of ${\mathcal {B}}$, which is ${\mathcal {A}}$ by 2.7. A symmetric argument shows that $y_i \in {\mathcal {A}}$ for all i. $\square$

      The following B.2 is a seed for central subalgebras of an invertible subalgebra, not necessarily on one dimension. Indeed, this lemma has nothing to do with locality. When we use it, the assumption of this lemma will be fulfilled always by 2.3.

Lemma B.2

(Lemma 3.8 of [2]). Let ${\mathcal {X}}\subseteq {\mathcal {Z}}$ be unital $*$-subalgebras of a full matrix algebra of dimension ${{\,\mathrm{{Tr}}\,}}({\textbf{1}}) < \infty$. Let $\{ \pi _\mu \}$ and $\{ \tau _\nu \}$ be the central minimal projectors of ${\mathcal {X}}$ and ${\mathcal {Z}}$, respectively. Suppose

$$\begin{aligned} \frac{{{\,\mathrm{{Tr}}\,}}(\pi _\mu \tau _\nu )}{{{\,\mathrm{{Tr}}\,}}({\textbf{1}})} = \frac{{{\,\mathrm{{Tr}}\,}}(\pi _\mu )}{{{\,\mathrm{{Tr}}\,}}({\textbf{1}})} \frac{{{\,\mathrm{{Tr}}\,}}(\tau _\nu )}{{{\,\mathrm{{Tr}}\,}}({\textbf{1}})} \end{aligned}$$

(74)

for all $\mu ,\nu$. Then, there exists a central (and hence simple) $*$-subalgebra ${\mathcal {Y}}$ such that ${\mathcal {X}}\subseteq {\mathcal {Y}}\subseteq {\mathcal {Z}}$.

      The proof below is the same as that in [2].

Proof

For the simple subalgebras $\pi _\mu {\mathcal {X}}$ and $\tau _\nu {\mathcal {Z}}$, let $[\pi _\mu {\mathcal {X}}]$ and $[\tau _\nu {\mathcal {Z}}]$ denote some full matrix algebras isomorphic to them so that we have ${{\,\mathrm{{Tr}}\,}}$-preserving isomorphisms for all $\mu , \nu$

$$\begin{aligned} \pi _\mu {\mathcal {X}}&\cong [\pi _\mu {\mathcal {X}}] \otimes {\textbf{1}}_{c^{\mathcal {X}}_\mu },&\tau _\nu {\mathcal {Z}}&\cong [\tau _\nu {\mathcal {Z}}] \otimes {\textbf{1}}_{c^{\mathcal {Z}}_\nu } ,\nonumber \\ {{\,\mathrm{{Tr}}\,}}(\pi _\mu )&= c^{\mathcal {X}}_\mu \dim [\pi _\mu {\mathcal {X}}],&{{\,\mathrm{{Tr}}\,}}(\tau _\nu )&= c^{\mathcal {Z}}_\nu \dim [\tau _\nu {\mathcal {Z}}]. \end{aligned}$$

(75)

Given a pair $(\nu ,\mu )$, consider a homomorphism

$$\begin{aligned} \varphi _{\nu ,\mu } : [\pi _\mu {\mathcal {X}}] \xrightarrow {\quad \otimes {\textbf{1}}_{c^{\mathcal {X}}_\mu }\quad } \pi _\mu {\mathcal {X}}\xrightarrow {\quad \tau _\nu \times \quad } \tau _\nu {\mathcal {Z}}\xrightarrow {\qquad } [\tau _\nu {\mathcal {Z}}]. \end{aligned}$$

(76)

Since $[\pi _\mu {\mathcal {X}}]$ is simple, this map $\varphi _{\nu ,\mu }$ can be either zero or an injection. Hence, there is a nonnegative integer $C_{\nu ,\mu }$ that counts the multiplicity of $[\pi _\mu {\mathcal {X}}]$ in the image. That is, $\varphi _{\nu ,\mu }([\pi _\mu {\mathcal {X}}]) \cong [\pi _\mu {\mathcal {X}}] \otimes {\textbf{1}}_{C_{\nu ,\mu }}$ where $\cong$ is a ${{\,\mathrm{{Tr}}\,}}$-preserving isomorphism. Tracking the traces of the unit, we have

$$\begin{aligned} C_{\nu , \mu } \frac{{{\,\mathrm{{Tr}}\,}}(\pi _\mu )}{c^{\mathcal {X}}_\mu } = \frac{{{\,\mathrm{{Tr}}\,}}(\tau _\nu \pi _\mu )}{c^{\mathcal {Z}}_\nu }. \end{aligned}$$

(77)

By assumption, this is equal to ${{\,\mathrm{{Tr}}\,}}(\tau _\nu ) {{\,\mathrm{{Tr}}\,}}(\pi _\mu )/{{\,\mathrm{{Tr}}\,}}({\textbf{1}}) c^{\mathcal {Z}}_\nu$. Rearranging, we have

$$\begin{aligned} C_{\nu ,\mu } = \frac{{{\,\mathrm{{Tr}}\,}}(\tau _\nu )}{c^{\mathcal {Z}}_\nu } \frac{c^{\mathcal {X}}_\mu }{{{\,\mathrm{{Tr}}\,}}({\textbf{1}})} = \frac{\dim [\tau _\nu {\mathcal {Z}}] \gcd }{{{\,\mathrm{{Tr}}\,}}({\textbf{1}})} \frac{c^{\mathcal {X}}_\mu }{\gcd } \end{aligned}$$

(78)

where $\gcd = \gcd (\{c^{\mathcal {X}}_\mu \}_\mu )$. If $n_\mu$ are integers such that $\sum _\mu n_\mu c^{\mathcal {X}}_\mu = \gcd$, we see that

$$\begin{aligned} N_\nu = \frac{(\dim [\tau _\nu {\mathcal {Z}}]) \gcd }{{{\,\mathrm{{Tr}}\,}}({\textbf{1}})} = \sum _\mu C_{\nu ,\mu } n_\mu \end{aligned}$$

(79)

is a positive integer for each $\nu$. It follows that

$$\begin{aligned} \bigoplus _\mu \varphi _{\nu ,\mu }([\pi _\mu {\mathcal {X}}]) \,\cong \, \bigoplus _\mu [\pi _\mu {\mathcal {X}}] \otimes {\textbf{1}}_{C_{\nu ,\mu }} \,\cong \, \bigoplus _\mu [\pi _\mu {\mathcal {X}}] \otimes {\textbf{1}}_{c^{\mathcal {X}}_{\mu }/\gcd } \otimes {\textbf{1}}_{N_\nu } \subseteq [\tau _\nu {\mathcal {Z}}] \end{aligned}$$

(80)

where all the isomorphisms are ${{\,\mathrm{{Tr}}\,}}$-preserving. Since ${\mathcal {X}}$ is a subalgebra of ${\mathcal {Z}}$, the $c^{\mathcal {Z}}_\nu$-weighted direct sum over $\nu$ of these images must be equal to ${\mathcal {X}}$ itself:

$$\begin{aligned} {\mathcal {X}}\cong \bigoplus _{\nu } \Big ( \bigoplus _\mu [\pi _\mu {\mathcal {X}}] \otimes {\textbf{1}}_{c^{\mathcal {X}}_{\mu }/\gcd } \Big ) \otimes {\textbf{1}}_{N_\nu c^{\mathcal {Z}}_\nu }. \end{aligned}$$

(81)

Now, let $d = \sum _\mu (c^{\mathcal {X}}_\mu / \gcd ) \dim [\pi _\mu {\mathcal {X}}]$. Observe that $d \cdot N_\nu c^{\mathcal {Z}}_\nu = \sum _\mu (c^{\mathcal {X}}_\mu / \gcd ) \dim [\pi _\mu {\mathcal {X}}] N_\nu c^{\mathcal {Z}}_\nu = c^{\mathcal {Z}}_\nu \sum _\mu C_{\nu ,\mu } \dim [\pi _\mu {\mathcal {X}}]$. The last sum $\sum _\mu C_{\nu ,\mu } \dim [\pi _\mu {\mathcal {X}}]$ is equal to $\dim [\tau _\nu {\mathcal {Z}}]$ because ${\mathcal {X}}$ is a unital subalgebra of ${\mathcal {Z}}$. Therefore, $d \cdot N_\nu c^{\mathcal {Z}}_\nu = {{\,\mathrm{{Tr}}\,}}(\tau _\nu )$. This dimension counting shows that a desired subalgebra ${\mathcal {Y}}$ can be defined as

$$\begin{aligned} {\mathcal {Y}}\cong \left\{ \bigoplus _{\nu } M \otimes {\textbf{1}}_{N_\nu c^{\mathcal {Z}}_\nu }~\bigg |~ M \text { is any }d \times d\text { matrix } \right\} \end{aligned}$$

(82)

$\square$

      The following lemma is another tool that we will find useful later. Namely, the commutant of certain local operators is locally generated.

Lemma B.3

Let ${\mathcal {D}}^\textrm{left} \subseteq {\mathcal {A}}_{[a-10\ell , a+10\ell ]}$ and ${\mathcal {D}}^\textrm{right} \subseteq {\mathcal {A}}_{[b-10\ell , b+10\ell ]}$ be $*$-subalgebras. For $b - a > 40\ell$, a $*$-subalgebra

$$\begin{aligned} {\mathcal {C}}&= {{\,\mathrm{{Comm}}\,}}({\mathcal {D}}^\textrm{left} {\mathcal {D}}^\textrm{right} \,|\, {\mathcal {A}}_{[a, b]} ) \end{aligned}$$

(83)

is $20\ell$-locally generated.

      The subsets ${\mathcal {D}}^\textrm{left,right}$ can be ${\mathbb {C}}$, in which case ${\mathcal {C}}= {\mathcal {A}}_{[a,b]}$.

Proof

${\mathcal {C}}$ contains ${\mathcal {A}}_{[a + 12\ell , b - 12\ell ]}$ and ${\mathcal {A}}_{[a+16\ell , b- 16\ell ]}$. By 2.3, the centers of these two algebras have disjoint support, and B.2 gives a central subalgebra ${\mathcal {C}}_0$ sandwiched by these two subalgebras. In the finite dimensional $*$-algebra ${\mathcal {C}}$, the central subalgebra ${\mathcal {C}}_0$ must be a tensor factor: ${\mathcal {C}}$ is unitarily isomorphic to $\bigoplus _\mu \pi _\mu {\mathcal {C}}\subseteq \mathop {\mathrm {{\textsf{Mat}}}}\limits ([a-10\ell ,b+10\ell ])$ for some minimal central projectors $\pi _\mu \in {\mathcal {C}}$, and each simple summand $\pi _\mu {\mathcal {C}}$ must have ${\mathcal {C}}_0$ as a tensor factor. So, we have ${\mathcal {C}}\cong {\mathcal {C}}_0 \otimes {\mathcal {M}}$ where ${\mathcal {M}}= {{\,\mathrm{{Comm}}\,}}({\mathcal {C}}_0 \,|\, {\mathcal {C}})$. We show that ${\mathcal {C}}_0$ is contained in a subalgebra of ${\mathcal {C}}$ generated by $20\ell$-local elements of ${\mathcal {C}}$, and that ${\mathcal {M}}$ is contained a tensor product of two subalgebras of ${\mathcal {C}}$, one supported on $[a,a+20\ell ]$ and the other on $[b-20\ell , b]$. This will complete the proof as ${\mathcal {C}}$ is generated by ${\mathcal {C}}_0$, that is covered by a $20\ell$-locally generated algebra, and the two subalgebras associated with ${\mathcal {M}}$, each of which can be taken as a subset of $20\ell$-local generators.

      If $x \in {\mathcal {C}}_0 \subseteq \mathop {\mathrm {{\textsf{Mat}}}}\limits ([a+12\ell , b- 12\ell ])$, we can write it as a sum of product of single-site operators, each of which is supported on $[a+12\ell , b-12\ell ]$. Since ${\mathcal {A}}$ is invertible, we can rewrite such a sum as another sum of products of elements of form yz where y is a $\ell$-local generator of ${\mathcal {A}}$ (2.6), and z is a $\ell$-local generator of ${\mathcal {B}}= {{\,\mathrm{{Comm}}\,}}({\mathcal {A}}, \mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}}))$. Applying ${{\,\mathrm{{{tr}}}\,}}_{\mathcal {B}}$ of 2.8, we express x as a sum of products of $\ell$-local elements of ${\mathcal {C}}$.

Next, let $w \in {\mathcal {M}}$. Since ${\mathcal {C}}_0$ contains ${\mathcal {A}}_{[a+16\ell , b- 16\ell ]}$, we have by 2.3 an unnormalized Schmidt decomposition $w = \sum _j w^L_j w^R_j$, where $w^L_j \in {\mathcal {A}}_{[a, a+19\ell ]}$ and $w^R_j \in {\mathcal {A}}_{[b-19\ell , b]}$ by B.1. Each of ${\mathcal {D}}^\textrm{left}$ and ${\mathcal {D}}^\textrm{right}$ cannot overlap with both $w^L_j$ and $w^R_j$, so we must have $w^L_j, w^R_j \in {\mathcal {C}}$ for all j.²$\square$

      The next lemma illustrates what we are going to do. The assumption of this lemma will hold always, as shown in the proof of B.5.

Lemma B.4

Assume that for any given site $s \in {\mathbb {Z}}$ there exist two $*$-subalgebras ${\mathcal {D}}_s^\textrm{right}, {\mathcal {D}}_s^\textrm{left}$ such that

• (i) ${\mathcal {D}}_s^\textrm{left}, {\mathcal {D}}_s^\textrm{right} \subseteq {\mathcal {A}}_{[s - 10\ell , s+10\ell ]}$,

• (ii) ${\mathcal {D}}_s^\textrm{right}$ commutes with ${\mathcal {A}}_{[s-30\ell , s-3\ell ]}$, ${\mathcal {D}}_s^\textrm{left}$ commutes with ${\mathcal {A}}_{[s+3\ell , s+30\ell ]}$,

• (iii) $\mathop {\mathrm {{Supp}}}\limits {{\,\mathrm{{Cent}}\,}}{{\,\mathrm{{Comm}}\,}}({\mathcal {D}}_s^\textrm{right} \,|\, {\mathcal {A}}_{[s - 30\ell , s + 3\ell ]} ) \subseteq [s-30\ell , s-7\ell ]$, and

  $\mathop {\mathrm {{Supp}}}\limits {{\,\mathrm{{Cent}}\,}}{{\,\mathrm{{Comm}}\,}}({\mathcal {D}}_s^\textrm{left} \,|\, {\mathcal {A}}_{[s - 3\ell , s + 30\ell ]} ) \subseteq [s + 7\ell , s+30\ell ]$.

Then, ${\mathcal {A}}$ is generated by mutually commuting central simple $*$-subalgebras ${\mathcal {A}}_i \subset {\mathcal {A}}$ indexed by $i \in {\mathbb {Z}}$ such that $\mathop {\mathrm {{Supp}}}\limits {\mathcal {A}}_i \subseteq [50\ell i - 49 \ell , 50 \ell i + 49 \ell ]$.

The subalgebras ${\mathcal {D}}^\textrm{left,right}$ in the assumption will control the position of potential central elements of their commutants. The annotation “left” and “right” means that they determine the left and right end of the commutant, respectively. As typical in this kind of proof, the constants such as 3, 10, 30, 50 are never meant to be optimal. They are chosen large enough to avoid certain overlaps.

Proof

For $a,b \in {\mathbb {Z}}$ such that $b - a > 40 \ell$, we define

$$\begin{aligned} {\mathcal {C}}(a,b)&= {{\,\mathrm{{Comm}}\,}}({\mathcal {D}}_{a}^\textrm{left} {\mathcal {D}}_{b}^\textrm{right} \,|\, {\mathcal {A}}_{[a, b]} ) . \end{aligned}$$

(84)

By B.3, ${\mathcal {A}}_{[a + 3\ell , b - 3\ell ]}$ is $20\ell$-locally generated. Then, (ii) implies that these local generators commute with both ${\mathcal {D}}_a^\textrm{left}$ and ${\mathcal {D}}_b^\textrm{right}$, so ${\mathcal {C}}(a,b)$ contains ${\mathcal {A}}_{[a + 3\ell , b - 3\ell ]}$. If z is a central element of ${\mathcal {C}}(a,b)$, then by 2.3, z must be supported on $L = [a, a + 5\ell ]$ union $R= [b - 5\ell , b]$. If $z = \sum _k z^L_k z^R_k$ is a Schmidt decomposition, then by B.1, $z^L_k, z^R_k \in {\mathcal {A}}_{[a,b]}$ for all k. By construction, we have $0 = [{\mathcal {D}}_{a}^\textrm{left}, z ] = \sum _k [{\mathcal {D}}_{a}^\textrm{left}, z^L_k] z^R_k$, where each commutator must vanish because $z^R_k$ are linearly independent, so $z^L_k \in {{\,\mathrm{{Comm}}\,}}({\mathcal {D}}_{a}^\textrm{left} \,|\, {\mathcal {A}}_{[a, a + 15\ell ]} ) \subseteq {{\,\mathrm{{Comm}}\,}}({\mathcal {D}}_{a}^\textrm{left} \,|\, {\mathcal {A}}_{[a, a + 30\ell ]} )$. Similarly, since ${\mathcal {C}}(a,b)$ is $20\ell$-locally generated by B.3, $z^L_k$ commutes with ${\mathcal {C}}(a,b)$ for each k. Since ${{\,\mathrm{{Comm}}\,}}({\mathcal {D}}_{a}^\textrm{left} \,|\, {\mathcal {A}}_{[a, a + 30\ell ]} ) \subseteq {\mathcal {C}}(a,b)$, it follows that $z^L_k \in {{\,\mathrm{{Cent}}\,}}{{\,\mathrm{{Comm}}\,}}({\mathcal {D}}_{a}^\textrm{left} \,|\, {\mathcal {A}}_{[a,a + 30\ell ]} )$. By (iii) we have $\mathop {\mathrm {{Supp}}}\limits (z^L_k) \subseteq [a + 7\ell , a + 30\ell ] \cap [a, a + 5 \ell ] = \emptyset$, so $z^L_k$ is a scalar for all k. A symmetric argument shows that $z^R_k$ is also a scalar for all k, and therefore ${\mathcal {C}}(a,b)$ is central.

Next, we define

$$\begin{aligned} {\mathcal {A}}_{2j}&= {\mathcal {C}}( 100 \ell j - 30\ell , 100 \ell j + 30 \ell )\nonumber \\ {\mathcal {A}}_{2j+1}&= {{\,\mathrm{{Comm}}\,}}( {\mathcal {A}}_{2j} {\mathcal {A}}_{2j+2} \,|\, {\mathcal {C}}(100 \ell j - 30 \ell , 100 \ell (j+1) + 30 \ell ) ) \end{aligned}$$

(85)

Since ${\mathcal {C}}(b,a)$ with $b - a > 40\ell$ is central and finite dimensional, it follows that ${\mathcal {A}}_{2j},{\mathcal {A}}_{2j+1}$ are all central. By construction, two consecutive subalgebras ${\mathcal {A}}_{i}$ and ${\mathcal {A}}_{i+1}$ are mutually commuting for all i. If we show the claim on the support of ${\mathcal {A}}_i$, the mutual commutativity will immediately follow.

Let us examine the support of ${\mathcal {A}}_{2j+1}$. Since the even-indexed ${\mathcal {A}}_{2j}$ contains ${\mathcal {A}}_{[100 \ell j - 20\ell , 100 \ell j + 20\ell ]}$, the odd index ${\mathcal {A}}_{2j+1}$ is supported on three disjoint intervals by 2.3: $L=[100 \ell j - 30\ell , 100\ell j - 18\ell ]$, $M=[100 \ell j + 18 \ell , 100 \ell (j+1) - 18 \ell ]$, and $R=[100 \ell (j+1) + 18 \ell , 100 \ell (j+1) + 30 \ell ]$. Let $x \in {\mathcal {A}}_{2j+1}$. Applying B.1 to $(L\cup R) \cup M$, we have a Schmidt decomposition $x = \sum _k x^{LR}_k x^M_k$ (unnormalized) where $x^{LR}_k \in {\mathcal {A}}_{[100\ell j - 30 \ell , 100 \ell j + 30\ell ]} {\mathcal {A}}_{[ 100 \ell (j+1) - 30 \ell , 100 \ell (j+1) + 30 \ell ]}$. Considering the commutation relation with ${\mathcal {D}}^\textrm{left}_{100\ell j-30\ell }$ and ${\mathcal {D}}^\textrm{right}_{100 \ell (j+1) + 30\ell }$, we see that $x^{LR}_k \in {\mathcal {A}}_{2j} {\mathcal {A}}_{2j+2}$ for all k. But x commutes with ${\mathcal {A}}_{2j} {\mathcal {A}}_{2j+2}$, so x must commute with all the $20\ell$-local generators of ${\mathcal {A}}_{2j} {\mathcal {A}}_{2j+2}$ (B.3), each of which can overlap with either $L\cup R$ or M, but not both. This implies that $x^{LR}_k$ commutes with every generator of ${\mathcal {A}}_{2j} {\mathcal {A}}_{2j+2}$. Since ${\mathcal {A}}_{2j}{\mathcal {A}}_{2j+2}$ is central, all $x^{LR}_k$ are scalars. Therefore, ${\mathcal {A}}_{2j+1}$ is supported on M, completing the proof for the claim on the support of ${\mathcal {A}}_{2j+1}$. The support of ${\mathcal {A}}_{2j}$ satisfies the claim by construction.

It remains to show that all ${\mathcal {A}}_i$’s generate the full ${\mathcal {A}}$. It suffices to check that every $\ell$-local generator g of ${\mathcal {A}}$ (2.6) is contained in the algebra generated by ${\mathcal {A}}_i$’s. Since g is $\ell$-local, we must have that $g \in {\mathcal {A}}_{[100\ell j - 10\ell , 100 \ell (j+1) + 10\ell ]}$ for some j. In particular, $g \in {\mathcal {C}}(100\ell j - 30\ell , 100 \ell j + 30 \ell ) = {\mathcal {A}}_{2j}{\mathcal {A}}_{2j+1} {\mathcal {A}}_{2j+2}$. This completes the proof. $\square$

Theorem B.5

The Brauer group of invertible subalgebras of $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}})$ is zero.

Proof

We will confirm that the assumption of B.4 is always true. Then, B.4 gives a decomposition of ${\mathcal {A}}$ into mutually commuting central simple $*$-subalgebras ${\mathcal {A}}_i$. Defining a local operator algebra $\mathop {\mathrm {{\textsf{Mat}}}}\limits ({\mathbb {Z}},q)$ with a local dimension assignment q by $q(s) = 1$ for all $s \in {\mathbb {Z}}$ except $q(s = 50\ell i) = \dim {\mathcal {A}}_i$ for $i \in {\mathbb {Z}}$, we have the Brauer triviality of ${\mathcal {A}}$. We will only construct ${\mathcal {D}}_s^\textrm{right}$; the construction for ${\mathcal {D}}_s^\textrm{left}$ will be completely parallel. The reference to a site s will only complicate the notation, so we work near a convenient point $0 \in {\mathbb {Z}}$.

      Let $\pi = \pi ^\dag \in {\mathcal {A}}_{[0,5\ell ]}$ be a nonzero minimal projector.³ That is, if $\tau = \tau ^\dag \ne \pi$ is any other projector such that $\pi - \tau$ is positive semidefinite, then $\tau = 0$. Such a minimal projector exists because ${\mathcal {A}}_{[0,5\ell ]}$ is finite dimensional. Define for any integer $j > 0$

$$\begin{aligned} {\mathcal {X}}_j = {{\,\mathrm{{Comm}}\,}}(\pi \,|\, {\mathcal {A}}_{[0,j \ell ]}). \end{aligned}$$

(86)

Clearly, ${\mathcal {X}}_j \subseteq {\mathcal {X}}_{j+1}$ for all j.

      For $j \ge 50$, consider the decomposition ${\mathcal {A}}_{[0,j\ell ]} = \bigoplus _{\mu } \xi _\mu {\mathcal {M}}_\mu$ where $\xi _\mu$ are minimal central projectors and ${\mathcal {M}}_\mu$ are central simple. The projector $\pi$ is represented in this decomposition as $\pi = \sum _\mu \xi _\mu \pi$ where $\xi _\mu \pi \in {\mathcal {M}}_\mu$ is a projector. Considering the commutant of $\pi$ in each simple subalgebra, we see that the center of ${\mathcal {X}}_j$ is in the ${\mathbb {C}}$-linear span of $\pi \xi _\mu , \xi _\mu$. But, by 2.3, any element $\xi \in {{\,\mathrm{{Cent}}\,}}({\mathcal {A}}_{[0,j \ell ]})$ is supported on $L = [0,2\ell ]$ union $R = [j\ell - 2\ell , j \ell ]$. Every Schmidt component of $\xi$ (B.1) on either L or R belongs to ${\mathcal {A}}_{[0, j\ell ]}$. Since ${\mathcal {A}}_{[0,j\ell ]}$ is $20\ell$-locally generated (B.3), we see that each Schmidt component must be central by itself. It follows that ${{\,\mathrm{{Cent}}\,}}( {\mathcal {A}}_{[0,j\ell ]} ) = \langle \xi _\nu ^L \xi _\sigma ^R \rangle$ is a tensor product of two commutative algebras generated by nonzero (possibly nonminimal) central projectors $\xi _\nu ^L$ and $\xi _\sigma ^R$. Since $\pi \xi _\nu ^L \preceq \pi$ and $\xi _\nu ^L \in {\mathcal {A}}_{[0,5\ell ]}$, we must have $\pi \xi _\nu ^L = \pi$ or $\pi \xi _\nu ^L = 0$ for the minimality of $\pi$ for all $\nu$. Therefore, the center of ${\mathcal {X}}_j$ is generated by $\pi$ and $\xi _\sigma ^R$ where all $\xi _\sigma ^R$ are supported on the right end $[j\ell - 2\ell , j\ell ]$.

      If $\pi \xi '$ is a central projector of $\pi {\mathcal {X}}_{53}$,⁴ then ${{\,\mathrm{{{tr}}}\,}}(\pi \xi _\sigma ^R \pi \xi ') {{\,\mathrm{{{tr}}}\,}}(\pi ) = {{\,\mathrm{{{tr}}}\,}}(\pi )^2 {{\,\mathrm{{{tr}}}\,}}(\xi _\sigma ^R) {{\,\mathrm{{{tr}}}\,}}(\xi ') = {{\,\mathrm{{{tr}}}\,}}(\pi \xi _\sigma ^R) {{\,\mathrm{{{tr}}}\,}}(\pi \xi ')$, which is the condition to apply B.2 to $\pi {\mathcal {X}}_{50} \subseteq \pi {\mathcal {X}}_{53}$. We obtain a central $*$-algebra ${\mathcal {C}}^\pi$ such that $\pi {\mathcal {X}}_{50} \subseteq {\mathcal {C}}^\pi \subseteq \pi {\mathcal {X}}_{53}$ where the superscript $\pi$ is to note that the unit $\pi$ of ${\mathcal {C}}^\pi$ is not equal to the unit ${\textbf{1}}\in {\mathcal {A}}$. Define

$$\begin{aligned} {\mathcal {D}}^\pi&= {{\,\mathrm{{Comm}}\,}}( {\mathcal {C}}^\pi \,|\, \pi {\mathcal {X}}_{53} ), \nonumber \\ {\mathcal {D}}&= \big \langle x^R_i ~|~ \sum _i x^L_i x^R_i \text { is a Schmidt decomposition of }x \in {\mathcal {D}}^\pi \text { w.r.t. } {\mathcal {A}}_{[0,20\ell ]} {\mathcal {A}}_{[40\ell ,53\ell ]} \big \rangle . \end{aligned}$$

(87)

Let us first check that ${\mathcal {D}}$ is well defined. Suppose $x = \pi x = x \pi \in {\mathcal {D}}^\pi$. If $t \in \mathop {\mathrm {{Supp}}}\limits (x) \cap [20\ell , 40\ell ]$, then the VS property of ${\mathcal {A}}$ gives $w \in {\mathcal {A}}_{[15\ell , 45\ell ]}$ such that $[w,x] \ne 0$. Then, $\pi w = w \pi \in \pi {\mathcal {A}}_{[15\ell , 45\ell ]} \subseteq \pi {\mathcal {X}}_{50} \subseteq {\mathcal {C}}^\pi$ and $[\pi w, x] = \pi w x - x \pi w = w x - x w \ne 0$, which is a contradiction. So, x is supported on the disjoint union of intervals that appear in the definition of ${\mathcal {D}}$, and thus ${\mathcal {D}}$ is well defined. We declare ${\mathcal {D}}_{50\ell }^\textrm{right} = {\mathcal {D}}$ and claim that ${\mathcal {D}}_{50\ell }^\textrm{right}$ satisfies all the properties in the assumption of B.4.

      First, ${\mathcal {D}}$ is supported on $[40\ell , 53\ell ] \subseteq [50\ell - 10\ell , 50 \ell + 10\ell ]$. This is (i). Second, we have to show that ${\mathcal {A}}_{[20\ell , 47\ell ]}$ commutes with ${\mathcal {D}}$. If $g \in {\mathcal {A}}_{[20\ell , 47\ell ]}$, then $\pi g = g \pi \in \pi {\mathcal {X}}_{50} \subseteq {\mathcal {C}}^\pi$, so $\pi g$ commutes with ${\mathcal {D}}^\pi$. If $\sum _i x_i^{\le 20} x^{\ge 40}_i$ is a Schmidt decomposed element of ${\mathcal {D}}^\pi$, we have $\sum _i x_i^{\le 20} \pi [g,x^{\ge 40}_i] = 0$ where $x_i^{\le 20} \pi = x_i^{\le 20}$ for any i, so $[g,x^{\ge 40}_i] =0$. This means that g commutes with ${\mathcal {D}}$, implying (ii). Finally, we have to show (iii) that $\mathop {\mathrm {{Supp}}}\limits {{\,\mathrm{{Cent}}\,}}{{\,\mathrm{{Comm}}\,}}({\mathcal {D}}\,|\, {\mathcal {A}}_{[20\ell ,53\ell ]} ) \subseteq [20\ell , 43\ell ]$. Since ${{\,\mathrm{{Comm}}\,}}({\mathcal {D}}\,|\, {\mathcal {A}}_{[20\ell ,53\ell ]} )$ contains ${\mathcal {A}}_{[20\ell , 47\ell ]}$ by (ii), any central element $y \in {{\,\mathrm{{Cent}}\,}}{{\,\mathrm{{Comm}}\,}}({\mathcal {D}}\,|\, {\mathcal {A}}_{[20\ell ,53\ell ]} )$ is supported on $L = [20\ell , 22\ell ]$ union $R= [45\ell ,53\ell ]$ by 2.3. Let $y = \sum _i y^L_i y^R_i$ be an unnormalized Schmidt decomposition. By B.1 we know $y^R_i \in {\mathcal {A}}_{[45\ell ,53\ell ]}$ for all i. We claim that $y^R_i \in {\mathbb {C}}{\textbf{1}}$, which completes the proof of (iii) and hence the theorem.

      Since y belongs to ${{\,\mathrm{{Comm}}\,}}({\mathcal {D}}| {\mathcal {A}}_{[20\ell ,53\ell ]})$ and ${\mathcal {D}}$ is supported on $[40\ell ,53\ell ]$, we see that $y^R_i$ must commute with ${\mathcal {D}}$ for all i, so $y^R_i \in {{\,\mathrm{{Comm}}\,}}({\mathcal {D}}\,|\, {\mathcal {A}}_{[20\ell ,53\ell ]} )$. Hence,

$$\begin{aligned} \pi y^R_i \in {{\,\mathrm{{Comm}}\,}}({\mathcal {D}}^\pi , \pi {\mathcal {X}}_{53}) = {\mathcal {C}}^\pi \end{aligned}$$

(88)

where the equality is because in any finite dimensional $*$-algebra ($\pi {\mathcal {X}}_{53}$) the bicommutant of a central $*$-subalgebra (${\mathcal {C}}^\pi$) is itself. Let us show that ${\mathcal {C}}^\pi$ admits a generating set of form $\{ \pi h\}$ where $h \in {\mathcal {A}}_{[0,53\ell ]}$ are $20\ell$-local; the argument is the same as in the proof of B.3. Note that ${\mathcal {X}}_{50}$ is $20\ell$-locally generated by B.3. Thus, $\pi {\mathcal {X}}_{50}$ is generated by $\pi$ times $20\ell$-local operators. By B.2 we find a central $*$-algebra ${\mathcal {E}}^\pi$ such that $\pi {\mathcal {X}}_{47} \subseteq {\mathcal {E}}^\pi \subseteq \pi {\mathcal {X}}_{50} \subseteq {\mathcal {C}}^\pi$. This ${\mathcal {E}}^\pi$ is contained in an algebra generated by $20\ell$-local operators. ${{\,\mathrm{{Comm}}\,}}({\mathcal {E}}^\pi \,|\, {\mathcal {C}}^\pi )$ is supported (modulo $\pi$) near the ends of the interval, which factorizes by B.1, and we can take each tensor factor as generators. Now, every $20\ell$-local generator $\pi h$ of ${\mathcal {C}}^\pi$ commutes with $\pi y^R_i$ trivially if h is supported on $[0,43\ell ]$. If h is supported on $[20\ell , 53\ell ]$, then by construction h belongs to ${{\,\mathrm{{Comm}}\,}}({\mathcal {D}}\,|\, {\mathcal {A}}_{[20\ell ,53\ell ]} )$. Such h commutes with $y \in {{\,\mathrm{{Cent}}\,}}{{\,\mathrm{{Comm}}\,}}({\mathcal {D}}\,|\, {\mathcal {A}}_{[20\ell ,53\ell ]} )$ and hence also with $y^R_i$. Therefore, $\pi y^R_i$ commutes with ${\mathcal {C}}^\pi$. Since $\pi y^R_i \in {\mathcal {C}}^\pi$ by (88), we have $\pi y^R_i \in {{\,\mathrm{{Comm}}\,}}({\mathcal {C}}^\pi , {\mathcal {C}}^\pi ) = {\mathbb {C}}\pi$. Since $\pi$ and $y^R_i$ are far apart in support, we conclude that $y^R_i \in {\mathbb {C}}{\textbf{1}}$. $\square$