$K_{m+\frac{1}{2}} \times \dots \times K_{n-\frac{1}{2}} \times K_{n-\frac{1}{2}} \times K_{n+\frac{1}{2}}$ in the following way: an element $\vec{k}$ is not related to anything if there exists a $x \in [[m+\frac{1}{2}, n-\frac{1}{2}]$ such that $\vec{k}$ does not have the same value in its two copies of $K_x$; and, otherwise, $\vec{k}$ is related to all the elements that have the same values as it in the two edge labels, $K_{m-\frac{1}{2}}$ and $K_{n+\frac{1}{2}}$. As a Boolean matrix, $\Delta_{\llbracket m, n \rrbracket}$ can be written as

$$ \begin{aligned} & (\Delta_{\llbracket m, n \rrbracket}) (k_x^L, k_x^R)_{x \in \llbracket m, n \rrbracket} \\ & = \delta_{k_m^L}^{l_m^L} \delta_{k_n^R}^{l_n^R} \prod_{x \in \llbracket m, n-1 \rrbracket} \delta_{k_x^R, k_{x+1}^L}^{l_x^R, l_{x+1}^L}. \end{aligned} \quad (132) $$

This route can be interpreted as ensuring the compliance with a single principle: for any $x \in \Gamma_N$, the indices $k_x^R$ and $k_{x+1}^L$ must always be equal. Thus, an operation on the sites of $\llbracket m, n \rrbracket$ can modify their value for any $x \in [m, n-1]$, as long as this value remains the same for both; while it cannot modify the values of $k_m^L$ and $k_n^R$, since they must remain equal to, respectively, $k_{m-1}^R$ and $k_{n+1}^L$, to which the operation does not have access. Note in particular that for the case of singletons, $\Delta_{\{n\}} = \delta \times \delta$. This corresponds to the index-matching circuits form in the main text's (44).

Finally, we accordingly denote the route for the whole algebra $\mathcal{A}_{\Gamma_N} = \omega$ as

$$ (\Delta_{\Gamma_N}) (k_x^L, k_x^R)_{x \in \Gamma_N} = \prod_{x \in \Gamma_N} \delta_{k_x^R, k_{x+1}^L} \delta_{k_x^R, k_{x+1}^L}^{l_x^R, l_{x+1}^R}. \quad (133) $$

Our representation theorem then takes the following form.

**Theorem F.2 (Representations of strongly connected 1D partitions).** With the previous notations, and defining $\tilde{\Pi} \in \text{Lin}(\otimes_{n \in X} \mathcal{H}_{A_n})$ through

$$ \tilde{\Pi} := \sum_{(k_n)} \left( \bigotimes_{n \in X} \Pi_{A_n}^{(k_{n-\frac{1}{2}}, k_{n+\frac{1}{2}})} \right), \quad (134) $$

there exists an isomorphism of $C^*$ algebras $\iota$ : $\omega \to \text{Lin}_{\Delta_{\Gamma_N}} (\otimes_{n \in \Gamma_N} \mathcal{H}_{A_n})$ such that

$$ \forall n \in X, \quad \iota(\mathcal{A}_n) = \left\{ \begin{array}{l} (f_{A_n} \otimes 1_{A_m}, m \neq n) \tilde{\Pi} \\ | \quad f \in \text{Lin}_{\delta \times \delta}(\mathcal{H}_{A_n}) \end{array} \right\}. \quad (135) $$

Furthermore, if this partition is fully representable, then $\iota$ can be chosen such that for any interval $[m, n] \subset \Gamma_N$,

$$ \iota(\mathcal{A}_{\llbracket m, n \rrbracket}) = \left\{ \begin{array}{l} (f_{A_m \dots A_n} \otimes 1_{A_{n+1} \dots A_{m-1}}) \tilde{\Pi} \\ | \quad f \in \text{Lin}_{\Delta_{\llbracket m, n \rrbracket}} (\mathcal{H}_{A_m} \otimes \dots \otimes \mathcal{H}_{A_n}) \end{array} \right\}. \quad (136) $$

Before proving this theorem, we note that combining it with Theorem F.1 directly yields the following, which is the technical version of the main text's Theorem 4.1.

**Theorem F.3.** Each of the $\Gamma_N^{\text{fine}}$-partitions at hand in Theorem D.2 is isomorphic to a partition of the form defined in Theorem F.2, with full representability, i.e. with (136) holding as well.

We now present the proof of Theorem F.2.

*Proof.* Ref. [21]'s Theorem 5.2 yields this result, but with (at least seemingly) different $\tilde{\Pi}$ and $\Delta$'s. What we have to show is thus that these can be replaced with the ones provided here.

Let us start with $\tilde{\Pi}$. In the original Theorem, it is defined from the Boolean tensor $\sigma$ encoding sectorial correlations, whose coefficients are given here by

$$ \begin{aligned} \sigma(k_n^L, k_n^R)_{n \in \Gamma_N} &= 0 \\ \iff & \prod_{n \in \Gamma_N} \pi_{n-\frac{1}{2}}^{k_n^L} \pi_{n+\frac{1}{2}}^{k_n^R} = 0. \end{aligned} \quad (137) $$

It is clear that the RHS condition is in particular satisfied whenever there exists a $n$ such that $k_n^R \neq k_{n+1}^L$. Otherwise, the condition reduces to $\prod_{n \in \Gamma_N} \pi_{n+\frac{1}{2}}^{k_n^R} = 0$. By Lemma F.1, this condition is equivalent to the existence of a $n \in \Gamma_n$ such that $\pi_{n-\frac{1}{2}}^{k_n^R} \pi_{n+\frac{1}{2}}^{k_n^R} = \pi_{n-\frac{1}{2}}^{k_n^L} \pi_{n+\frac{1}{2}}^{k_n^R} = 0$. Thus, defining for any $n \in \Gamma_N$ the Boolean tensor $\sigma(n)$ on $K_{n-\frac{1}{2}} \times K_{n+\frac{1}{2}}$ by

$$ \sigma(n)_{k_n^L k_n^R} = 0 \iff \pi_{n-\frac{1}{2}}^{k_n^L} \pi_{n+\frac{1}{2}}^{k_n^R} = 0, \quad (138) $$

we can write $\sigma$'s coefficients as

$$ \sigma(k_n^L, k_n^R)_{n \in \Gamma_N} = \prod_{n \in \Gamma_N} \delta_{k_n^R, k_{n+1}^L} \sigma(n)_{k_n^L k_n^R}. \quad (139) $$

Now, $\tilde{\Pi}$ is defined from $\sigma$ as
$$ \tilde{\Pi} := \sum_{(k_n^L, k_n^R)_n | \sigma_{(k_n^L, k_n^R)_n} \neq 0} \left( \bigotimes_{n \in \Gamma_N} \Pi_{A_n}^{(k_n^L, k_n^R)} \right). \quad (140) $$

However, one can always add to this formula some null projectors, and, from the way we defined the $\mathcal{H}_{A_n}^{(k_n^L, k_n^R)}$, we know that the ones labelled by indices for which $\sigma(n)_{(k_n^L, k_n^R)} = 0$ — i.e. $\pi_{n-\frac{1}{2}}^{k_n^L} \pi_{n+\frac{1}{2}}^{k_n^R} = 0$ — are null, and thus that the corresponding $\Pi$'s are null projectors. Thus, if we define

$$ \sigma_{(k_n^L, k_n^R)_n}^{\text{comp}} = \prod_{n \in \Gamma_n} \delta_{k_n^R, k_{n+1}^L} \overline{\prod_{n \in \Gamma_n} \sigma(n)_{k_n^L, k_n^R}}, \quad (141) $$

where the bar stands for Boolean negation ($\bar{0} := 1$ and $\bar{1} := 0$), then for any value assignment on which the previous is non-null, there is a $n \in \Gamma_n$ such that $\Pi_{A_n}^{(k_n^L, k_n^R)} = 0$, and thus $\bigotimes_{n \in \Gamma_N} \Pi_{A_n}^{(k_n^L, k_n^R)} = 0$.

Therefore, by adding only null projectors, we can replace the $\sigma_{(k_n^L, k_n^R)_n} \neq 0$ summation condition in (140) with $0 \neq \sigma_{(k_n^L, k_n^R)_n} + \sigma_{(k_n^L, k_n^R)_n}^{\text{comp}} = \prod_{n \in \Gamma_n} \delta_{k_n^R, k_{n+1}^L}$, which, after a relabelling of indices, yields (134).

Let us now turn to $\Delta_{\llbracket m, n \rrbracket}$. Ref. [21]'s Theorem 5.1 tells us that it is linked to AtomProj($\mathcal{Z}_{\llbracket m, n \rrbracket}$) in the following way: a certain tuple $(k_x^L, k_x^R)_{x \in \llbracket m, n \rrbracket}$ is in a non-empty equivalence class of the partial equivalence relation $\Delta_{\llbracket m, n \rrbracket}$ if and only if the corresponding $\prod_{x \in \llbracket m, n \rrbracket} \pi_n^{(k_x^L, k_x^R)}$ is non null; and in that case, there is a unique $\pi \in \text{AtomProj}(\mathcal{Z}_{\llbracket m, n \rrbracket})$ such that $\prod_{x \in \llbracket m, n \rrbracket} \pi_n^{(k_x^L, k_x^R)} \ll \pi$ (i.e. the former absorbs composition with the latter), and the tuple's equivalence class is that of the other tuples leading to the same inequality with the same $\pi$.

Now, similarly to before, it is direct that $\prod_{x \in \llbracket m, n \rrbracket} \pi_n^{(k_x^L, k_x^R)} = \prod_{x \in \llbracket m, n \rrbracket} \pi_{x-\frac{1}{2}}^{k_x^L} \pi_{x+\frac{1}{2}}^{k_x^R} \neq 0 \Rightarrow \forall x \in \llbracket m, n-1 \rrbracket, k_x^R = k_{x-1}^L$; and, assuming the latter (which we call (i)), we then have by Lemma F.2 that the former is equivalent to the statement that (ii) for every $x \in \llbracket m, n \rrbracket$, $\pi_{x-\frac{1}{2}}^{k_x^L} \pi_{x+\frac{1}{2}}^{k_x^R} \neq 0$ — i.e. $\sigma(x)_{(k_x^L, k_x^R)} \neq 0$ — and (iii) $\pi_{m-\frac{1}{2}}^{k_m^L} \pi_{n+\frac{1}{2}}^{k_n^R} \neq 0$. Thus, a certain tuple $(k_x^L, k_x^R)_{x \in \llbracket m, n \rrbracket}$ is in a non-empty equivalence

class if and only if conditions (i), (ii) and (iii) hold.

Assuming these, we now have to figure out what this equivalence class is. Since $\mathcal{Z}_{\llbracket m, n \rrbracket} = (\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \lor (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})$, its atomic projectors are the $\pi_{m-\frac{1}{2}}^{q_m^L} \pi_{n+\frac{1}{2}}^{q_n^R}$'s that are non-null. Obviously, we have $\prod_{x \in \llbracket m, n \rrbracket} \pi_n^{(k_x^L, k_x^R)} = \prod_{x \in \llbracket m, n \rrbracket} \pi_{x-\frac{1}{2}}^{k_x^L} \pi_{x+\frac{1}{2}}^{k_x^R} \ll \pi_{m-\frac{1}{2}}^{q_m^L} \pi_{n+\frac{1}{2}}^{q_n^R} \Leftrightarrow q_m^L = k_m^L \land q_n^R = k_n^R$. Thus, $\Delta_{\llbracket m, n \rrbracket}$ relates those assignments that agree on the value of $k_m^L$ and on that of $k_n^R$.

This leads us to the following formula for $\Delta_{\llbracket m, n \rrbracket}$, where we define $\sigma(m, n)_{k_m^L, k_n^R} = 0 \Leftrightarrow \pi_{m-\frac{1}{2}}^{k_m^L} \pi_{n+\frac{1}{2}}^{k_n^R} = 0$:

$$ \begin{aligned} & \left(\Delta_{\llbracket m, n \rrbracket}\right)_{\substack{l_x^L, l_x^R \\ (k_x^L, k_x^R)}}_{x \in \llbracket m, n \rrbracket} \\ &= \delta_{k_m^L}^{l_m^L} \delta_{k_n^R}^{l_n^R} \left( \prod_{x \in \llbracket m, n-1 \rrbracket} \delta_{k_x^R, k_{x+1}^L}^{l_x^R} \delta_{l_x^L, l_{x+1}^L}^{l_x^R} \right) \\ & \left( \prod_{x \in \llbracket m, n \rrbracket} \sigma(x)_{k_x^L, k_x^R} \right) \left( \prod_{x \in \llbracket m, n \rrbracket} \sigma(x)_{l_x^L, l_x^R} \right) \\ & \left(\sigma(m, n)_{k_m^L, k_n^R}\right). \end{aligned} \quad (142) $$

We see that the difference with (132) lies in the two last lines. Let us start by showing how we can get rid of the penultimate one, using the following Lemma.

**Lemma F.7.** Let $\mathcal{H} = \bigoplus_{k \in K} \mathcal{H}^k$ be a sectorised Hilbert space, whose sectors can be null. Suppose $\mu$ is a relation on $K$ satisfying: $\forall k, l, \mu_k^l = 1 \Rightarrow \dim(\mathcal{H}^k) = 0$ or $\dim(\mathcal{H}^l) = 0$. Then $\text{Lin}_{\Delta}(\mathcal{H}) = \text{Lin}_{\Delta+\mu}(\mathcal{H})$.

The proof is direct from the fact that for any $k, l$ such that $\mu_k^l = 1$, the $\mathcal{H}^k \to \mathcal{H}^l$ component of any operator on $\mathcal{H}$ is necessarily null, since at least one of these has null dimension.

Here, let us define $\mu$ through the formula (142) with the penultimate line replaced with its barred version. Then, for any choice of index values making $\mu$ non-null, there exists a $x$ such that $\sigma(x)_{k_x^L, k_x^R}$ or $\sigma(x)_{l_x^L, l_x^R}$ is null; thus one of $\mathcal{H}_{A_n}^{k_x^L, k_x^R}$ or $\mathcal{H}_{A_n}^{l_x^L, l_x^R}$ is of dimension 0, therefore so is one of $\otimes_{x \in \llbracket m, n \rrbracket} \mathcal{H}_{A_n}^{k_x^L, k_x^R}$ or $\otimes_{x \in \llbracket m, n \rrbracket} \mathcal{H}_{A_n}^{l_x^L, l_x^R}$, so Lemma F.7 can be applied, and we can replace $\Delta$ with $\Delta+\mu$ in (136).
$\Delta + \mu$ is given by the formula (142) without its RHS's second line, so we can redefine $\Delta_{[[m,n]]}$ as this:

$$ \begin{aligned} & (\Delta_{[[m,n]]})_{(k_x^L, k_x^R) \in [[m,n]]}^{(l_x^L, l_x^R) \in [[m,n]]} \\ & = \delta_{k_m^L}^{l_m^L} \delta_{k_n^R}^{l_n^R} \left( \prod_{x \in [[m,n-1]]} \delta_{k_x^R, k_{x+1}^L} \delta_{l_x^R, l_{x+1}^L} \right) \quad (143) \\ & (\sigma(m,n))_{k_m^L, k_n^R}; \end{aligned} $$

we still have to get rid of the last line, by adding a $\nu$ similarly defined as featuring this line's barred version. We can do it because of the following Lemma.

**Lemma F.8.** If $\nu$ is a relation on $K_m \times \dots \times K_n$ such that $\forall f \in \text{Lin}_\nu(\mathcal{H}_{A_m} \otimes \dots \otimes \mathcal{H}_{A_n})$, $\tilde{\Pi}(f \otimes 1_{A_{n+1}\dots A_{m-1}}) = (f \otimes 1_{A_{n+1}\dots A_{m-1}})\tilde{\Pi} = 0$, then (136) can be rewritten with $\Delta_{[[m,n]]} + \nu$ instead of $\Delta_{[[m,n]]}$.

The proof is direct from the fact that a $f$ in $\text{Lin}_{\Delta+\nu}(\mathcal{H})$ can be written as $f_0 + f_1$ where $f_0 \in \text{Lin}_\Delta(\mathcal{H})$ and $f_1 \in \text{Lin}_\nu(\mathcal{H})$, so that, in (136), the set of the $(f \otimes 1)\tilde{\Pi}$'s for $f \in \text{Lin}_{\Delta+\nu}(\mathcal{H})$ is the same as the set of the $(f_0 \otimes 1)\tilde{\Pi}$'s for $f \in \text{Lin}_\Delta(\mathcal{H})$.

Taking $\nu$ as being equal to (143) with its last line barred, let us check that it satisfies Lemma F.8's assumptions. For a choice of indices for which it is non-null, one has $\sigma(m,n)_{k_m^L, k_n^R} = \sigma(m,n)_{l_m^L, l_n^R} = 0$ and therefore $\pi_{m-\frac{1}{2}}^{k_m^L} \pi_{n+\frac{1}{2}}^{k_n^R} = 0$. This entails that, for any choice of index values $(k_x)_{x \in \Gamma_N^{+1/2}}$ such that $k_{m-\frac{1}{2}} = k_m^L$ and $k_{n+\frac{1}{2}} = k_n^R$, $\prod_x \pi_x = 0$, so that Lemma F.1 yields that there exists a $x \in \Gamma_N$ such that $\pi_{x-\frac{1}{2}}^{k_{x-\frac{1}{2}}} \pi_{x+\frac{1}{2}}^{k_{x+\frac{1}{2}}} = 0$, and consequently $\prod_{A_x}^{(k_{x-\frac{1}{2}}, k_{x+\frac{1}{2}})} = 0$.

Therefore, for a given $f \in \text{Lin}_\nu(\mathcal{H}_{A_m} \otimes \dots \otimes \mathcal{H}_{A_n})$, using the formula (134) for $\tilde{\Pi}$ and the fact

that $f$ follows $\nu$, we can write

$$ \begin{aligned} & = \sum_{k_m^L, k_n^R | \sigma(m,n)_{k_m^L, k_n^R} = 0} (f_{A_m \dots A_n} \otimes 1_{A_{n+1} \dots A_{m-1}}) \tilde{\Pi} \\ & \qquad \sum_{(k_x)_{x \in \Gamma_N^{+1/2}} | k_{m-\frac{1}{2}} = k_m^L, k_{n+\frac{1}{2}} = k_n^R} (f_{A_m \dots A_n} \otimes 1_{A_{n+1} \dots A_{m-1}}) \\ & \qquad \left( \bigotimes_{x \in \Gamma_N} \prod_{A_x}^{(k_{x-\frac{1}{2}}, k_{x+\frac{1}{2}})} \right) \\ & = 0, \end{aligned} \quad (144) $$

since for every term in the sum, at least one of the $\prod_{A_x}^{(k_{x-\frac{1}{2}}, k_{x+\frac{1}{2}})}$'s in the last line is null. Since the same is true symmetrically for the $l_x$'s, yielding $\tilde{\Pi}(f_{A_m \dots A_n} \otimes 1_{A_{n+1} \dots A_{m-1}}) = 0$, the assumptions of Lemma F.8 are satisfied and we can therefore replace $\Delta_{[[m,n]]}$ with $\Delta_{[[m,n]]} + \nu$ in (143), thus yielding the desired form (132). $\square$

## F.4 Unitary-product form of local automorphisms

Now that we have shown that each of the intermediate partitions in the causal decompositions is representable, we have to ensure that the mappings between them themselves take a local form when seen as maps between Hilbert spaces. This requires to investigate the conditions under which a local automorphism on a partition corresponds to conjugation by a product of unitaries.

**Definition F.1 (Local automorphisms).** We say that a $C^*$-algebra automorphism $\mathcal{U}$ on a partitioned $C^*$ algebra $(\mathcal{A}_S)_{S \subseteq X} \vdash \omega$ is *local* if it satisfies:

$$ \forall S \subseteq X, \quad \mathcal{U}(\mathcal{A}_S) = \mathcal{A}_S. \quad (145) $$

Labelling each of the individual algebras' centres' atomic projectors as AtomProj($\mathcal{Z}_x$) = $\{\pi_x^{k_x}\}_{k_x \in K_x}$ for any $x \in X$, we say that a local $\mathcal{U}$ is *inner-local*<sup>24</sup> if it satisfies:

$$ \forall x \in X, \forall k_x \in K_x, \quad \mathcal{U}(\pi_x^{k_x}) = \pi_x^{k_x}, \quad (146) $$

or in other words if it restricts to the identity on each of the $\mathcal{Z}_x$'s.

<footnote n="24"><sup>24</sup>This terminology stems from the fact that (146) is equivalent to the requirement that $\mathcal{U}$'s restrictions to each of the $\mathcal{A}_x$'s be inner automorphisms.</footnote>
**Proposition F.2.** Let $(\mathcal{A}_S)_{S \subseteq X} \vdash \omega$ be a partition. Let $\mathcal{U}$ be an inner-local $\mathcal{C}^*$ automorphism on it.

Then there exist unitaries $U_n \in \mathcal{A}_n$ for every $n \in X$, and a dephasing unitary $\phi \in \mathrm{Unit}(\bigvee_n \mathcal{Z}_n)$, such that, denoting

$$U = \phi \prod_{n \in X} U_n, \qquad (147)$$

$\mathcal{U}$ acts as conjugation by $U$, i.e.

$$\mathcal{U} : f \mapsto U f U^\dagger. \qquad (148)$$

*Proof.* If $\omega$ is factor, then an automorphism of it is necessarily of the form $\mathcal{U} : f \mapsto U f U^\dagger$ for some unitary map $U \in \omega$. If it is not, then, since $\mathcal{U}$ acts as the identity on $\mathcal{Z}(\omega) \subseteq \bigvee_{n \in X} \mathcal{Z}_n$, it stabilises each of $\omega$'s blocks, and thus also has the form $\mathcal{U} : f \mapsto U f U^\dagger$. Furthermore, $\mathcal{U}(\pi_n^{k_n}) = U \pi_n^{k_n} U^\dagger = \pi_n^{k_n}$ is equivalent to $U \pi_n^{k_n} = \pi_n^{k_n} U$, so we find $U = \sum_{(k_n)_{n \in X}} \prod_n \pi_n^{k_n} U = \sum_{(k_n)_{n \in X}} (\prod_n \pi_n^{k_n}) U (\prod_n \pi_n^{k_n})$. Denoting $\pi^k := \prod_n \pi_n^{k_n}$ and $U^k := \pi^k U \pi^k$, this becomes $U = \sum_{\vec{k}} U^{\vec{k}}$, a block decomposition for $U$.

Fixing $n \in X$, and an atomic projector $\pi_n^{k_n}$ of $\mathcal{Z}_n$, $\pi_n^{k_n} \mathcal{A}_n$ is a factor $\mathcal{C}^*$ algebra, preserved by $\mathcal{U}$, so $\mathcal{U}|_{\pi_n^{k_n} \mathcal{A}_n}$ is of the form $f \mapsto U_n^{k_n} f (U_n^{k_n})^\dagger$, with $U_n^{k_n}$ a unitary operator of $\pi_n^{k_n} \mathcal{A}_n$. Fixing $\vec{k} = (k_m)_{m \in X}$ and using $\pi^k = \prod_m \pi_m^{k_m} = \prod_m U_m^{k_m} (U_m^{k_m})^\dagger = (\prod_m U_m^{k_m}) (\prod_m (U_m^{k_m})^\dagger)$, this leads, for a $f \in \pi^k \mathcal{A}_n$, to

$$\begin{aligned}
\mathcal{U}(f) &= \mathcal{U}(\pi^{\vec{k}} f) \\
&= \mathcal{U}(\pi^{\vec{k}} \pi_n^{k_n} f) \\
&= \mathcal{U}(\pi^{\vec{k}}) \mathcal{U}(\pi_n^{k_n} f) \\
&= \pi^{\vec{k}} U_n^{k_n} f (U_n^{k_n})^\dagger \qquad (149) \\
&= (\prod_m U_m^{k_m}) (\prod_m (U_m^{k_m})^\dagger) U_n^{k_n} f (U_n^{k_n})^\dagger \\
&= (\prod_m U_m^{k_m}) f (\prod_m (U_m^{k_m})^\dagger).
\end{aligned}$$

Thus, writing $\tilde{U}^{\vec{k}} := \prod_m U_m^{k_m}$, $\mathcal{U}$ and $f \mapsto \tilde{U}^{\vec{k}} f (\tilde{U}^{\vec{k}})^\dagger$ coincide on each of the $\pi^{\vec{k}} \mathcal{A}_n$ for varying $n$. Since the latter algebraically span $\pi^{\vec{k}}(\bigvee_n \mathcal{A}_n)$, the two morphisms also coincide on this space, and therefore $U^{\vec{k}}$ and $\tilde{U}^{\vec{k}}$ differ only by a phase $\alpha$, which might in general depend on $\vec{k}$:

$$U^{\vec{k}} = e^{i\alpha(\vec{k})} \tilde{U}^{\vec{k}} = e^{i\alpha(\vec{k})} \prod_n U_n^{k_n}. \qquad (150)$$

Denoting $\phi := \sum_{\vec{k}} e^{i\alpha(\vec{k})} \prod_n \pi_n^{k_n} \in \mathrm{Unit}(\bigvee_n \mathcal{Z}_n)$ and $U_n := \sum_{k_n} U_n^{k_n}$, the block decomposition $U = \sum_{\vec{k}} U^{\vec{k}}$ turns into (147). $\square$

Proposition F.2 shows that not all inner-local automorphisms are only a conjugation by a $\prod_n U_n$; they might include a dephasing $\phi$ as well, and we don't necessarily have $\phi = \prod_n \phi_n$ with $\phi_n \in \mathrm{Unit}(\mathcal{Z}_n)$ (in which case we could absorb it into the $U_n$'s).$^{25}$

It is non-trivial to determine the form of the possible dephasings $\phi$ in (147): $\hat{\phi}$ itself has to be local (since $\mathcal{U}$ and conjugation by $\prod_n U_n$ are local), a condition that not all dephasings satisfy.$^{26}$ The following Lemma yields some constraints on local dephasings.

**Lemma F.9.** Let $(\mathcal{A}_S)_{S \subseteq X} \vdash \omega$ be a partition of a factor. Let $\phi \in \mathrm{Unit}(\bigvee_{n \in X} \mathcal{Z}_n)$ be a dephasing such that $\hat{\phi}$ is an inner-local $\mathcal{C}^*$ automorphism.

$^{25}$A concrete counterexample is given by the (strongly connected) 3-partition of $\mathrm{Lin}(\mathbb{C}^8)$ given by $\mathcal{A}_0 := \{\mathrm{Diag}(\alpha, \alpha, \beta, \beta, \gamma, \gamma, \delta, \delta)\}$, $\mathcal{A}_1 := \{\mathrm{Diag}(\alpha, \beta, \alpha, \beta, \gamma, \delta, \gamma, \delta)\}$, $\mathcal{A}_2 := \{\mathrm{Diag}(\alpha, \beta, \gamma, \delta, \alpha, \beta, \gamma, \delta)\}$ ('Diag' stands for the diagonal matrix with the specified list of diagonal entries), with other algebras given by taking commutants: conjugation by the dephasing $\mathrm{Diag}(1, 1, 1, 1, 1, 1, 1, -1)$ is inner-local on this partition, but not of the form $\prod_n U_n$ with $U_n \in \mathrm{Unit}(\mathcal{A}_n)$.

$^{26}$For instance, in the (strongly connected) 4-partition of $\mathrm{Lin}(\mathbb{C}^4)$ defined by $\mathcal{A}_0 = \mathcal{A}_1 = \begin{Bmatrix} \alpha & 0 & 0 & 0 \\ 0 & \beta & 0 & 0 \\ 0 & 0 & \alpha & 0 \\ 0 & 0 & 0 & \beta \end{Bmatrix}$, $\mathcal{A}_2 = \mathcal{A}_3 = \begin{Bmatrix} \alpha & 0 & 0 & 0 \\ 0 & \alpha & 0 & 0 \\ 0 & 0 & \beta & 0 \\ 0 & 0 & 0 & \beta \end{Bmatrix}$, $\mathcal{A}_{\{0,1\}} = \begin{Bmatrix} \alpha & \gamma & 0 & 0 \\ \delta & \beta & 0 & 0 \\ 0 & 0 & \alpha & \gamma \\ 0 & 0 & \delta & \beta \end{Bmatrix}$, with other algebras defined by taking spans or commutants, conjugation by the dephasing $\phi = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$ is not local, as $\hat{\phi}(\mathcal{A}_{\{0,1\}}) = \begin{Bmatrix} \alpha & \gamma & 0 & 0 \\ \delta & \beta & 0 & 0 \\ 0 & 0 & \alpha & -\gamma \\ 0 & 0 & -\delta & \beta \end{Bmatrix} \neq \mathcal{A}_{\{0,1\}}$.
Then for any $S \subseteq X$, there exist $\phi_S \in \mathrm{Unit}(\bigvee_{n \in S} \mathcal{Z}_n)$, $\phi_{\bar{S}} \in \mathrm{Unit}(\bigvee_{n \in \bar{S}} \mathcal{Z}_n)$ such that

$$ \phi = \phi_S \phi_{\bar{S}}. \qquad (151) $$

*Proof.* We fix $S \subseteq X$. By Proposition F.2 applied to the coarse-graining of the partition into the bipartition $(\mathcal{A}_S, \mathcal{A}_{\bar{S}}) \vdash \omega$, $\hat{\phi}$ can be seen as a conjugation by $\tilde{\phi}V_S V_{\bar{S}}$, with $V_S \in \mathrm{Unit}(\mathcal{A}_S)$, $V_{\bar{S}} \in \mathrm{Unit}(\mathcal{A}_{\bar{S}})$, and $\tilde{\phi} \in \mathrm{Unit}(\mathcal{Z}_S \vee \mathcal{Z}_{\bar{S}})$. $\mathcal{Z}_S = \mathcal{Z}_{\bar{S}}$, so $\tilde{\phi}$ is in $\mathrm{Unit}(\mathcal{A}_S)$ and can be absorbed into $V_S$. Thus, we find $\phi = V_S V_{\bar{S}}$.

Furthermore, since $\hat{\phi}$'s restriction to $\mathcal{A}_S$ (given by $\hat{V}_S$) is a local C*-automorphism with respect to the restricted partition $(\mathcal{A}_T)_{T \subseteq S} \vdash \mathcal{A}_S$, Proposition F.2 yields $V_S = \phi_S \prod_{n \in S} V_n$ for some $\phi_S \in \mathrm{Unit}(\bigvee_{n \in S} \mathcal{Z}_n)$ and $V_n \in \mathrm{Unit}(\mathcal{A}_n)$. The $V_n$'s are defined (as in the proof of Proposition F.2) from the action of $\hat{\phi}$ on each individual $\pi_n^{k_n} \mathcal{A}_n$; but given the form of $\phi$, this action is the identity, so the $V_n$'s can be taken to be identities. Applying the same reasoning to the case of $\bar{S}$, we find the form (151). $\square$

Using this Lemma, we can prove that in the case of strongly connected 1D partitions, of a size sufficiently larger than their correlation length, the inner-local automorphisms involve no dephasing.

**Proposition F.3.** Let $(\mathcal{A}_S)_{S \subseteq \Gamma_N} \vdash \omega$ be a strongly connected partition of a factor $\omega$, with correlation length at most $l$, where $N \ge 3l + 1$. Let $\mathcal{U}$ be an inner-local C* automorphism on it.

Then there exists, for every $n \in \Gamma_N$, a unitary $U_n \in \mathcal{A}_n$ such that, denoting $U := \prod_n U_n$,

$$ \forall f \in \omega, \quad \mathcal{U}(f) = U f U^\dagger. \qquad (152) $$

*Proof.* Leaning on Proposition F.2, what is left to show is that $\phi$ can be made to disappear. This amounts to proving that any dephasing $\phi$ such that $\hat{\phi}$ is local is of the form $\phi = \prod_n \phi_n$, with $\phi_n \in \mathrm{Unit}(\mathcal{Z}_n)$, so that it can simply be absorbed into the $U_n$'s.

We fix two antipodes $p_+$ and $p_-$ to 0, as in Lemma D.9. Applying Lemma F.9 for $S = \{0, 1\}$ and for $S = \llbracket 1, p_+ \rrbracket$ yields $\phi = \phi_{\{0,1\}} \phi_{\llbracket 2,-1 \rrbracket} = \phi_{\llbracket 1, p_+ \rrbracket} \phi_{\llbracket p_-, 0 \rrbracket}$, and thus

$$ \phi_{\{0,1\}} = \phi_{\llbracket 2,-1 \rrbracket}^{\dagger} \phi_{\llbracket 1, p_+ \rrbracket} \phi_{\llbracket p_-, 0 \rrbracket}. \qquad (153) $$

Moreover, from strong connectedness, we have that for any $m, n \in \Gamma_N$, $\phi_{\llbracket m, n \rrbracket} \in$

$\mathrm{Unit}(\bigvee_{x=m-\frac{1}{2}}^{n+\frac{1}{2}} \mathcal{Z}_{x-\frac{1}{2}} \cap \mathcal{Z}_{x+\frac{1}{2}})$; so there exists a real function $\alpha_{\llbracket m, n \rrbracket}(k_{m-\frac{1}{2}}, \dots, k_{n+\frac{1}{2}})$ (defined modulo $2\pi$) such that

$$ \phi_{\llbracket m, n \rrbracket} = \sum_{k_{m-\frac{1}{2}}, \dots, k_{n+\frac{1}{2}}} e^{i \alpha_{\llbracket m, n \rrbracket} (k_{m-\frac{1}{2}}, \dots, k_{n+\frac{1}{2}})} \prod_{x=m-\frac{1}{2}}^{n+\frac{1}{2}} \pi_x^{k_x}. \qquad (154) $$

We now fix compatible values (in the sense that the product of their corresponding projectors is not null) $k_{-\frac{1}{2}}, k_{\frac{1}{2}}, k_{\frac{3}{2}}$ in the corresponding sets of labels. We then fix a $k_{p_+ + \frac{1}{2}}$; note that it is necessarily compatible with the three previous ones by correlation length, so its choice does not have to depend on them. Finally we fix values of $k$ for all other half-indices such that the list of values $\vec{k} = (k_n)_{n \in \Gamma_N^{+1/2}}$ is compatible. Importantly, Lemma F.1 ensures that imposing compatibility makes the choice of $(k_{\frac{5}{2}}, \dots, k_{p_+ - \frac{1}{2}})$ dependent only on the values of $k_{\frac{3}{2}}$ and $k_{p_+ + \frac{1}{2}}$ (but not on those of $k_{-\frac{1}{2}}$ and $k_{\frac{1}{2}}$); while symmetrically the choice $(k_{p_- + \frac{1}{2}}, \dots, k_{-\frac{3}{2}})$ depends only on the values of $k_{-\frac{1}{2}}$ and $k_{p_+ + \frac{1}{2}}$. Multiplying (153) by $\prod_{n \in \Gamma_N^{+1/2}} \pi_n^{k_n}$ then yields

$$ \alpha_{\{0,1\}}(k_{-\frac{1}{2}}, k_{\frac{1}{2}}, k_{\frac{3}{2}}) = \alpha_{\llbracket 1, p_+ \rrbracket}(k_{\frac{1}{2}}, \dots, k_{p_+ + \frac{1}{2}}) \\ + \alpha_{\llbracket p_-, 0 \rrbracket}(k_{p_+ + \frac{1}{2}}, \dots, k_{\frac{1}{2}}) \\ - \alpha_{\llbracket 2, -1 \rrbracket}(k_{\frac{3}{2}}, \dots, k_{-\frac{1}{2}}). \qquad (155) $$

The previous considerations on choice dependencies entail that $\alpha_{\llbracket 1, p_+ \rrbracket}$, $\alpha_{\llbracket p_-, 0 \rrbracket}$ and $\alpha_{\llbracket 2, -1 \rrbracket}$ only depend on $(k_{\frac{1}{2}}, k_{\frac{3}{2}})$, $(k_{-\frac{1}{2}}, k_{\frac{1}{2}})$, and $(k_{-\frac{1}{2}}, k_{\frac{3}{2}})$, respectively. It follows that
$$
\begin{align}
\phi_{\{0,1\}} &= \sum_{k_{-\frac{1}{2}}, k_{\frac{1}{2}}, k_{\frac{3}{2}}} e^{i\alpha_{\{0,1\}}(k_{-\frac{1}{2}}, k_{\frac{1}{2}}, k_{\frac{3}{2}})} \prod_{x=-\frac{1}{2}}^{\frac{3}{2}} \pi_x^{k_x} \nonumber \\
&= \sum_{k_{-\frac{1}{2}}, k_{\frac{1}{2}}, k_{\frac{3}{2}}} \left( e^{i\alpha_{\llbracket 1, p+1 \rrbracket}(k_{\frac{1}{2}}, k_{\frac{3}{2}})} \pi_{\frac{1}{2}}^{k_{\frac{1}{2}}} \pi_{\frac{3}{2}}^{k_{\frac{3}{2}}} \right. \nonumber \\
&\qquad \left. e^{i\alpha_{\llbracket p-, 0 \rrbracket}(k_{-\frac{1}{2}}, k_{\frac{1}{2}})} \pi_{-\frac{1}{2}}^{k_{-\frac{1}{2}}} \pi_{\frac{1}{2}}^{k_{\frac{1}{2}}} \right. \nonumber \\
&\qquad \left. e^{i\alpha_{\llbracket 2, -1 \rrbracket}(k_{-\frac{1}{2}}, k_{\frac{3}{2}})} \pi_{\frac{1}{2}}^{k_{-\frac{1}{2}}} \pi_{\frac{3}{2}}^{k_{\frac{3}{2}}} \right) \tag{156} \\
&= \left( \sum_{k_{\frac{1}{2}}, k_{\frac{3}{2}}} e^{i\alpha_{\llbracket 1, p+1 \rrbracket}(k_{\frac{1}{2}}, k_{\frac{3}{2}})} \pi_{\frac{1}{2}}^{k_{\frac{1}{2}}} \pi_{\frac{3}{2}}^{k_{\frac{3}{2}}} \right) \nonumber \\
&\quad \left( \sum_{k_{-\frac{1}{2}}, k_{\frac{1}{2}}} e^{i\alpha_{\llbracket p-, 0 \rrbracket}(k_{-\frac{1}{2}}, k_{\frac{1}{2}})} \pi_{-\frac{1}{2}}^{k_{-\frac{1}{2}}} \pi_{\frac{1}{2}}^{k_{\frac{1}{2}}} \right) \nonumber \\
&\quad \left( \sum_{k_{-\frac{1}{2}}, k_{\frac{3}{2}}} e^{i\alpha_{\llbracket 2, -1 \rrbracket}(k_{-\frac{1}{2}}, k_{\frac{3}{2}})} \pi_{\frac{1}{2}}^{k_{-\frac{1}{2}}} \pi_{\frac{3}{2}}^{k_{\frac{3}{2}}} \right). \nonumber
\end{align}
$$

Thus $\phi_{\{0,1\}}$ can be written as the product of a unitary operator of $(\mathcal{Z}_0 \cap \mathcal{Z}_1) \vee (\mathcal{Z}_1 \cap \mathcal{Z}_2) = \mathcal{Z}_1$, a unitary operator of $(\mathcal{Z}_{-1} \cap \mathcal{Z}_0) \vee (\mathcal{Z}_0 \cap \mathcal{Z}_1) = \mathcal{Z}_0$, and a unitary operator of $(\mathcal{Z}_{-1} \cap \mathcal{Z}_0) \vee (\mathcal{Z}_1 \cap \mathcal{Z}_2)$.

Therefore, up to unitary operators of individual $\mathcal{Z}_n$'s (which can be absorbed in the $\prod_n U_n$ part), $\phi = \phi_{\{0,1\}}\phi_{\llbracket 2,-1 \rrbracket}$ is in Unit$(\bigcup_{n \neq \frac{1}{2}} \mathcal{Z}_{n-\frac{1}{2}} \cap \mathcal{Z}_{n+\frac{1}{2}})$ (i.e. it 'does not depend on $k_{\frac{1}{2}}$'). We can repeat this procedure to erase the centres one by one and find that $\phi$ can eventually be reduced to the identity.

<figure><img src="image_18.png" /></figure>

We can now apply this to the case in which the partition is in its represented form given by Theorem F.2, i.e. seen as a space of routed maps over a tensor product of Hilbert spaces.

**Corollary F.1.** Let $(\mathcal{A}_S)_{S \subseteq \Gamma_N} \vdash \omega$ be a strongly connected 1D partition of a factor $\omega$ with correlation length at most $l$, where $N \ge 3l + 1$. We suppose it is in its canonical represented form expressed by Theorem F.2. Let $\mathcal{U}$ be an inner-local $C^*$ automorphism on it.

Then there exists, for every $n \in \Gamma_N$, a unitary $U_n \in \text{Lin}_{\delta \times \delta}(\mathcal{H}_{A_n})$ such that

$$
\forall f \in \omega, \mathcal{U}(f) = \left( \bigotimes_n U_n \right) f \left( \bigotimes_n U_n^\dagger \right) . \quad (157)
$$

*Proof.* Proposition F.3 and (135) yield the form $\mathcal{U}(f) = (\otimes_n U_n) \tilde{\cap} f (\otimes_n U_n^\dagger) \tilde{\cap}$. Since we have $\forall f \in \omega, f = \tilde{\cap}f$, we can absorb the $\tilde{\cap}$'s into $f$ and get the desired form. $\square$

<figure><img src="image_19.png" /></figure>

F.5 Representing the causal decompositions

We are finally ready to express the $C^*$-isomorphisms in the causal decompositions in terms of unitary maps between Hilbert spaces.

Let us go back to Theorem F.2 and look at the sets of indices of a partition $(\mathcal{A}_S^{\text{in}})_{S \subseteq \Gamma_N} \vdash \omega^{\text{in}}$ over $\Gamma_N$, and those of a fine-graining $(\mathcal{A}_S^{\text{out}})_{S \subseteq \Gamma_N^{\text{fine}}} \vdash \omega^{\text{out}}$ of it over $\Gamma_N^{\text{fine}}$. For the $\Gamma_N$ partition, each node $n \in \Gamma_N$ bears two indices $k_n^L \in K_{n-\frac{1}{2}}$ and $k_n^R \in K_{n+\frac{1}{2}}$, where, for $m \in \Gamma_N^{+1/2}$, $K_m$ indexes AtomProj($\mathcal{Z}_{m-\frac{1}{2}} \cap \mathcal{Z}_{m+\frac{1}{2}}$).

For its fine-graining, consider, for instance, the node $-\frac{1}{4}$; it bears two indices $k_{-\frac{1}{4}}^L$ and $k_{-\frac{1}{4}}^R$. The first belongs to a set indexing AtomProj($\mathcal{Z}_{-\frac{3}{4}} \cap \mathcal{Z}_{-\frac{1}{4}}$). Yet $\mathcal{Z}_{-\frac{3}{4}} \cap \mathcal{Z}_{-\frac{1}{4}} = \mathcal{Z}_{\{-\frac{5}{4}, -\frac{3}{4}\}} \cap \mathcal{Z}_{\{-\frac{1}{4}, +\frac{1}{4}\}}$ by strong connectedness; and $\mathcal{Z}_{\{-\frac{5}{4}, -\frac{3}{4}\}}$ is isomorphic to the $\mathcal{Z}_{-1}$ of the $\Gamma_N$-partition, while $\mathcal{Z}_{\{-\frac{1}{4}, +\frac{1}{4}\}}$ is isomorphic to $\mathcal{Z}_0$. In other words, the set of indices for $k_{-\frac{1}{4}}^L$ is precisely $K_{-\frac{1}{2}}$, as defined for the $\Gamma_N$-partition. On the other hand, if we denote as $\tilde{K}_0$ a set of indices for AtomProj($\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}$), $k_{-\frac{1}{4}}^R$ takes value in $\tilde{K}_0$.

Generalising this, we see that in a representation of the fine-grained partition by routed maps as in Theorem F.2, any node of the form $n - \frac{1}{4}$ for $n \in \Gamma_N$ bears an index $k_{n-\frac{1}{4}}^L \in K_{n-\frac{1}{2}}$, and an index $k_{n-\frac{1}{4}}^R \in \tilde{K}_n$, where the latter is a set of indices for AtomProj($\mathcal{Z}_{n-\frac{1}{4}} \cap \mathcal{Z}_{n+\frac{1}{2}}$). It will thus be represented over a sectorised Hilbert space

$$
\mathcal{H}_{n-\frac{1}{4}} = \bigoplus_{\substack{k_{n-\frac{1}{4}}^{\text{L}} \in K_{-\frac{1}{2}} \\ k_{n-\frac{1}{4}}^{\text{R}} \in \tilde{K}_n}} \mathcal{H}_{n-\frac{1}{4}}^{(k_{n-\frac{1}{4}}^{\text{L}}, k_{n-\frac{1}{4}}^{\text{R}})} . \quad (158)
$$

For nodes of the form $n + \frac{1}{4}$ where $n \in \Gamma_N$, the situation is symmetric.

The fact that one partition is a fine-graining of the other entails that each block of $\mathcal{A}_n^{\text{in}}$ is isomorphic to the corresponding block of $\mathcal{A}_{\{n-\frac{1}{4}, n+\frac{1}{4}\}}^{\text{out}}$, and thus at the level of Hilbert spaces, for every $(k_n^L, k_n^R) \in K_{n-\frac{1}{2}} \times K_{n+\frac{1}{2}}$,
$$ \mathcal{H}_n^{(k_n^L, k_n^R)} \cong \bigoplus_{\tilde{k}_n \in \tilde{K}_n} \mathcal{H}_{n-\frac{1}{4}}^{(k_n^L, \tilde{k}_n)} \otimes \mathcal{H}_{n+\frac{1}{4}}^{(\tilde{k}_n, k_n^R)}. \quad (159) $$

A practical way to represent these isomorphisms is through a routed unitary. Defining the relation $\lambda_n : K_{n-\frac{1}{2}} \times K_{n+\frac{1}{2}} \rightarrow K_{n-\frac{1}{2}} \times \tilde{K}_n \times \tilde{K}_n \times K_{n+\frac{1}{2}}$ by

$$ (\lambda_n)_{k_n^L, k_n^R}^{k_n^L, \tilde{k}_n, \tilde{k}'_n, k_n^R} = \delta_{k_n^L}^{k_n^L} \delta_{k_n^R}^{k_n^R} \delta_{\tilde{k}_n, \tilde{k}'_n}, \quad (160) $$

a choice of unitary maps instantiating the isomorphisms (159) corresponds to a choice of a routed unitary $(\lambda_n, U)$ from $\bigoplus_{k_n^L, k_n^R} \mathcal{H}_n^{(k_n^L, k_n^R)}$ to $(\bigoplus_{k_n^L, \tilde{k}_n} \mathcal{H}_{n-\frac{1}{4}}^{(k_n^L, \tilde{k}_n)} \otimes (\bigoplus_{\tilde{k}'_n, k_n^R} \mathcal{H}_{n+\frac{1}{4}}^{(\tilde{k}'_n, k_n^R)})$.

The next Proposition states that, if the algebras are in the represented form specified by Theorem F.2, then the fine-graining isomorphism $\omega^{\text{in}} \rightarrow \omega^{\text{out}}$ acts as a conjugation by a tensor product of such routed unitaries.

**Proposition F.4.** Let $\mathcal{U} : \omega^{\text{in}} \rightarrow \omega^{\text{out}}$ be a fine-graining from the strongly connected 1D partition $(A_S^{\text{in}})_{S \subseteq \Gamma_N} \vdash \omega^{\text{in}}$ to the strongly connected 1D partition $(A_S^{\text{out}})_{S \subseteq \Gamma_N^{\text{fine}}} \vdash \omega^{\text{out}}$. We suppose both are partitions of factors and have correlation length at most $l$ where $N \ge 3l + 1$. We also suppose $\omega^{\text{in}}$ is a representation as an operator algebra adapted to its partition, with the form specified by Theorem F.2, and similarly for $\omega^{\text{out}}$.

With the notations laid out in the previous paragraphs, there exists, for every $n \in \Gamma_N$, a routed unitary $(\lambda_n, U_n)$ from $\bigoplus_{k_n^L, k_n^R} \mathcal{H}_n^{(k_n^L, k_n^R)}$ to $(\bigoplus_{k_n^L, \tilde{k}_n} \mathcal{H}_{n-\frac{1}{4}}^{(k_n^L, \tilde{k}_n)} \otimes (\bigoplus_{\tilde{k}'_n, k_n^R} \mathcal{H}_{n+\frac{1}{4}}^{(\tilde{k}'_n, k_n^R)})$ such that

$$ \mathcal{U} : f \mapsto \left( \bigotimes_n U_n \right) f \left( \bigotimes_n U_n \right)^{\dagger}. \quad (161) $$

*Proof.* For every $n \in \Gamma_N$, we define an arbitrary routed unitary $(\lambda_n, V_n)$ from $\bigoplus_{k_n^L, k_n^R} \mathcal{H}_n^{(k_n^L, k_n^R)}$ to $(\bigoplus_{k_n^L, \tilde{k}_n} \mathcal{H}_{n-\frac{1}{4}}^{(k_n^L, \tilde{k}_n)} \otimes (\bigoplus_{\tilde{k}'_n, k_n^R} \mathcal{H}_{n+\frac{1}{4}}^{(\tilde{k}'_n, k_n^R)})$. Denoting $V = \bigotimes_n V_n$, $\hat{V}$ is then a fine-graining isomorphism from $(A_S^{\text{in}})_{S \subseteq \Gamma_N} \vdash \omega^{\text{in}}$ to $(A_S^{\text{out}})_{S \subseteq \Gamma_N^{\text{fine}}} \vdash \omega^{\text{out}}$. Thus $\hat{V}^{-1} \circ \mathcal{U}$ is an inner-local$^{27}$ automorphism of $(A_S^{\text{in}})_{S \subseteq \Gamma_N}$.

By Corollary F.1, we can find unitaries $W_n \in \text{Lin}_{\delta \times \delta}(\mathcal{H}_n)$ such that $\hat{V}^{-1} \circ \mathcal{U}(f) = (\otimes_n W_n) f (\otimes_n W_n^\dagger)$. Defining $U_n := V_n W_n$ then yields (161). $\square$

Note that Proposition F.4 applies symmetrically to coarse-grainings, and to fine- or coarse-grainings of partitions over $\Gamma_N^{+1/2}$. We can now provide the technical proof of the main text’s Theorem 4.2.

**Theorem F.4.** Let $N, r$ be such that $N > 4r$. We define collections of Hilbert spaces and a unitary map between them,

$$ U : \bigotimes_n \mathcal{H}_{A_n^{\text{in}}} \rightarrow \bigotimes_n \mathcal{H}_{A_n^{\text{out}}}, \quad (162) $$

where (depending on whether $r$ is a half-integer) the output factorisation might be indexed by $\Gamma_N^{+1/2}$.

$\hat{U}$ is a 1D QCA of radius $r$, with respect to the natural factorisations of its input and output spaces, if and only if $U$ admits a decomposition of the form represented in Figure 1.

*Proof.* The ‘if’ part is a direct application of Proposition 4.2. Turning to the ‘only if’ part, we suppose $\hat{U}$ is a 1D QCA of radius $r$. By Theorem D.2, there exist $2r$ fine-grainings $\mathcal{U}_1^{\text{fine}}, \dots, \mathcal{U}_{2r}^{\text{fine}}$ and $2r$ coarse-grainings $\mathcal{U}_1^{\text{coarse}}, \dots, \mathcal{U}_{2r}^{\text{coarse}}$ such that

$$ \mathcal{U} = \prod_{i=1}^{2r} \mathcal{U}_i^{\text{coarse}} \mathcal{U}_i^{\text{fine}}. \quad (163) $$

By Theorem F.3, all of the intermediate partitions involved are isomorphic to adapted representations as operator algebras. By inserting $\iota^{-1}_l$'s between each of the isomorphisms in (163) and redefining the $\mathcal{U}_i^{\text{coarse}}$'s and $\mathcal{U}_i^{\text{fine}}$'s through absorbing them, we obtain a version of (163) in which all of the fine- and coarse-grainings obey the conditions of Proposition F.4.

Thus, by this Proposition F.4, there exist routed unitaries $(\lambda_n^{i, \text{fine}}, U_n^{i, \text{fine}})$ and $(\lambda_n^{i, \text{coarse}}, U_n^{i, \text{coarse}})$ such that for any $i$, $\mathcal{U}_i^{\text{fine}}$ acts as conjugation by $(\otimes_n U_n^{i, \text{fine}})$ and $\mathcal{U}_i^{\text{coarse}}$ acts as conjugation by $(\otimes_n U_n^{i, \text{coarse}})$. $\hat{U}$ and conjugation by their sequential composition are therefore equal as isomorphisms of $C^*$ algebras, so $U$ and the circuit describing this sequential composition are equal, up to a global phase that

<footnote>$^{27}$Innerness stems from the fact that we precisely used $\mathcal{U}$ to identify blocks of the two partitions.</footnote>
can be absorbed into any of the linear maps. By seeing each $U_n^{i+1, \text{fine}} \circ U_n^{i, \text{coarse}}$ in this circuit as a single map, we obtain the form given in Figure 1. $\square$

## F.6 The translation-invariant case

We now turn to the specific case of translation-invariant QCAs. Note that when a strongly connected 1D partition is represented in the canonical way (given by Theorem F.2) as an operator algebra over a tensor product of sectorised Hilbert spaces, and when each sectorised Hilbert space $\mathcal{H}_{A_n}$ is identical to $\mathcal{H}_{A_{n+1}}$, there is a natural shift automorphism Sh of this partition given by conjugation by the shift unitary map:

$$ \mathrm{Sh} = \hat{\mathrm{S}} \qquad (164) $$

with

$$ S = \bigotimes_n S_n, \qquad (165) $$

where each of the $S_n$'s is an identity map from $\mathcal{H}_{A_n}$ to $\mathcal{H}_{A_{n+1}}$. This includes, as a special instance, the case of algebras of operators over tensor products of Hilbert spaces.

(It is important to remark that if the partition is indexed by $\Gamma_N^{\text{fine}}$, then this does not entail that all of the Hilbert spaces at hand are identical, since one has $\mathcal{H}_{A_n} = \mathcal{H}_{A_{n+1}}$ but not necessarily $\mathcal{H}_{A_n} = \mathcal{H}_{A_{n+1/2}}$. Thus, in this case there are two repeated Hilbert spaces involved: $\mathcal{H}_{A_{1/4}} = \mathcal{H}_{A_{5/4}} = \dots$ and $\mathcal{H}_{A_{3/4}} = \mathcal{H}_{A_{7/4}} = \dots$. Similarly, in the case of partitions over $\Gamma_N$ or $\Gamma_N^{+1/2}$, all Hilbert spaces are indeed identical, as well as the index sets $K_n = K_n^L \times K_n^R$ for their sectorisations; but that does not entail that the left-index sets and right-index sets are identical: one can have $K_0^L = K_1^L = \dots \neq K_0^R = K_1^R = \dots$)

First, let us prove that any strongly connected 1D partition with a preferred shift automorphism can be represented in this canonical way.

**Theorem F.5.** When applying Theorem F.2 to a partition equipped with a preferred shift automorphism Sh, the sectorised Hilbert spaces can be picked such that $\mathcal{H}_{A_n} = \mathcal{H}_{A_{n+1}}$ for every n,

and the isomorphism $\iota$ can be picked such that (using the definition (165) of S)

$$ \iota \mathrm{Sh} = \hat{\mathrm{S}} \iota. \qquad (166) $$

*Proof.* For any $n$, $A_n$ is isomorphic to $\mathrm{Sh}(A_n) = A_{n+1}$, so the sectorised Hilbert spaces used to represent them can be picked to be identical.

We then have that $\hat{\mathrm{S}}^{-1} \iota \mathrm{Sh} \iota^{-1}$ is an inner-local automorphism of the partition's represented form $(\iota(A_S))_{S \subseteq \Gamma_N}$, so by Corollary F.1 it is a conjugation by some tensor product of unitaries $\otimes_n U_n$, with (34b) amounting to (up to a phase which we can absorb into one of the $U_n$'s)

$$ U_{N-1} \dots U_0 = 1. \qquad (167) $$

We then denote

$$ \forall n, \quad V_n := (U_{n-1} \dots U_0)^{\dagger}, \qquad (168) $$

and define a new isomorphism $\tilde{\iota} := \widehat{\otimes_n V_n} \iota$. Since $\widehat{\otimes_n V_n}$ is an inner-local automorphism of $(\iota(A_S))_{S \subseteq \Gamma_N}$, $\tilde{\iota}$ is also a representation isomorphism. We can then compute

$$ \begin{aligned} \hat{S}^{-1} \tilde{\iota} \mathrm{Sh} \tilde{\iota}^{-1} & = \hat{S}^{-1} \widehat{\otimes_n V_n} \iota \mathrm{Sh} \iota^{-1} \widehat{\otimes_n V_n^\dagger} \\ & = \widehat{\otimes_n V_{n+1}} \hat{S}^{-1} \iota \mathrm{Sh} \iota^{-1} \widehat{\otimes_n V_n^\dagger} \\ & = \widehat{\otimes_n V_{n+1}} \widehat{\otimes_n U_n} \widehat{\otimes_n V_n^\dagger} \\ & = \widehat{\otimes_n V_{n+1} U_n V_n^\dagger}, \end{aligned} \qquad (169) $$

and it is straightforward to compute that $V_{n+1} U_n V_n^\dagger = 1 \forall n$, yielding (166). $\square$

Next, let us make Corollary F.1 more specific in the translation-invariant case.

**Proposition F.5.** In Corollary F.1, we additionally suppose that the partition is equipped with the canonical shift (164), and that $U$ is translation-invariant. Then the $U_n$'s for different $n$'s can be taken to be identical.

*Proof.* $U$, which is a conjugation by $\otimes_n U_n$, is by translation invariance equal to $\hat{\mathrm{S}}^{-1} U \hat{\mathrm{S}}$, which is a conjugation by $\otimes_n U_{n+1}$; thus the two are equal up to a global phase. Possibly redefining the $U_n$'s

<footnote>$^{28}$Note that this is also requires that the spaces' sectorisations, as well as the sets indexing them, are identical.</footnote>
by changing their phase (which does not change
the action by conjugation), this yields

$$
\forall n, \quad U_n = U_{n+1}. \qquad (170)
$$

<figure><img src="image_20.png" /></figure>

We can then similarly adapt Proposition F.4.

**Proposition F.6.** In Proposition F.4, we additionally suppose that the two partitions are equipped with the canonical shift (164) and that $\mathcal{U}$ is translation-invariant. Then the $U_n$'s for different $n$'s can be taken to be identical.

Proof. Going through the proof of Proposition F.4, we take the arbitrary routed unitaries $(\lambda_n, V_n)$ to be identical (note that the $\lambda_n$'s, which are fixed, are identical since the indexing sets for different $n$'s are the same). Then $\hat{V}^{-1} \circ \mathcal{U}$ is also translation-invariant, so we can apply Proposition F.5 to take the $W_n$'s to be identical, so that the $U_n := V_n W_n$'s are identical as well. $\square$

<figure><img src="image_21.png" /></figure>

Finally, we obtain the translation-invariant
version of our main theorem.

**Theorem F.6.** In Theorem F.4, we suppose that the $\mathcal{H}_{A_n^{\text{in}}}$'s are all identical, and that so are the $\mathcal{H}_{A_n^{\text{out}}}$'s.

$\hat{U}$ is a translation-invariant 1D QCA if and only $U$ admits a decomposition of the form represented in Figure 1, with all the unitary maps in any given layer being identical.

Proof. For the ‘if’ part, one can directly infer
from the translation-invariance of the circuit that
$\hat{U}$ is itself translation-invariant. Turning to the
‘only if’ part, Theorem 3.2 ensures that all the
intermediate isomorphisms in (163) can be taken
to be translation-invariant as well.

Following through our proof of Theorem F.4,
we can use Theorem F.5 to ensure that the repre-
sentation isomorphisms are themselves adapted
to the shifts, so that our redefinitions of the
$U_i^{\text{fine}}$'s and $U_i^{\text{coarse}}$ are translation-invariant as
well. One can then apply Proposition F.6 to en-
sure that, for any given $i$, the $U_n^{i,\text{fine}}$'s are all iden-
tical, and that so are the $U_n^{i,\text{coarse}}$'s. $\square$

<figure><img src="image_22.png" /></figure>