# Causal decompositions of one-dimensional quantum cellular automata

Augustin Vanrietvelde, Octave Mestoudjian, Pablo Arrighi

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Augustin Vanrietvelde, Octave Mestoudjian, Pablo Arrighi. Causal decompositions of one-dimensional quantum cellular automata. 2025. hal-05236387

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# Causal decompositions of one-dimensional quantum cellular automata

Augustin Vanrietvelde<fnref n="1" />, Octave Mestoudjian<fnref n="2" />, and Pablo Arrighi<fnref n="2" />

Understanding quantum theory's causal structure stands out as a major matter, since it radically departs from classical notions of causality. We present advances in the research program of *causal decompositions*, which investigates the existence of an equivalence between the causal and the compositional structures of unitary channels. Our results concern one-dimensional Quantum Cellular Automata (1D QCAs), i.e. unitary channels over a line of N quantum systems (with or without periodic boundary conditions) that feature a causality radius *r*: a given input cannot causally influence outputs at a distance more than *r*. We prove that, for $N \ge 4r+1$, 1D QCAs all admit causal decompositions: a unitary channel is a 1D QCA if and only if it can be decomposed into a unitary routed circuit of nearest-neighbour interactions, in which its causal structure is compositionally obvious. This provides the first constructive form of 1D QCAs with causality radius one or more, fully elucidating their structure. In addition, we show that this decomposition can be taken to be translation-invariant for the case of translation-invariant QCAs. Our proof of these results makes use of innovative algebraic techniques, leveraging a new framework for capturing partitions into non-factor sub-C* algebras.

## Contents

1 **Introduction** 2
Relationship with previous work . 4

2 **Partitions and QCAs over them** 6

Augustin Vanrietvelde: vanrietvelde@telecom-paris.fr

2.1 Reminders about C*-algebras . . . . 6

2.2 Partitions . . . . . . . . . . . . . . . . 7

2.3 1D QCAs over partitions . . . . . . 7

3 **Causal decompositions of 1D QCAs** 8

3.1 The general case . . . . . . . . . . . . 8

3.2 The translation-invariant case . . . 10

4 **Causal decompositions in routed unitary circuits** 11

4.1 Routed maps and index-matching circuits . . . . . . . . . . . . . . . . . . 11

4.2 Representations of 1D partitions . 12

4.3 The general case . . . . . . . . . . . . 13

4.4 The translation-invariant case . . . 14

5 **Conclusion and outlook** 14

**Acknowledgements** 15

**Data availability statement** 15

**References** 15

A **Notations** 17

B **General results on C* algebras** 18

B.1 Homomorphisms of C* algebras and their kernels . . . . . . . . . . . . . 18

B.2 Sub-C* algebras of a centre are coarse-grainings . . . . . . . . . . . . . . 19

B.3 Double commutants . . . . . . . . . . . 20

B.4 Some results on partitions . . . . . . 20

B.5 Proof of Proposition 2.1 . . . . . . . . 21

C **Further notions on partitions** 21

C.1 Correlations between subalgebras and correlation length . . . . . . . . . 22

C.2 Connected partitions over a graph 23

C.3 Strong connectedness in 1D . . . . . . 24

D **Proof of Theorem 3.1** 25

D.1 A few lemmas . . . . . . . . . . . . . . . 25

<footnote n="1">¹Télécom Paris, Institut Polytechnique de Paris, Inria Saclay, Palaiseau, France</footnote>

<footnote n="2">²Université Paris-Saclay, Inria, CNRS, LMF, 91190 Gif-sur-Yvette, France</footnote>
<table><tbody><tr><td>D.2</td><td>Vertex-splitting</td><td>28</td></tr><tr><td>D.3</td><td>Fine-graining</td><td>34</td></tr><tr><td>D.4</td><td>Envoi</td><td>36</td></tr><tr><td>E</td><td>Proof of Theorem 3.2</td><td>37</td></tr><tr><td>F</td><td>Proofs for Section 4</td><td>37</td></tr><tr><td>F.1</td><td>Some Lemmas</td><td>37</td></tr><tr><td>F.2</td><td>Full representability of the partitions</td><td>38</td></tr><tr><td>F.3</td><td>Representations of strongly connected 1D partitions</td><td>40</td></tr><tr><td>F.4</td><td>Unitary-product form of local automorphisms</td><td>43</td></tr><tr><td>F.5</td><td>Representing the causal decompositions</td><td>46</td></tr><tr><td>F.6</td><td>The translation-invariant case</td><td>48</td></tr></tbody></table>

# 1 Introduction

Quantum theory's *causal structure* stands out as one of its most defining features, dramatically encapsulating its departure from classical notions. While the violation of Bell inequalities demonstrates the inadequacy of classical causal modelling in a quantum context, unitary quantum theory is endowed with its own, very robust, notion of causal structure, as articulated e.g. by the framework of quantum causal models [1, 2, 3]. One of many equivalent definitions for (the absence of) quantum causal influence is the following. A unitary channel $U$ with two inputs $A$ and $B$ and two outputs $C$ and $D$

<figure><img src="image_2.png" /></figure>

features no causal influence from $A$ to $D$ iff discarding $C$ turns it into just a discarding of $A$ and a quantum channel from $B$ to $D$:

$$ \exists C, \quad \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.3] \draw (-0.5,0) -- (0.5,0); \node[above] at (-0.5,0) {$C$}; \node[above] at (0.5,0) {$D$}; \node[left] at (-0.5,0) {$A$}; \node[right] at (0.5,0) {$B$}; \node[above] at (0,0.5) {$U$}; \node[below] at (0,-0.5) {$C$}; \node[below] at (0,-0.5) {$D$}; \end{tikzpicture} = \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.3] \draw (-0.5,0) -- (0.5,0); \node[above] at (-0.5,0) {$C$}; \node[above] at (0.5,0) {$D$}; \node[left] at (-0.5,0) {$A$}; \node[right] at (0.5,0) {$B$}; \node[above] at (0,0.5) {$U$}; \node[below] at (0,-0.5) {$C$}; \node[below] at (0,-0.5) {$D$}; \end{tikzpicture}. \quad (2) $$

This definition can be readily extended to channels with arbitrary numbers of inputs and outputs.

A promising direction for better understanding the causal structure of unitary quantum theory

is to explore its ties with a closely related but a priori distinct notion — with which it is, in fact, routinely confused — that of its *compositional structure*. A statement about our previous $U$'s compositional structure would, for instance, be that it can be built out of unitary channels $U_1$ and $U_2$ in the following way:

<figure><img src="image_3.png" /></figure>

To better express the respective significance of causal and compositional structures, it is enlightening to emphasise their different conceptual foundations. In broad terms, the absence of causal influence, expressed in (2) is an *operational* statement, which can for instance be framed in terms of the possible signalling between agents related by the unitary channel $U$. The decomposition into local channels expressed in (3), on the other hand, has no obvious operational meaning; rather, it concerns our ability to fine-grain the dynamics at hand, with the identification of a causal mediator $X$. What (3) lacks in operational meaning compared to (2), it gains in mathematical intelligibility, telling us valuable information about the inner structure of $U$ and the ways in which it can be built from elementary unitary channels.

In (finite-dimensional) unitary quantum theory, (2) and (3) are equivalent [4]. This striking fact has two major consequences. First, the two conceptually distinct notions of causal and compositional structures turn out — at least for the simple case at hand — to be two faces of the same coin. Second, it unlocks a *constructive form* of the unitary channels satisfying the no-influence relation (2): they are all and only the channels that can be built in the form (3). For this reason, we call the decomposition (3) a *causal decomposition*.

A natural research direction, put forward in Ref. [5], is then to investigate whether the above equivalence generalises to the case of other, more involved causal structures, relating more input and more outputs. Ref. [5] showed that all causal structures relating three inputs to three outputs
indeed admit such causal decompositions (some
of which have to be written in the framework
of unitary routed circuits [6, 7], an extension of
standard unitary quantum circuits). That every
causal structure is equivalent to a corresponding
causal decomposition in unitary quantum theory
remains, to this date, an open conjecture.<fnref n="1" /> Pro-
ving it would unveil a remarkable correspondence
between the informational and the structural as-
pects of the theory.

*Quantum Cellular Automata* (QCAs) [9, 10] form an important class of quantum dynamics singled out by their causal structure (see the reviews [11, 12]). In the most general definition, a QCA consists of a unitary dynamics defined on a given graph of quantum systems (or sites), and displaying, with respect to this graph’s notion of distance, a certain causality radius *r*: a given input site cannot causally influence output systems at a distance *r* from it.<fnref n="2" /> QCAs have been proposed as a model of distributed quantum computing [13, 14], and as models of discretised spacetimes, with the causality radius enforcing relativistic causality, e.g. in order to cast quantum simulation algorithms for quantum field theory [15, 16, 17, 18].

A question of crucial importance to better un-
derstand QCAs is that of their classification:
what are the possible ones? Of equal importance
is that of their construction: how to build them?
Answering these questions means fully figuring
out (in a simplified, discretised case) the exact

scope of quantum dynamics consistent with the
causal structure of a Minkowski spacetime. The
most satisfactory result, solving both problems
at once, would be to obtain a *constructive form*
analogous to (3): a way to specify QCAs as all
and only those *U*'s that can be built in a certain
way from elementary unitary channels.

So far, a fully constructive form has only been
found for QCAs of radius $\frac{1}{2}$ over a 1D lattice (i.e.
a line): they are all and only the unitary channels
of the form

<figure><img src="image_4.png" /></figure>

where each of the smaller boxes is a unitary chan-
nel [10, 19, 20]. We can precisely recognise this,
in hindsight, as an instance of a causal decom-
position. However, even in 1D, the existence
of causal decompositions (or of any constructive
form) had so far not been proven for radii greater
than $\frac{1}{2}$ (see the discussion of previous results at
the end of this Section).

In this paper, we prove the existence of causal
decompositions for 1D QCAs, in the case of fi-
nite length segments or finite diameter loops.
Namely, we show that for any *r*, any 1D QCA
of radius *r* over a lattice of length strictly greater
than *4r*, admits a causal decomposition general-
ising (4). For instance, our result yields that 1D
QCAs of radius 1 are all and only the unitary
operators of the form

<figure><img src="image_5.png" /></figure>

while higher radii are simply given by stacking up
more layers (see Figure 1 for the general form). In
the case of a translation-invariant QCA, we prove
that the decomposition itself can be written in a
translation-invariant way.

One should note that such decompositions can-
not be expressed in terms of standard unitary

<footnote n="1">¹One might also wonder whether this holds for the the-
ory of quantum channels. This is indeed the case for sim-
ple instances: Ref. [4]’s result in fact applies to channels
in general (see also Ref. [8] for a related result about the
equivalence of commutation and factorisation of a Hilbert
space for general quantum channels). However, this can-
not be the case in general, due to the fact that quantum
channels feature no causal structure but a mere signalling
structure, that does not satisfy the important property
of atomicity [3]: a quantum channel may exhibit no sig-
nalling from A to B and from A to C, while exhibiting
signalling from A to the joint system BC. Such a sig-
nalling structure cannot find any compositional counter-
part in terms of absences of paths in a circuit. Thus, a
decomposition encoding all of the information present in
a given signalling structure cannot be provided in general
for quantum channels.</footnote>

<footnote n="2">²A condition of homogeneity is also usually imposed,
but as we shall see, it is not important for this paper’s
results. Unless otherwise specified, we take the various
local Hilbert spaces to be of possibly different dimensions,
and the QCAs not to be translation-invariant.</footnote>
quantum circuits; as is (5), they are written in
terms of the more general *unitary routed circuits*,
a more flexible, fully formalised framework aimed
at better capturing representations of non-factor
C\* algebras [6, 7]. These feature distinctive float-
ing labels *i*, *j*, ..., which serve to denote the use of
blends of direct sums and tensor products when
combining Hilbert spaces. For sufficiently com-
plex causal structures, causal decompositions do
not exist in standard unitary circuits without
routes (see Ref. [5] for a counterexample).

Relatedly, we emphasise that, just like (4) in
the case of radius $\frac{1}{2}$, layers in such a decomposi-
tion do not have to map between the same fac-
torisations: it is possible that the dimensions of
local Hilbert spaces change from a layer to an-
other, as long as unitarity is maintained. This
includes the possibility that the decomposition
of a QCA on a homogeneous array (i.e. one in
which every system has the same dimension) in-
volves non-homogeneous arrays as its intermedi-
ate systems.

We thus have a fully constructive form, and
one of dramatic physical significance: in discre-
tised 1D, a unitary dynamics is causal if and only
if it can be described as a sequence of nearest-
neighbour interactions. This also yields a full
understanding of the structure of 1D QCAs, of
any radius.

Our result is also noteworthy for its novel proof
technique, which relies on the introduction of a
(concurrently developed) theory of partitions of
(finite-dimensional) C\* algebras into an arbitrary
number of subsystems that, crucially, account for
the latter's potential non-factorness [21]. Our
proofs unveil many interesting structural features
of the partitions involved in the decompositions,
in particular the interplay between algebras' cen-
tres and the 1D lattice's metric and topological
structures. We expect the insights provided by
this innovative structural perspective to prove
useful for future extensions of these results.

Before closing this introduction with a discus-
sion of our results' relationships with previous
work, let us present this paper's structure. We
start with an introduction to the theory of parti-
tions and a definition of QCAs over them, gene-
ralising the standard QCAs defined over factori-
sations (Section 2). We then prove our main
result in the C\* algebraic picture (Section 3),
before showing how, through the use of routed

quantum circuits, it can be expressed in terms
of (routed) maps between Hilbert spaces as well
(Section 4). We conclude in Section 5.

Relationship with previous work

A certain number of results have been found that
partly elucidate the structure of QCAs. In par-
ticular, Ref. [10] proved that 1D QCAs of radius
$\frac{1}{2}$ are all and only the unitaries of the form (4).
A source of confusion is that it was claimed in
Ref. [10] that the same result also held for spa-
tial dimensions 2 and more; but the proof in those
cases turned out to be broken. The proof was cu-
rated for dimension 1 in Ref. [19], where a coun-
terexample was also given for higher dimension.
The same ideas lead to the theory of a 1D QCA's
index [20].

Because a 1D QCA of arbitrary causality ra-
dius *r* can always be coarse-grained (by grouping
sites) into a 1D QCA of radius $\frac{1}{2}$, all 1D QCAs
can be classified by their index, and put into the
circuit form (4). However, resorting to coarse-
graining then applying the radius $\frac{1}{2}$ decomposi-
tion fails to provide a causal decomposition, or a
constructive form, for QCAs of radius *r*: it yields
a decomposition that corresponds to a causality
radius $2r - \frac{1}{2}$ instead. This failure translates into
a practical obstacle to constructing QCAs of ra-
dius *r*: using this circuit form, one will also pro-
duce QCAs of larger radius.

Another important structural result is that
QCAs over arbitrary graphs, and with an arbi-
trary causality radius, can all be implemented by
a local circuit of finite depth [22]. Albeit appeal-
ing for its robustness and generality, this result
again fails to provide a causal decomposition, for
two reasons. First, the unitaries used in such
a decomposition also act on ancillary systems,
with the promise that the total action amounts
to an identity on these ancillary systems, some-
thing that cannot be checked easily when start-
ing from an arbitrary decomposition of that form.
Second, these unitaries are also promised to com-
mute pairwise, a condition which, here again,
cannot be readily checked. Clearly, these two
departures from fully constructive form lead to
practical obstacles towards constructing QCAs of
a given radius.
<figure><img src="image_6.png" /><figcaption>Figure 1: Causal decomposition of a 1D QCA of radius $r$. $r$ is either an integer or a half-integer. (In the latter case, the decomposition would end with shifted outputs $B_{1/2}, B_{3/2},$...) This uses the language of routed unitary quantum circuits, and more specifically here, of index-matching circuits [6]. All boxes represent routed unitaries.

Theorem 4.2 states that a unitary over a (possibly looped) 1D graph of $N$ finite-dimensional Hilbert spaces, with $N > 4r$, is a 1D QCA of radius $r$ if and only if it admits a decomposition of this form. Theorem 4.3 states that it is a translation-invariant such QCA if and only if it admits a decomposition of this form in which all boxes of a same layer are identical.</figcaption></figure>
## 2 Partitions and QCAs over them

The most important ingredient of this paper — essential to the statement and proof of its main result — is the theory of partitions of a quantum system. These generalise factorisations in order to accommodate the potential non-factorness of the C* algebras corresponding to some parts of the system. Here, we provide basic definitions and facts: further discussions, as well as proofs of the theorems, can be found in Ref. [21].

### 2.1 Reminders about C*-algebras

Partitions leverage the theory and structure of C* algebras; let us remind the basics. All our further definitions and theorems pertain solely to finite dimension (and for the sake of brevity, we won't recall it every time).<fnref n="3" />

**Definition 2.1** (C* algebras). A (finite-dimensional) *algebra* $\omega$ is a (finite-dimensional) algebra over complex numbers, equipped with a dagger, i.e. an antilinear involution satisfying $(fg)^{\dagger} = g^{\dagger}f^{\dagger}$.

A *sub-* algebra $\mathcal{A}$ of $\omega$ is a (unital) subalgebra of $\omega$, closed under the dagger. Given two *sub-* algebras $A_1$ and $A_2$ we denote by $A_1 \vee A_2$ their algebraic and $\dagger$ closure, which is also a *sub-* algebra.

We say that $\omega$ is a *C*\* algebra if it is embeddable as a *sub-* algebra of some space of complex matrices, equipped with the standard adjoint as its dagger. Any *sub-* algebra of $\omega$ is then a C\* algebra as well, and can thus also be called a *sub-C*\* algebra of $\omega$.

In finite dimension, typical examples of C* algebras are spaces $\text{Lin}(\mathcal{H})$ of linear operators over a certain finite-dimensional Hilbert space. If $\omega \cong \text{Lin}(\mathcal{H}_{\Omega})$ for a certain $\mathcal{H}_{\Omega}$, we call $\omega$ a factor. Not all C* algebras are factors, however; the general case is for them to be a direct sum of factors. To understand how this plays out, we need the notion of a centre.

**Definition 2.2** (Centre). The centre of a C\* algebra $\omega$ is

$$ \mathcal{Z}(\omega) := \{f \in \omega | \forall g \in \omega, fg = gf\}. \quad (6) $$

The centre of a C\* algebra is a commutative C\* algebra, and as such satisfies the following theorem.

**Theorem 2.1.** Let $\mathcal{Z}$ be a commutative C\* algebra. Then there exists a unique finite set $\text{AtomProj}(\mathcal{Z}) = \{\pi^k\}_{k \in K}$ of non-zero projectors of $\mathcal{Z}$, called $\mathcal{Z}$'s atomic projectors, satisfying $\pi^k = (\pi^k)^{\dagger}$, $\pi^k \pi^l = \delta^{kl}\pi^k$ and $\sum_k \pi^k = 1$, and forming a basis of $\mathcal{Z}$.

Centres are crucial tools to deal with C\* algebras. In particular, the non-factorness of $\omega$ is witnessed by the existence of a non-trivial centre (i.e. one not reduced to scalar multiples of the identity 1), and this centre's atomic projectors then precisely designate each of $\omega$'s summand factors.

**Theorem 2.2.** Let $\omega$ be a C\* algebra. We denote its centre's atomic projectors as $\text{AtomProj}(\mathcal{Z}(\omega)) = \{\pi^k\}_{k \in K}$. Then each of the $\pi^k\omega := \{\pi^k f = f\pi^k | f \in \omega\}$ is a factor, i.e. $\pi^k\omega \cong \text{Lin}(\mathbb{C}^{d_k})$ and

$$ \begin{align} \omega &= \biguplus_k \pi^k \omega \notag \\ &\cong \bigoplus_k \pi^k \omega \tag{7} \\ &\cong \bigoplus_k \text{Lin}(\mathbb{C}^{d_k}), \notag \end{align} $$

where the $\biguplus$ denotes the linear span of the union. Consequently, $\omega$ is a factor if and only if $\text{AtomProj}(\mathcal{Z}(\omega)) = \{1\}$.

Thus, any non-factor C\* algebra can concretely be seen as (i.e. is isomorphic to) an algebra of block-diagonal matrices; its centre's atomic projectors are the projectors onto each of its blocks.

Finally, let us introduce commutants, which are a convenient way of defining a *sub-C*\* algebra's complement.

**Definition 2.3** (Commutant). Let $\omega$ be a C\* algebra and $\mathcal{A}$ a *sub-C*\* algebra of it. $\mathcal{A}$'s *commutant* within $\omega$ is

$$ \mathcal{A}' := \{f \in \omega | \forall g \in \omega, fg = gf\}. \quad (8) $$

If $\omega$ is a factor, then $\mathcal{A}'' = \mathcal{A}$ (see Proposition B.2), which also implies that $\mathcal{Z}(\mathcal{A}) := \mathcal{A} \cap \mathcal{A}' = \mathcal{A}'' \cap \mathcal{A}' =: \mathcal{Z}(\mathcal{A}')$.

<footnote n="3">³In the context of this paper, we could equivalently use *Von Neumann algebras*, which in finite dimension are the same thing as C* algebras.</footnote>
the former to the latter is an isomorphism of $C^*$
algebras $\mathcal{U}^{\text{fine}} : \omega^{\text{in}} \to \omega^{\text{out}}$ satisfying

$$
\forall S \subseteq \Gamma, \quad \mathcal{U}^{\text{fine}}(\mathcal{A}_S^{\text{in}}) = \mathcal{A}_{S \pm \frac{1}{4}}^{\text{out}}. \qquad (23)
$$

A *coarse-graining* is the inverse of a *fine-graining*.

In our informal graphical notation, we can de-
note a fine-graining of a partition over $\Gamma_N$ as

<figure><img src="image_7.png" /></figure>

and a coarse-graining symmetrically as

<figure><img src="image_8.png" /></figure>

while those over $\Gamma_N^{+1/2}$ take a symmetric form.
In this graphical notation, performing a $\Gamma_N$-fine-
graining then a $\Gamma_N^{+1/2}$-coarse-graining looks like

<figure><img src="image_9.png" /></figure>

in which we can intuitively see that this yields
a QCA of radius $\frac{1}{2}$ (which is indeed straightfor-
ward to prove formally). This yields the tools
to express this paper's main theorem: 1D QCAs
are decomposable as sequences of such radius-$\frac{1}{2}$
QCAs.

**Theorem 3.1.** Let $\mathcal{U}$ be a 1D QCA between factorisations over $\Gamma_N$ or $\Gamma_N^{+1/2}$, with causality radius $r$, where $N > 4r$.

There exist $2r$ (generalised) 1D QCAs
$\mathcal{U}_1, \dots, \mathcal{U}_{2r}$, with causality radius $\frac{1}{2}$, such that

$$
\mathcal{U} = \mathcal{U}_{2r} \dots \mathcal{U}_1. \qquad (27)
$$

Furthermore, in that case, each of the $\mathcal{U}_i$'s decomposes as

$$
\mathcal{U}_i = \mathcal{U}_i^{\text{coarse}} \mathcal{U}_i^{\text{fine}}, \qquad (28)
$$

where $\mathcal{U}_i^{\text{fine}}$ is a fine-graining and $\mathcal{U}_i^{\text{coarse}}$ is a coarse-graining.

Note that in the previous theorem, the $\mathcal{U}_i$'s al-
ternately map from a partition over $\Gamma_N$ to a par-
tition over $\Gamma_N^{+1/2}$ and back. Each time, this is
done by fine-graining the partition over (say) $\Gamma_N$
to one over $\Gamma_N^{\text{fine}}$, then coarse-graining the pre-
vious to a partition over $\Gamma_N^{+1/2}$. This becomes
clearer using our graphical notation, in which the
decomposition for an integer $r$ becomes

<figure><img src="image_10.png" /></figure>

with a similar form for $r$ a half-integer.

The proof of Theorem 3.1 is presented in Appendix D. It requires the use of further general notions about partitions on graphs, namely correlation length, connectedness, and strong connectedness. We leave the presentation of these notions to Appendix C, as they are not essential to Theorem 3.1’s statement, but we do encourage the interested reader to get acquainted with them.

It can be seen in (29) that causal decomposi-
tions could be written in a more concise way by
combining one step’s coarse-graining map with
the next step’s fine-graining map. This yields
maps between partitions on $\Gamma_N^{\text{fine}}$ that preserve
the algebras respective to either the $\Gamma_N$ or the $\Gamma_N^{+1/2}$ structure.

**Definition 3.2.** Let $(A_S^{\text{in}})_{S \subseteq \Gamma_N^{\text{fine}}} \vdash \omega^{\text{in}}$ and $(A_S^{\text{out}})_{S \subseteq \Gamma_N^{\text{fine}}} \vdash \omega^{\text{out}}$ be connected partitions. An isomorphism of $C^*$ algebras $\mathcal{U}: \omega^{\text{in}} \to \omega^{\text{out}}$ is $\underline{\Gamma}_N$-local (resp. $\Gamma_N^{+1/2}$-local) from the former to the latter partitions if for every $S \subseteq \Gamma_N$ (resp. $\subseteq \Gamma_N^{+1/2}$),

$$ \mathcal{U}(A_{S \pm \frac{1}{4}}^{\text{in}}) = A_{S \pm \frac{1}{4}}^{\text{out}}. \qquad (30) $$

A $\Gamma_N$-local isomorphism, for instance, can be represented as

<figure><img src="image_11.png" /></figure>

while a $\Gamma_N^{+1/2}$-local one has the same form, shifted to the right by $\frac{1}{2}$.

**Corollary 3.1.** Let $\mathcal{U}$ be a 1D QCA between factorisations over $\Gamma_N$, with causality radius $r$, with $N > 4r$.

There exist a fine-graining $\mathcal{U}^{\text{fine}}$, a coarse-graining $\mathcal{U}^{\text{coarse}}$, and $2r-1$ isomorphisms of $C^*$-algebras $\mathcal{V}_1, \dots, \mathcal{V}_{2r-1}$, alternately $\Gamma_N^{+1/2}$-local and $\Gamma_N$-local, such that

$$ \mathcal{U} = \mathcal{U}^{\text{coarse}}\mathcal{V}_{2r-1} \dots \mathcal{V}_1\mathcal{U}^{\text{fine}}. \qquad (32) $$

This decomposition is simply obtained by taking $\mathcal{V}_i := \mathcal{U}_{i+1}^{\text{fine}} \mathcal{U}_i^{\text{coarse}}$ in Theorem 3.1. It can be represented as

<figure><img src="image_12.png" /></figure>

## 3.2 The translation-invariant case

QCAs are often assumed to satisfy, on top of the causality condition, a condition of translation invariance. As we saw, this additional assumption is not necessary to derive our result. However, in the translation-invariant case, we will now show that the elements in the causal decomposition can themselves be taken to be translation invariant.

Let us start by formally introducing translation-invariance in the context of our presentation of QCAs. Because we view subsystems in a top-down way using partitions, there is no a priori, natural way to see translations on these; one has to specify by hand a partition's preferred translation structure, given by a shift automorphism. For $S$ a subset of $[0, N]$, we define the shifted set $S^{\text{Sh}} := \{n+1 \mod N \mid n \in S\}$.

**Definition 3.3** (Shift automorphism). Let $(A_S)_{S \subseteq \Gamma} \vdash \omega$ be a connected partition, with $\Gamma \in \{\Gamma_N, \Gamma_N^{+1/2}, \Gamma_N^{\text{fine}}\}$. A *shift automorphism for it is a C* automorphism *Sh* of *ω* satisfying

$$ \forall S \subseteq \Gamma, \quad \mathrm{Sh}(A_S) = A_{S^{\mathrm{Sh}}}, \qquad (34a) $$

$$ \mathrm{Sh}^N = \mathcal{I}, \qquad (34b) $$

where $\mathcal{I}$ is the identity automorphism.

Note that no shift automorphism can exist for a partition whose individual algebras are not all isomorphic.
**Definition 3.4** (Translation-Invariance). Let $\mathcal{U}$ be a quantum cellular automaton, or a fine- or coarse-graining, or a $\Gamma_N$- or $\Gamma_N^{+1/2}$- local isomorphism. We suppose its input and output partitions are equipped with preferred shift automorphisms, respectively $\mathrm{Sh}^{\mathrm{in}}$ and $\mathrm{Sh}^{\mathrm{out}}$. $\mathcal{U}$ is translation-invariant if

$$ \mathcal{U} \mathrm{Sh}^{\mathrm{in}} = \mathrm{Sh}^{\mathrm{out}} \mathcal{U}. \qquad (35) $$

**Theorem 3.2.** Let $\mathcal{U}$ be a translation-invariant 1D QCA between factorisations over $\Gamma_N$, with causality radius $r$, with $N > 4r$. Then all of the isomorphisms in the decompositions of Theorem 3.1 and Corollary 3.1 can be chosen to be translation-invariant.

# 4 Causal decompositions in routed unitary circuits

## 4.1 Routed maps and index-matching circuits

Our causal decompositions were so far expressed as compositions of $C^*$ algebra isomorphisms, which can be an unwieldy level of abstraction. Moreover, these isomorphisms in the decomposition were required to satisfy somewhat opaque conditions, such as (30). To make the decomposition wieldier, we need to map it back to the level of unitary maps between Hilbert spaces. This can be achieved, in spite of the non-factorness involved, using the language of index-matching circuits [6, 7]. We provide a basic introduction; detailed discussions, focused on the link with non-factor $C^*$-algebras, are available in Ref. [21].

We call a space $\mathcal{H}_A = \bigoplus_{k \in K} \mathcal{H}_A^k$ equipped with a preferred (finite) sectorisation a *sectorised Hilbert space*.<fnref n="5" /> Sectorised Hilbert spaces can be tensored together in the standard way. A relation $\lambda: K \to L$, where $K$ and $L$ are two sets, is a function from $K$ to the powerset of $L$; it should be thought of as a generalisation of a function to being possibly multi- or empty-valued on each input, and can equivalently be represented as a Boolean matrix $(\lambda_k^l)_{k \in K}^{l \in L}$, where each coefficient

$\lambda_k^l$ is equal to 1 if $\lambda$ relates $k$ to $l$, and 0 otherwise (we denote input indices as subscripts and output ones as superscripts).

A routed map $(\lambda, f)$ from $\bigoplus_{k \in K} \mathcal{H}_A^k$ to $\bigoplus_{l \in L} \mathcal{H}_B^l$ consists of a relation $\lambda: K \to L$ and a linear map $f: \mathcal{H}_A \to \mathcal{H}_B$ such that $f$ follows the route $\lambda$, i.e. such that for every $k$ and $l$, $\lambda_k^l = 0$ implies that the $\mathcal{H}_A^k \to \mathcal{H}_B^l$ block of $f$ is null. This routed map is represented diagrammatically as

$$ \lambda_k^l \boxed{\begin{array}{c} l \\ f \\ \hline k \end{array}} . \qquad (36) $$

In the cases of interest to this paper, all routes can be built out of Kronecker deltas. This allows for an Einstein-notation-like graphical depiction: using repetitions of indices to denote the Kronecker deltas. For instance, a routed map $(\delta, f)$ from $\bigoplus_{k \in K} \mathcal{H}_A^k$ to itself, where $\delta$ is the identity relation on $K$, is represented as

$$ \boxed{\begin{array}{c} k \\ f \\ \hline k \end{array}} . \qquad (37) $$

For a more involved example, taking $\mathcal{H}_X = \bigoplus_{m \in M} \mathcal{H}_X^{km}$, $\mathcal{H}_Y = \bigoplus_{l \in L} \mathcal{H}_Y^{kl}$ and $\mathcal{H}_Z = \bigoplus_{m \in M} \mathcal{H}_Z^{lm}$, and the route $\Delta: K \times M \to K \times L \times L \times M$ defined by $\Delta_{km}^{k'l'm'} = \delta_k^{k'} \delta_{l'm'}^{k'l'm'}$, a routed map $(\Delta, f): \mathcal{H}_X \to \mathcal{H}_Y \otimes \mathcal{H}_Z$ is represented as

$$ \boxed{\begin{array}{c|c} k & l \\ \hline l & m \end{array}} ; \qquad (38) $$

note how we write two indices next to a wire to indicate that the corresponding Hilbert space's sectorisation is indexed by their Cartesian product.

The same notation is used for circuit diagrams representing sequential and parallel compositions of several routed maps; these diagrams are called *index-matching circuits*. Standard quantum circuits are accommodated into index-matching circuits, with un-indexed wires denoting Hilbert spaces with a trivial sectorisation.

A final diagrammatic simplification is that in an index-matching circuit, the repetition of an

<footnote n="5">⁵A subtle difference with Ref. [6] is that we allow for some of these sectors to be null; as we explain in Appendix F.3, this is a structurally innocuous modification that allows for a simplification of the representations of the causal decompositions.</footnote>
index over global input (resp. output) wires that are not otherwise connected through the graph implicitly means that the process is pre- (resp. post-) composed with the orthogonal projector enforcing equality of the values of that index. For instance, the index-matching circuit

<figure><img src="image_13.png" /></figure>

is to be interpreted as the depicted tensor product of maps, pre-composed with the projector onto $\bigoplus_{k,l,m} \mathcal{H}_Y^{kl} \otimes \mathcal{H}_Z^{lm} \subsetneq \mathcal{H}_Y \otimes \mathcal{H}_Z$.

In a routed map $(\lambda, f)$, $\lambda$'s empty-valued inputs (resp. never-reached outputs) indicate sectors on which (resp. to which) $f$ is null. For instance, in the example of (38), the form of $\lambda$ indicates that $f$ never reaches the sectors of the form $\mathcal{H}_Y^{kl} \otimes \mathcal{H}_Z^{lm}$ for $l \neq l'$. We say that $(\lambda, f)$ is a routed unitary if $f$ is unitary once we have excluded such sectors from its input and output spaces; for instance (38) is a routed unitary if it is unitary from $\mathcal{H}_X$ to $\bigoplus_{k,l,m} \mathcal{H}_Y^{kl} \otimes \mathcal{H}_Z^{lm} \subsetneq \mathcal{H}_Y \otimes \mathcal{H}_Z$.

While routed maps can be composed in sequence and in parallel via pairwise composition, and thus form a circuit theory [6], this is not necessarily the case for routed unitaries. In the case of index-matching, however, there is a direct characterisation of the well-behaved circuits in which routed unitarity is preserved. In the graph of an index-matching circuit diagram, we say that a vertex is a starting point for a certain repeated index if that index appears on at least one of its output wires but on none of its input wires; and that it is an endpoint for the index in the symmetric case. We say that the diagram is consistent if every index in it has exactly one starting point and one endpoint.<fnref n="6" />

**Proposition 4.1.** A consistent index-matching circuit in which all of the individual maps are routed unitaries represents a unitary map.

This is formally proven in Ref. [6]'s Theorem 20. For instance, the following index-matching diagram is consistent.

<figure><img src="image_14.png" /></figure>

Finally, consistent index-matching unitary circuits are sound for causal reasoning, in the following sense.

**Proposition 4.2.** If, in a unitary $U$ decomposed as a consistent index-matching unitary circuit, there is no forward path between an input and an output, then $U$ features no causal influence from this input to this output.

This is proven in Ref. [5]'s Appendix A.10.

## 4.2 Representations of 1D partitions

(Although important to understand where the causal decomposition of the next subsection comes from, the results of this subsection are not directly needed for their statement; so the hurried reader can skip it.)

Using routed maps, we can craft specific non-factor partitions by taking algebras of operators over a tensor product of sectorised Hilbert spaces. For the cases of interest to this paper, we do it in the following way. Fixing an $N$, we define for every $n \in \Gamma_N$ a sectorised Hilbert space whose sectors are labelled with two indices: a 'left' index $k_n^L \in K_n^L$ and a 'right' index $k_n^R \in K_n^R$:

$$ \mathcal{H}_{A_n} = \bigoplus_{\substack{k_n^L \in K_n^L \\ k_n^R \in K_n^R}} \mathcal{H}_{A_n}^{k_n^L k_n^R}, \quad (41) $$

with the requirement that $K_n^L = K_{n-1}^R$ for every $n$; in other words, each site's right-index takes value in the same set as the left-index of its neighbour to the right. In fact, we will in practice only use the subspace of $\otimes_n \mathcal{H}_{A_n}$ in which these values are equal:<fnref n="7" />

<footnote n="6">⁶Note that this enforces that the diagram's global input and output wires feature no indices, which is sufficient for our needs here (hence why we simply say it yields a unitary map). However the notion can be extended to the case in which these bear indices [6].</footnote>

<footnote n="7">⁷We denote the common value of $k_n^R$ and $k_{n+1}^L$ as $k_{n+\frac{1}{2}}$.</footnote>
$$
\begin{equation}
\begin{aligned}
\mathcal{H}_{\Omega} &:= \bigoplus_{\substack{k_{\frac{1}{2}} \in K_0^{\mathrm{R}} \\ k_{\frac{3}{2}} \in K_1^{\mathrm{R}} \\ \dots}}} \mathcal{H}_{A_0}^{k_{-\frac{1}{2}} k_{\frac{1}{2}}} \otimes \mathcal{H}_{A_1}^{k_{\frac{1}{2}} k_{\frac{3}{2}}} \otimes \dots \\
&\subseteq \mathcal{H}_{\Omega}^{\mathrm{ext}} := \bigotimes_{n \in \Gamma_N} \mathcal{H}_{A_n}.
\end{aligned}
\tag{42}
\end{equation}
$$

Using routed maps, we can then define the (factor) algebra $\omega$ of operators over $\mathcal{H}_{\Omega}$ as living within $\mathrm{Lin}(\mathcal{H}_{\Omega}^{\mathrm{ext}})$: it is the set of operators of the form

$$
\begin{equation}
\left[
\begin{array}{ccccccc}
k'_{-\frac{3}{2}} & k'_{-\frac{1}{2}} & k'_{-\frac{1}{2}} & k'_{\frac{1}{2}} & k'_{\frac{1}{2}} & k'_{\frac{3}{2}} & k'_{\frac{5}{2}} \\
\hline
\multicolumn{7}{c}{f} \\
\hline
k_{-\frac{3}{2}} & k_{-\frac{1}{2}} & k_{-\frac{1}{2}} & k_{\frac{1}{2}} & k_{\frac{1}{2}} & k_{\frac{3}{2}} & k_{\frac{5}{2}}
\end{array}
\right]
\begin{array}{c}
\text{---} \\
\text{---} \\
\text{---} \\
A_{-1} \quad A_0 \quad A_1 \quad A_2
\end{array}
, \quad (43)
\end{equation}
$$

where the route depicted by the index repetitions precisely enforces that $f$, though formally acting on $\mathcal{H}_{\Omega}^{\mathrm{ext}}$, has null components on and to the orthogonal complement of $\mathcal{H}_{\Omega}$.

We can then define sub-C\* algebras of $\omega$, each corresponding to a collection of sites $S \subseteq \Gamma_N$, in the following way. Given an interval $[[m, n]] \subseteq \Gamma_N$, we define $\mathcal{A}_{[[m,n]]}$ as the set of maps on $\mathcal{H}_{A_m} \otimes \dots \otimes \mathcal{H}_{A_n}$ that

1. have to preserve the value of $A_m$'s left-index and $A_n$'s right-index;

2. for any $l \in [[m, n - 1]]$, can change the common value of $A_l$'s right-hand index and $A_{l+1}$'s left-hand index, as long as this value remains the same for both indices.

Index-matching circuits make this graphically natural: $\mathcal{A}_0, \mathcal{A}_{\{0,1\}}, \mathcal{A}_{\{0,2\}}$, for instance, respectively take the form

$$
\left\{
\begin{array}{c}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\node (A0) at (0,0) {$A_0$};
\node (A1) at (1,0) {$A_1$};
\node (A2) at (2,0) {$A_2$};
\draw[->] (A0) -- (A1);
\draw[->] (A0) -- (A2);
\draw[->] (A1) -- (A2);
\node at (0.5,0.5) {$k_{-\frac{1}{2}}$};
\node at (0.5,-0.5) {$k_{\frac{1}{2}}$};
\node at (1.5,0.5) {$k_{-\frac{3}{2}}$};
\node at (1.5,-0.5) {$k_{\frac{3}{2}}$};
\node at (2.5,0.5) {$k_{\frac{1}{2}}$};
\node at (2.5,-0.5) {$k_{-\frac{3}{2}}$};
\node at (0.5,0.5) {$f$};
\node at (0.5,-0.5) {};
\end{tikzpicture} \\
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\node (A0) at (0,0) {$A_0$};
\node (A1) at (1,0) {$A_1$};
\node (A2) at (2,0) {$A_2$};
\draw[->] (A0) -- (A1);
\draw[->] (A0) -- (A2);
\draw[->] (A1) -- (A2);
\node at (0.5,0.5) {$k_{-\frac{1}{2}}$};
\node at (0.5,-0.5) {$k_{\frac{1}{2}}$};
\node at (1.5,0.5) {$k_{-\frac{3}{2}}$};
\node at (1.5,-0.5) {$k_{\frac{3}{2}}$};
\node at (2.5,0.5) {$k_{\frac{1}{2}}$};
\node at (2.5,-0.5) {$k_{-\frac{3}{2}}$};
\node at (0.5,0.5) {$f$};
\node at (0.5,-0.5) {};
\end{tikzpicture}
\end{array}
\right\} , \quad (44a)
$$

$$
\left\{
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\node (A0) at (0,0) {$A_0$};
\node (A1) at (1,0) {$A_1$};
\node (A2) at (2,0) {$A_2$};
\draw[->] (A0) -- (A1);
\draw[->] (A0) -- (A2);
\draw[->] (A1) -- (A2);
\node at (0.5,0.5) {$k_{-\frac{1}{2}}$};
\node at (0.5,-0.5) {$k_{\frac{1}{2}}$};
\node at (1.5,0.5) {$k_{-\frac{3}{2}}$};
\node at (1.5,-0.5) {$k_{\frac{3}{2}}$};
\node at (2.5,0.5) {$k_{\frac{1}{2}}$};
\node at (2.5,-0.5) {$k_{-\frac{3}{2}}$};
\node at (0.5,0.5) {$f$};
\node at (0.5,-0.5) {};
\end{tikzpicture}
\right\} , \quad (44b)
$$

$$
\left\{
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
\node (A0) at (0,0) {$A_0$};
\node (A1) at (1,0) {$A_1$};
\node (A2) at (2,0) {$A_2$};
\draw[->] (A0) -- (A1);
\draw[->] (A0) -- (A2);
\draw[->] (A1) -- (A2);
\node at (0.5,0.5) {$k_{-\frac{1}{2}}$};
\node at (0.5,-0.5) {$k_{\frac{1}{2}}$};
\node at (1.5,0.5) {$k_{-\frac{3}{2}}$};
\node at (1.5,-0.5) {$k_{\frac{3}{2}}$};
\node at (2.5,0.5) {$k_{\frac{1}{2}}$};
\node at (2.5,-0.5) {$k_{-\frac{3}{2}}$};
\node at (0.5,0.5) {$f$};
\node at (0.5,-0.5) {};
\end{tikzpicture}
\right\} , \quad (44c)
$$

etc. Defining subalgebras of non-connected subsets of $\Gamma_N$ as the algebraic span of their connected components’ subalgebras, we obtain a specification of sub-C\* algebras $(\mathcal{A}_S)_{S \subseteq \Gamma_N} \vdash \omega$. Furthermore, all of the partitions involved in Theorem 3.1 are isomorphic to partitions of this type.

**Theorem 4.1.** *The previous type of specification defines a 1D partition $(\mathcal{A}_S)_{S \subseteq \Gamma_N} \vdash \omega$. Furthermore, all of the partitions involved in Theorem 3.1 are isomorphic to partitions of this type.*

The proof can be found in Appendix F, with the technical version of this theorem being Theorem F.3. Note that this also applies to the partitions over $\Gamma_N^{\mathrm{fine}}$.

## 4.3 The general case

We are now in a position to state a version of Theorem 3.1 that involves (routed) unitary maps between (sectorised) Hilbert spaces.

**Definition 4.1.** Let

$$ U : \bigotimes_n \mathcal{H}_{A_n^{\mathrm{in}}} \to \bigotimes_n \mathcal{H}_{A_n^{\mathrm{out}}}, \quad (45) $$

for some collections of Hilbert spaces $(\mathcal{H}_{A_n^{\mathrm{in}}})_n$ and $(\mathcal{H}_{A_n^{\mathrm{out}}})_n$ (where $n$ takes value in either $\Gamma_N$ or $\Gamma_N^{+1/2}$), be a unitary map.

We say that $U$ is a QCA of radius $r$ if $\hat{U}$ is one with respect to the factorisations (in the sense of Example 1) of $\mathrm{Lin}(\bigotimes_n \mathcal{H}_{A_n^{\mathrm{in}}})$ and $\mathrm{Lin}(\bigotimes_n \mathcal{H}_{A_n^{\mathrm{out}}})$ arising naturally from the tensor product structures of the Hilbert spaces.

**Theorem 4.2.** Let $U$ be a unitary map between tensor products of Hilbert spaces over $\Gamma_N$
or $\Gamma_N^{+1/2}$. We define a causality radius $r$ where $N > 4r$.

U is a 1D QCA of radius $r$ if and only if it admits a decomposition as a routed unitary circuit of the form given in Figure 1.

For the proof, see Theorem F.4 in the Appendices.

## 4.4 The translation-invariant case

In the translation-invariant case, we obtain a translation-invariant circuit decomposition, in which all unitary maps in any given layer are identical.

**Theorem 4.3.** Let $U$ be a unitary map between tensor products of identical Hilbert spaces over $\Gamma_N$ or $\Gamma_N^{+1/2}$. (The input and output Hilbert spaces do not have to be identical.) We define a causality radius $r$ where $N > 4r$.

U is a translation-invariant 1D QCA of radius $r$ if and only if it admits a decomposition as a routed unitary circuit of the form given in Figure 1, in which:

* each of the intermediate Hilbert spaces are translation-invariant, i.e. satisfy $\mathcal{H}_{A_n} = \mathcal{H}_{A_{n+1}}$;

* in each layer of routed unitaries, all of the unitaries are identical.

For the proof, see Theorem F.6. Note that the translation-invariance is with respect to a shift by +1, so that in the intermediate Hilbert spaces, which are indexed by $\Gamma_N^{\text{fine}}$, one has in general $\mathcal{H}_{A_1} = \mathcal{H}_{A_2} = \dots \neq \mathcal{H}_{A_3} = \mathcal{H}_{A_4} = \dots$. The identity of Hilbert spaces here also means that their sectorisations are the same, and indexed by the same set.

# 5 Conclusion and outlook

Our result fully elucidates the structure of QCAs with causality radius $r$, over a finite 1D lattice of size $N > 4r$, in both the general and the translation-invariant cases. In doing so, it demonstrates the close kinship between quantum theory's causal and compositional structures in that case.

A consequential lesson of our result lies in how it reveals the crucial importance of non-factor

sub-C* algebras for studying causality and its consequences — an importance which we suspect extends far beyond that specific subject, and into many structural aspects of quantum theory. As can be seen in the proofs, these algebras' interplay (and in particular the behaviour of their centres, encapsulating their non-factorness) displays a singularly rich structure, whose handling requires particular care but ultimately bears stimulating fruit. In particular, this exemplifies the use of the partition framework laid down in Ref. [21]. At the syntactical level, our results also demonstrate the benefit of routed quantum circuits for situations involving non-factorness: indeed, the causal decompositions we found do not, in general, exist within standard unitary circuits.

Given the deeply compositional flavour of our constructive form, it would be natural to frame it in the framework of matrix-product unitaries [23, 24, 25, 26], a specific case of tensor networks which has been shown to be tightly connected to QCAs, and in particular equivalent to them in the translation-invariant 1D case [23, 24]. Still at the syntactical level, one could also explore how our routed circuit decompositions may be written using the framework of shaded tangles [27], which bears many similarities with the index-matching case of routed circuits.

We expect our algebraic methods to prove useful for future extensions of these results, first and foremost to QCAs over an infinite 1D lattice. In that case, these methods would have to be refined to incorporate some of the important subtleties of infinite dimension — possibly topological considerations — e.g. using the wrapping lemma of Ref. [10]. Another natural extension would be to generalise our results to the case of fermionic 1D QCAs [28, 29, 25, 30], or other types of 1D QCAs constrained by symmetries [31, 32, 33, 34] (where, in both cases, the references we cite already provide partial structural elucidation).

An extension to QCAs beyond 1D would, of course, be another very important direction for future research. In that case, we expect causal decompositions, if they exist, to display a very different structure, in parallel with the profound qualitative differences between QCAs in 1D and in higher dimensions, typically in terms of their index theory [35, 36, 37, 38, 39, 40]. A related problem would be to obtain a constructive form for 'narrow' 1D QCAs, by which we mean those
satisfying $2r + 2 \le N \le 4r$, which our result
does not cover. (For $N \le 2r + 1$, the causality
condition becomes trivial.) Indeed, for instance,
specific QCAs of radius 1 over a 2D lattice can
be built by dividing the lattice into individual
squares of 4 sites, then defining on each of these
a 1D QCA of radius $r = 1$, which is precisely the
simplest example of a narrow 1D QCA, satisfying
$2r + 2 = 4 = N = 4r$.<fnref n="8" /> This seems to indicate
that elucidating the structure of 2D QCAs would
necessitate, in particular, a good understanding
of the structure of narrow 1D QCAs.

Acknowledgements

It is a pleasure to thank Robin Lorenz and Tein
Van der Lugt for helpful discussions and com-
ments.

AV is supported by the STeP2 grant (ANR-22-
EXES-0013) of Agence Nationale de la Recherche
(ANR), the PEPR integrated project EPiQ
(ANR-22-PETQ-0007) as part of Plan France
2030, the ANR grant TaQC (ANR-22-CE47-
0012), and the ID #62312 grant from the John
Templeton Foundation, as part of the ‘The
Quantum Information Structure of Spacetime’
Project (QISS). OM and PA are partially funded
by the European Union through the MSCA
SE project QCOMICAL, by the French Na-
tional Research Agency (ANR): projects TaQC
ANR-22-CE47-0012 and within the framework of
‘Plan France 2030’, under the research projects
EPIQ ANR-22-PETQ-0007, OQULUS ANR-23-
PETQ-0013, HQI-Acquisition ANR-22-PNCQ-
0001 and HQI-R&D ANR-22-PNCQ-0002, and
by the ID #62312 grant from the John Templeton
Foundation, as part of the ‘The Quantum Infor-
mation Structure of Spacetime’ Project (QISS).
The opinions expressed in this publication are
those of the authors and do not necessarily reflect
the views of the John Templeton Foundation.

Data availability statement

This paper has no associated data.

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# A Notations

In this Appendix, we recap the notations used through the paper. All algebras are finite-dimensional, and all sets of indices are finite.

**Algebras** $C^*$-algebras are denoted using curly letters $\omega, A, B$, etc. Given a $C^*$-algebra $\omega$,

• $\mathcal{Z}(\omega)$ is its centre (Definition 2.2), denoted as $\mathcal{Z}_S$ if $\omega$ is denoted as a $A_S$;

• AtomProj($\mathcal{Z}(\omega)$) is the set of atomic projectors of its centre (Theorem 2.1);

• $\omega \cong \omega'$ denotes that there exists an isomorphism of $C^*$-algebras between $\omega$ and $\omega'$.

Given two sub-$C^*$-algebras $A, B$ of $\omega$,

• $A \cap B$ is their intersection;

• $A \vee B$ is their algebraic span;

• $A \bowtie B$ is their linear span;

• $A'$ is the commutant of $A$ within $\omega$ (Definition 2.3);

• $(A, B) \vdash \omega$ denotes that $A$ and $B$ bipartition $\omega$ (Definition 2.4);

• if $A$ and $B$ commute, $A \perp B$ denotes that $A$ and $B$ are uncorrelated (Definition C.1);

• $\pi_A$, where $\pi$ is an orthogonal projector in $\omega$ that commutes with $A$, denotes the $C^*$ algebra $\{\pi f = f\pi | f \in A\}$.

In addition,

• $\text{Lin}(\mathcal{H})$ is the $C^*$ algebra of linear operators over the Hilbert space $\mathcal{H}$;

• for $f$ an element of $\omega$, $\hat{f}$ denotes its action on $\omega$ by conjugation, $h \mapsto f h f^\dagger$;

• for $\pi$ an orthogonal projector in $\omega$, $\pi := 1 - \pi$;

• Unit($A$) is the set of unitary elements of the $C^*$ algebra $A$.
**Sets** For two subsets $S, T \subseteq X$,

• $S \cap T$ is their intersection;

• $S \cup T$ is their union, also denoted as $S \sqcup T$ (disjoint union) when $S$ and $T$ are disjoint;

• $\bar{S} := X \setminus S$ is $S$'s complement in $X$.

**Graphs** The graphs explicitly used in this paper are all 1D, and defined in a cyclic way: the last vertex in the specification is connected to the first, and their labels are modulo $N$ ($-1$ is the same vertex as $N-1$, etc). They are<fnref n="9" />

• $\Gamma_N = \{0, 1, \dots, N-1\}$;

• $\Gamma_N^{+1/2} = \{\frac{1}{2}, \frac{3}{2}, \dots, N - \frac{1}{2}\}$;

• $\Gamma_N^{\text{fine}} = \{\frac{1}{4}, \frac{3}{4}, \dots, N - \frac{1}{4}\}$;

• $\Gamma^{\text{split}} = \{\frac{1}{4}, 1, 2, \dots, N-2, N-1, N-\frac{1}{4}\}$.

In a given graph,

• $[m, n]$ is the set of vertices of the graph located between $m$ and $n$ (including $m$ and $n$ themselves);<fnref n="10" />

• $d(m, n) := \min(|n - m|, N - |n - m|)$ is the (cyclic) distance between $m$ and $n$;<fnref n="11" />

• for $S$ a subset of vertices and $l \ge 0$, $S \pm l$ is the set of vertices at a distance lesser or equal than $l$ to $S$.<fnref n="12" />

# B General results on C\* algebras

## B.1 Homomorphisms of C\* algebras and their kernels

Here, we show that the kernel of a homomorphism of C\* algebras $h: \mathcal{F} \to \mathcal{G}$ is made of entire blocks $\pi^k \mathcal{F}$ of $\mathcal{F}$.

**Definition B.1.** Let $\mathcal{F}$ and $\mathcal{G}$ be (finite-dimensional) C\*-algebras. A homomorphism of C\* algebras $h: \mathcal{F} \to \mathcal{G}$ is a homomorphism of algebras (i.e. a linear map preserving the product) that, in addition, preserves the adjoint, i.e. satisfies $\forall f \in \mathcal{F}, h(f^\dagger) = h(f)^\dagger$.

Its kernel $\ker h$ is the linear subspace of $\mathcal{F}$ defined by the condition $\forall f \in \ker h, h(f) = 0$.

**Proposition B.1.** Let $h: \mathcal{F} \to \mathcal{G}$ be a homomorphism of C\* algebras. We denote $\text{AtomProj}(\mathcal{Z}(\mathcal{F})) = \{\pi^k\}_{k \in K}$. Then, taking $K_h := \{k \in K | h(\pi^k) \neq 0\}$ and denoting the orthogonal projectors $\mu := \sum_{k \in K_h} \pi^k$ and $\bar{\mu} := 1 - \mu = \sum_{k \in K \setminus K_h} \pi^k$,

$$ \mathcal{F} = \mu\mathcal{F} \uplus \bar{\mu}\mathcal{F}; \quad (46a) $$

$$ \ker h = \bar{\mu}\mathcal{F}; \quad (46b) $$

$$ h|_{\mu\mathcal{F}} \text{ is injective.} \quad (46c) $$

*Proof.* (46a) comes from the fact that $\mathcal{F} = \biguplus_{k \in K} \pi^k \mathcal{F} = (\biguplus_{k \in K_h} \pi^k \mathcal{F}) \uplus (\biguplus_{k \in K \setminus K_h} \pi^k \mathcal{F})$; it is easy to compute that the first term is $\mu\mathcal{F}$ and the second is $\bar{\mu}\mathcal{F}$.

For the reverse inclusion in (46b) consider $f \in \bar{\mu}\mathcal{F}$, $h(f) = h(\bar{\mu}f) = h(\bar{\mu})h(f) = 0h(f) = 0$, so $\bar{\mu}\mathcal{F} \subseteq \ker h$. Let us suppose that inclusion is strict and reach a contradiction.

Then, $f \in \ker h \setminus \bar{\mu}\mathcal{F}$ can be decomposed as $\mu f + \bar{\mu} f$ with $\mu f \neq 0$; as $0 = h(f) = h(\mu f)$, this leads to $\ker h \cap \mu\mathcal{F} \neq \{0\}$.

Let us thus take such a non-null $f \in \mu\mathcal{F} \cap \ker h$. First, by Theorem 2.2, we have $\mu\mathcal{F} \cong \bigoplus_k \text{Lin}(\mathbb{C}^{d_k})$, in which $ff^\dagger$ is a non-null positive element, which can therefore spectrally decomposed as $\sum_s p_s \nu^s$, where the $\nu^s$'s are pairwise orthogonal projectors of rank 1, and the $p_s$'s are strictly positive reals. This leads to

$$ \begin{aligned} 0 &= h(f)h(f^\dagger) = h(ff^\dagger) \\ &= h\left(\sum_s p_s \nu^s\right) = \sum_s p_s h(\nu^s). \end{aligned} \quad (47) $$

The $h(\nu^s)$'s are themselves pairwise orthogonal projectors, as they satisfy $h(\nu^s)h(\nu^s) = h(\nu^s\nu^s) = h(\nu^s)$, $h(\nu^s)^\dagger = h((\nu^s)^\dagger) = h(\nu^s)$, and, for $s \neq s'$, $h(\nu^s)h(\nu^{s'}) = h(\nu^s\nu^{s'}) = h(0) = 0$. Thus, (47) leads to $h(\nu^s) = 0$ for all $s$. There is therefore an orthogonal projector of rank 1 $\nu \in \mu\mathcal{F}$ satisfying $h(\nu) = 0$. As $\nu = \mu\nu = \sum_{k \in K_h} \pi^k \nu$

<footnote n="9">⁹See also Figures 2 and 3.</footnote>

<footnote n="10">¹⁰By cyclicity of the graph, an interval might go ‘over the edge’: for instance, in $\Gamma_N$, $[N-1, 1] = \{N-1, 0, 1\} = \{-1, 0, 1\} = [-1, 1]$.</footnote>

<footnote n="11">¹¹Note that distance is defined in terms of the graph's labels, not in terms of adjacency in the graph. For instance, in $\Gamma_N^{\text{fine}}$, $d(\frac{1}{4}, \frac{3}{4}) = \frac{1}{2}$, even though $\frac{1}{4}$ and $\frac{3}{4}$ are 'one node away' from each other.</footnote>

<footnote n="12">¹²With a slight abuse of notation, we sometimes use this notation across different graphs: for instance, taking $S \subseteq \Gamma_N$, we denote as $S \pm \frac{1}{4}$ the set of vertices of $\Gamma_N^{\text{fine}}$ that are at a distance lesser or equal than $\frac{1}{4}$ to $S$.</footnote>
and $\nu$ has rank 1, there exists $k \in K_h$ such that $\pi^k\nu = \nu$, i.e. $\nu \in \pi^k\mathcal{F}$.

Again since $\pi^k\mathcal{F} \cong \text{Lin}(\mathbb{C}^{d_k})$ is a factor algebra by Th.2.2, we can assume without loss of generality that in this representation, $\nu$ is of the form $|0\rangle\langle 0|$ for some unit vector $|0\rangle$. Completing the orthonormal basis as $\{|n\rangle\}_n$, $\pi^k\mathcal{F}$ is then represented by the span of $\{|m\rangle\langle n|\}_{m,n}$. Yet we have, for arbitrary $m$ and $n$,

$$ 
\begin{aligned}
h(|m\rangle\langle n|) &= h(|m\rangle\langle 0| \ |0\rangle\langle 0| \ |0\rangle\langle n|) \\
&= h(|m\rangle\langle 0|) h(|0\rangle\langle 0|) h(|0\rangle\langle n|) \quad (48) \\
&= h(|m\rangle\langle 0|) 0 h(|0\rangle\langle n|) \\
&= 0.
\end{aligned}
$$

Therefore $\pi^k\mathcal{F} \subseteq \text{ker } h$, so in particular $\pi^k \in \text{ker } h$, which contradicts $k \in K_h$. Thus our assumption $\bar{\mu}\mathcal{F} \subsetneq \text{ker } h$ led to a contradiction, so $\bar{\mu}\mathcal{F} = \text{ker } h$. In addition, we then have $\text{ker}(h|_{\mu\mathcal{F}}) = \mu\mathcal{F} \cap \text{ker } h = \mu\mathcal{F} \cap \bar{\mu}\mathcal{F} = \{0\}$, so $h|_{\mu\mathcal{F}}$ is injective. $\square$

In particular, this allows us to say that if an orthogonal projector commutes with $\mathcal{F}$, then its action is null on some of $\mathcal{F}$'s blocks and injective on the others.

**Corollary B.1.** Let $\omega$ and $\mathcal{F} \subseteq \omega$. Let $\pi$ be an orthogonal projector of $\omega$ that commutes with the elements of $\mathcal{F}$. We denote $\text{AtomProj}(\mathcal{Z}(\mathcal{F})) = \{\nu^k\}_{k \in K}$.

Then, defining $K_\pi = \{k|\pi\nu^k \neq 0\}$, $\mu := \sum_{k \in K_\pi} \nu^k$ and $\bar{\mu} := 1 - \mu = \sum_{k \in K \setminus K_\pi} \nu^k$, we have

$$ \mathcal{F} = \mu\mathcal{F} \oplus \bar{\mu}\mathcal{F}; \quad (49a) $$

$$ \pi = \pi\mu; \quad (49b) $$

$$ \hat{\pi} : f \mapsto \pi f = f\pi \text{ is injective on } \mu\mathcal{F}. \quad (49c) $$

In particular, $\mu$ is the minimal projector in $\mathcal{Z}(\mathcal{F})$ such that $\pi\mu = \pi$.

**Proof.** Using the idempotency and self-adjointness of $\pi$ and the fact that it commutes with elements of $\mathcal{F}$, it is direct to check that $\hat{\pi} : f \mapsto \pi f = f\pi$ is a homomorphism of $\mathcal{C}^*$ algebras from $\mathcal{F}$ to $\omega$. Applying Proposition B.1 then yields the desired $\mu$ and $\bar{\mu}$. (49b) comes from the fact that $\pi = \pi(\mu + \bar{\mu}) = \pi\mu + \pi\bar{\mu} = \pi\mu$.

For the minimalness of $\mu$, let $\mu'$ be another projector in $\mathcal{Z}(\mathcal{F})$ satisfying $\pi\mu' = \pi$. It is also of

the form $\mu' = \sum_{k \in K' \subseteq K} \nu^k$. Suppose that there exists $l \in K_\pi \setminus K'$ then $0 \neq \pi\nu^l = \pi\mu'\nu^l = \pi \sum_{k \in K'} \nu^k\nu^l = 0$ because $l \notin K'$. This implies that $K_\pi \subseteq K'$ and thus that $\mu$ is minimal. $\square$

A useful consequence is that taking centres and projecting by a certain $\pi$ are commuting operations, in the following sense.

**Lemma B.1.** Let $\omega$ and $\mathcal{F} \subseteq \omega$ be $\mathcal{C}^*$ algebras, and let $\pi \in \omega$ be an orthogonal projector commuting with $\mathcal{F}$. Then

$$ \mathcal{Z}(\pi\mathcal{F}) = \pi\mathcal{Z}(\mathcal{F}). \quad (50) $$

*Proof.* We use the notations of Corollary B.1. By (49b), $\mathcal{Z}(\pi\mathcal{F}) = \mathcal{Z}(\pi\mu\mathcal{F})$. Since $\pi$ is injective on $\mu\mathcal{F}$ by (49c), it preserves centres, so $\mathcal{Z}(\pi\mu\mathcal{F}) = \pi\mathcal{Z}(\mu\mathcal{F})$. Furthermore, since $\mu\mathcal{F} = \biguplus_{k \in K_\pi} \nu^k\mathcal{F}$, where the $\nu^k\mathcal{F}$'s are factors, we have $\mathcal{Z}(\mu\mathcal{F}) = \text{LinSpan}(\{\nu^k\}_{k \in K_\pi})$, so $\pi\mathcal{Z}(\mu\mathcal{F}) = \text{LinSpan}(\{\pi\nu^k\}_{k \in K_\pi}) = \text{LinSpan}(\{\pi\nu^k\}_{k \in K}) = \pi\text{LinSpan}(\{\nu^k\}_{k \in K}) = \pi\mathcal{Z}(\mathcal{F})$; we were able to add the $\pi\nu^k$'s for $k \notin K_\pi$ in the linear span as they are null. $\square$

**Lemma B.2.** Let $\mathcal{Z}_1, \mathcal{Z}_2$ and $\mathcal{Z}_3$ be three commutative $\mathcal{C}^*$ algebras such that $\mathcal{Z}_3 = \mathcal{Z}_1 \vee \mathcal{Z}_2$. Let $\text{AtomProj}(\mathcal{Z}_1) = \{\pi_1^k\}_{k \in K_1}$ and $\text{AtomProj}(\mathcal{Z}_2) = \{\pi_2^k\}_{k \in K_2}$, then $\text{AtomProj}(\mathcal{Z}_3) = \{\pi_1^k\pi_2^l \neq 0 | k \in K_1, l \in K_2\}$.

*Proof.* Let us prove that the $\pi_1^k\pi_2^l$'s satisfy the properties of a family of atomic projectors in $\mathcal{Z}_3$; by unicity of this family for a given $\mathcal{C}^*$ algebra, $\{\pi_1^k\pi_2^l \neq 0 | k \in K_1, l \in K_2\}$ will be the atomic projectors of $\mathcal{Z}_3$. We first remark that all the $\pi_1^k$ and $\pi_2^l$ are elements of $\mathcal{Z}_3$ and thus commute with each other. It follows that the $\pi_1^k\pi_2^l$ are projectors, as $(\pi_1^k\pi_2^l)^2 = \pi_1^k\pi_1^k\pi_2^l\pi_2^l = \pi_1^k\pi_2^l$. The family $\{\pi_1^k\pi_2^l \neq 0 | k \in K_1, l \in K_2\}$ is finite because $K_1$ and $K_2$ are finite, and its elements are non zero by definition. Now let $\pi_3^i = \pi_1^k\pi_2^l$ and $\pi_3^j = \pi_1^m\pi_2^n$ be elements of $\{\pi_1^k\pi_2^l \neq 0 | k \in K_1, l \in K_2\}$, because the $\pi_1^k$ and $\pi_2^l$ are atomic projectors we have that $(\pi_3^i)^\dagger = (\pi_2^l)^\dagger(\pi_1^k)^\dagger = \pi_2^l\pi_1^k = \pi_3^i$, and that $\pi_3^i\pi_3^j = \pi_1^k\pi_1^m\pi_2^l\pi_2^n = \delta^{km}\delta^{ln}\pi_1^k\pi_2^l = \delta^{ij}\pi_3^i\pi_3^j$. Finally, $\sum_i \pi_3^i = \sum_{k,l} \pi_1^k\pi_2^l = \sum_k \pi_1^k\sum_l \pi_2^l = 1$. $\square$

## B.2 Sub-$\mathcal{C}^*$ algebras of a centre are coarse-grainings

The following lemma shows that any sub-$\mathcal{C}^*$ algebra of a centre can essentially be seen as a coarse-graining, in the sense that its atomic projectors
are lumping together atomic projectors of that centre.

**Lemma B.3.** Let $\mathcal{F}$ be a $C^*$ algebra, and $\mathcal{Z}_1 \subseteq \mathcal{Z}(\mathcal{F})$ be a (necessarily commutative) sub-$C^*$ algebra of its centre. Then $\mathcal{Z}_1$'s atomic projectors are coarse-grainings of those of $\mathcal{Z}(\mathcal{F})$ in the following sense: denoting AtomProj($\mathcal{Z}_1$) = {$\mu^l$}$_{l \in K}$ and AtomProj($\mathcal{Z}(\mathcal{F})$) = {$\pi^k$}$_{k \in K}$, there exists a partition (in the sense of sets) of $K$ into a disjoint union of subsets, $K = \bigsqcup_{l \in L} K_l$, such that

$$ \forall l, \quad \mu^l = \sum_{k \in K_l} \pi^k. \qquad (51) $$

Consequently,

$$ \mathcal{F} = \bigsqcup_{l \in L} \mu^l \mathcal{F}, \qquad (52) $$

and, for every $\pi^k \in \text{AtomProj}(\mathcal{Z}(\mathcal{F}))$, there exists a $l$ (given by the condition $k \in K_l$) such that $\pi^k \in \mu^l \mathcal{F}$, or in other words, $\pi^k = \mu^l \pi^k$.

*Proof.* By Theorem 2.1, any $\mu^l$ is of the form $\mu^l = \sum_k \alpha_{kl} \pi^k$. Idempotency leads to $\alpha_{kl} \in \{0, 1\}$, so one can rewrite $\mu^l = \sum_{k \in K_l} \pi^k$. The fact that the $\mu^l$'s are pairwise orthogonal and sum to the identity yields the fact that the $K_l$'s form a partition of $K$.

$\pi^k = \mu^l \pi^k$ for $k \in K_l$ then follows directly, and it is then easy to prove, using (7), that $\mu^l \mathcal{F} = \bigsqcup_{k \in K_l} \pi^k \mathcal{F}$, so (7) can be rewritten as (52) by bunching terms. $\square$

## B.3 Double commutants

**Proposition B.2.** For $\mathcal{F} \subseteq \omega C^*$ algebras, with $\omega$ factor,

$$ \mathcal{F}'' = \mathcal{F}. \qquad (53) $$

*Proof.* Since it is a factor, $\omega$ can without loss of generality be taken to be a certain $\text{Lin}(\mathcal{H}_\Omega)$. By Von Neumann's bicommutant theorem [41], for $\mathcal{F}$ a sub-$C^*$ algebra of an algebra of bounded operators on a Hilbert space, $\mathcal{F}$'s bicommutant within the latter is equal to $\mathcal{F}$'s closure under the strong operator topology. In finite dimension, all subspaces are closed under any topology, so $\mathcal{F}'' = \mathcal{F}$. $\square$

If $\omega$ is not a factor, the situation is slightly more involved.

**Proposition B.3.** For $\mathcal{F} \subseteq \omega C^*$ algebras,

$$ \mathcal{F}'' = \mathcal{Z}(\omega) \vee \mathcal{F}. \qquad (54) $$

*Proof.* Since $\mathcal{Z}(\omega)$ commutes with every operator in $\omega$, it is in $\mathcal{F}'$, and more specifically in $\mathcal{Z}(\mathcal{F}'')$. Thus, by Lemma B.3, $\mathcal{F}'' = \bigsqcup_{l \in L} \mu^l \mathcal{F}''$, where the $\mu^l$'s are $\mathcal{Z}(\omega)$'s atomic projectors. We will show that commutation can be studied at the level of each $\mu^l$ subspace, so as to then be able to apply Proposition B.3.

Indeed we have that for any $\mathcal{G} \subseteq \omega$, $\mathcal{G}' \cap \text{Im}(\hat{\mu}^l) = (\mu^l \mathcal{G})' \cap \text{Im}(\hat{\mu}^l)$. The direct inclusion comes from the fact that $\mathcal{G}' \subseteq (\mu^l \mathcal{G})'$ since $\mu^l$ commutes with every element of $\omega$, and then intersecting with $\text{Im}(\hat{\mu}^l)$. The reverse inclusion that for an element $f$ of the RHS, we have for any $g \in \mathcal{G}$, $[f, g] = [\mu^l f, g] = [f, \mu^l g] = 0$, where we again used the fact that $\mu^l$ commutes with every element of $\omega$.

This allows us to rewrite $\mu^l \mathcal{F}'' = \mu^l \mathcal{F}'' \cap \text{Im}(\hat{\mu}^l) = (\mu^l \mathcal{F}')' \cap \text{Im}(\hat{\mu}^l) = (\mu^l \mathcal{F}' \cap \text{Im}(\hat{\mu}^l))' \cap \text{Im}(\hat{\mu}^l) = ((\mu^l \mathcal{F})' \cap \text{Im}(\hat{\mu}^l))' \cap \text{Im}(\hat{\mu}^l)$. The latter is the double commutant of $\mu^l \mathcal{F}$ within the factor $C^*$ algebra $\text{Im}(\hat{\mu}^l)$; by Proposition B.2, it is equal to $\mu^l \mathcal{F}$. We thus find $\mu^l \mathcal{F}'' = \mu^l \mathcal{F}$, which allows us to conclude that $\mathcal{F}'' = \bigsqcup_l \mu^l \mathcal{F} \subseteq \mathcal{Z}(\omega) \vee \mathcal{F}$. The reverse inclusion is direct because $\mathcal{F} \subseteq \mathcal{F}''$ and $\mathcal{Z}(\omega) \subseteq \mathcal{F}''$. $\square$

## B.4 Some results on partitions

**Proposition B.4.** If $(A_1, A_2) \vdash \omega$, then

$$ A_1' = \mathcal{Z}(\omega) \vee A_2. \qquad (55) $$

In particular, this implies that $A_1$ and $A_2$ commute, and that in the case of $\omega$ factor,

$$ A_1' = A_2. \qquad (56) $$

*Proof.* Since the elements of $\mathcal{Z}(\omega)$ commute with all elements of $\omega$, they commute in particular with all elements of $A_1$, so $\mathcal{Z}(\omega) \subseteq A_1'$. Furthermore, for the same reason, elements of $\mathcal{Z}(\omega)$ commute with all elements of $A_1'$, so $\mathcal{Z}(\omega) \subseteq \mathcal{Z}(A_1')$. Therefore, by Lemma B.3, denoting $\mathcal{Z}(\omega)$'s atomic projectors as $\pi^k$'s, we have $A_1' = \bigsqcup_k \pi^k A_1'$. Yet using Lemma D.1 then the assumption $(A_1, A_2) \vdash \omega$, we have for every $k$: $\pi^k A_1' = A_1' \cap \text{Im}(\hat{\pi}^k) = A_1' \cap \pi^k \omega = \pi^k A_2$, so $A_1' = \bigsqcup_k \pi^k A_2$. Lemma B.3 shows that $\mathcal{Z}(\omega) \vee A_2$ is equal to the latter's RHS, so $A_1' = \mathcal{Z}(\omega) \vee A_2$. $\square$
**Proposition B.5.** If $(\mathcal{A}_1, \mathcal{A}_2) \vdash \omega$, then

$$ \mathcal{Z}_1 \subseteq \mathcal{Z}(\omega) \lor \mathcal{Z}_2. \qquad (57) $$

In particular, when $\omega$ is factor, $\mathcal{Z}_1 = \mathcal{Z}_2$.

*Proof.* We fix $\mu \in \text{AtomProj}(\mathcal{Z}_1)$. Let us prove that, for any $\pi \in \text{AtomProj}(\mathcal{Z}(\omega))$, $\pi\mu \in \mathcal{Z}(\omega) \lor \mathcal{Z}_2$; indeed this implies that $\mu = \sum_{\pi} \pi\mu \in \mathcal{Z}(\omega) \lor \mathcal{Z}_2$, and thus (57). By Lemma B.1, $\pi\mu$ is in $\mathcal{Z}(\pi\mathcal{A}_1)$. Yet $(\pi\mathcal{A}_1, \pi\mathcal{A}_2)$ is a partition of the factor $\pi\omega$, so by Proposition B.4 $\mathcal{Z}(\pi\mathcal{A}_1) = \mathcal{Z}(\pi\mathcal{A}_2) \subseteq \mathcal{Z}(\omega) \lor \mathcal{Z}_2$, so $\pi\mu \in \mathcal{Z}(\omega) \lor \mathcal{Z}_2$. $\square$

**Proposition B.6.** If $(\mathcal{A}_1, \mathcal{A}_2) \vdash \omega$, then:

$$ \omega = \mathcal{A}_1 \lor \mathcal{A}_2 \iff \mathcal{Z}(\omega) = \mathcal{Z}_1 \lor \mathcal{Z}_2. \qquad (58) $$

*Proof.* Supposing the LHS, (10) yields $\mathcal{Z}(\omega) = \mathcal{Z}(\mathcal{A}_1 \lor \mathcal{A}_2) \subseteq \mathcal{Z}_1 \lor \mathcal{Z}_2$. The reverse inclusion $\mathcal{Z}_1 \lor \mathcal{Z}_2 \subseteq \mathcal{Z}(\mathcal{A}_1 \lor \mathcal{A}_2)$ comes from the fact that the elements of $\mathcal{Z}_1 \lor \mathcal{Z}_2$ of the form $\pi_1\pi_2$ commute with the elements of $\mathcal{A}_1 \lor \mathcal{A}_2$ of the form $a_1a_2$, either because $\pi_k$ is in the centre of $\mathcal{A}_k$ or because $\mathcal{A}_1$ and $\mathcal{A}_2$ commute by (55); so the LHS implies the RHS.

Conversely, the RHS implies that $\mathcal{Z}(\omega)$'s atomic projectors are all of the form $\pi = \pi_1\pi_2$ with $\pi_n \in \text{AtomProj}(\mathcal{Z}_n)$. Fixing such a $\pi$, $\pi\mathcal{A}_1$ and $\pi\omega$ are then isomorphic to $\text{Lin}(\mathbb{C}^p) \otimes \mathbb{1}_q$ and $\text{Lin}_{pq}(\mathbb{C})$ respectively. Since $\pi\mathcal{A}_2$ is the commutant of $\pi\mathcal{A}_1$ in $\pi\omega$ by (9), it is isomorphic to $\mathbb{1}_p \otimes M_q(\mathbb{C})$. This all implies $\pi\omega = \pi\mathcal{A}_1 \lor \pi\mathcal{A}_2$. Moreover $(\pi\mathcal{A}_1) \lor (\pi\mathcal{A}_2) = \pi(\mathcal{A}_1 \lor \mathcal{A}_2)$ because $\pi_1\pi_2a_1\pi_1\pi_2a_2 = \pi_1\pi_2\pi_1\pi_2a_1a_2$ as $\pi_1 \in \mathcal{Z}_1$ and $\mathcal{A}_1$ and $\mathcal{A}_2$ commute by Prop. 55. Hence $\omega = \bigvee_{\pi \in \text{AtomProj}(\mathcal{Z}(\omega))} \pi\omega = \bigvee_{\pi} \pi(\mathcal{A}_1 \lor \mathcal{A}_2) = \mathcal{A}_1 \lor \mathcal{A}_2$, where the first and last equalities come from the fact any sub-$C^*$ algebra of $\omega$ decompose as such by Theorem 2.2. $\square$

**Proposition B.7.** Let $\mathcal{A}_1, \mathcal{A}_2 \subseteq \mathcal{F}$ be $C^*$ algebras. We suppose that there exist commutative $C^*$ algebras $\tilde{\mathcal{Z}}_1 \subseteq \mathcal{Z}(\mathcal{A}_1)$, $\tilde{\mathcal{Z}}_2 \subseteq \mathcal{Z}(\mathcal{A}_2)$ such that $\mathcal{Z}(\mathcal{F}) = \tilde{\mathcal{Z}}_1 \lor \tilde{\mathcal{Z}}_2$. Then

$$ \begin{array}{l} (\mathcal{A}_1, \mathcal{A}_2) \vdash \mathcal{F} \\ \iff \forall \pi_1 \in \text{AtomProj}(\tilde{\mathcal{Z}}_1), \qquad (59) \\ \qquad \pi_1 \mathcal{A}'_1 = \pi_1 \mathcal{A}_2. \end{array} $$

*Proof.* Since $\mathcal{Z}(\mathcal{F}) = \tilde{\mathcal{Z}}_1 \lor \tilde{\mathcal{Z}}_2$, by Lemma B.2, any atomic projector $\pi$ of $\mathcal{Z}(\mathcal{F})$ is of the form

$\pi_1\pi_2$, where $\pi_1$ and $\pi_2$ are atomic projectors of $\tilde{\mathcal{Z}}_1$ and $\tilde{\mathcal{Z}}_2$ respectively.

Suppose $(\mathcal{A}_1, \mathcal{A}_2) \vdash \mathcal{F}$; by Proposition B.4 then Lemma B.3, $\mathcal{A}'_1 = \mathcal{Z}(\mathcal{F}) \lor \mathcal{A}_2 = \bigvee_{\pi \in \text{AtomProj}(\mathcal{Z}(\mathcal{F}))} \pi\mathcal{A}_2 = \bigvee_{\pi_i \in \text{AtomProj}(\tilde{\mathcal{Z}}_i)} \pi_1\pi_2\mathcal{A}_2$. Therefore, for a given $\pi_1 \in \text{AtomProj}(\tilde{\mathcal{Z}}_1)$, $\pi_1\mathcal{A}'_1 = \pi_1 \bigvee_{\pi_2 \in \text{AtomProj}(\tilde{\mathcal{Z}}_2)} \pi_2\mathcal{A}_2$. But, by Lemma B.3, $\mathcal{A}_2 = \bigvee_{\pi_2 \in \text{AtomProj}(\tilde{\mathcal{Z}}_2)} \pi_2\mathcal{A}_2$, hence $\pi_1\mathcal{A}'_1 = \pi_1\mathcal{A}_2$.

Conversely, suppose the RHS of (59). Then $\pi_1\mathcal{A}'_1 = \pi_2\pi_1\mathcal{A}'_1 = \pi_2\pi_1\mathcal{A}_2 = \pi\mathcal{A}_2$. Furthermore $\mathcal{Z}(\mathcal{F}) = \tilde{\mathcal{Z}}_1 \lor \tilde{\mathcal{Z}}_2 \subseteq \mathcal{Z}(\mathcal{A}_1) \lor \mathcal{Z}(\mathcal{A}_2)$, so $(\mathcal{A}_1, \mathcal{A}_2) \vdash \mathcal{F}$. $\square$

## B.5 Proof of Proposition 2.1

*Proof.* We have to prove that for any $S \subseteq \Gamma_{\text{out}}^N$, $\mathcal{U}^{-1}(\mathcal{B}_S) \subseteq \mathcal{A}_{S\pm r}$, or equivalently, $\mathcal{B}_S \subseteq \mathcal{U}(\mathcal{A}_{S\pm r})$. We pick such an $S$; then we have $S \subseteq \overline{S \pm r} \pm r$, as picking $n \in S \cap \overline{S \pm r} \pm r$ would yield an $m \in \overline{S \pm r}$ such that $d(m, S) \le d(m, n) \le r$, which is a contradiction.

Thus we have the inclusion

$$ \mathcal{B}_S \subseteq \mathcal{B}_{\overline{S \pm r \pm r}} = \mathcal{B}'_{\overline{S \pm r \pm r}} \qquad (60) $$

where we have used the fact that, as $\omega$ is a factor, $\mathcal{B}_{\overline{T}} = \mathcal{B}'_T$ by Proposition B.4 (taking here $T = \overline{S \pm r} \pm r$). Moreover, by (19), we have that for any $T$, $\mathcal{U}(\mathcal{A}_T) \subseteq \mathcal{B}_{T\pm r}$ and hence $\mathcal{B}'_{T\pm r} \subseteq \mathcal{U}(\mathcal{A}_T)' = \mathcal{U}(\mathcal{A}'_T)$, where the equality stems from the fact that $\mathcal{U}$ is an isomorphism of $C^*$ algebras. Therefore, taking $T = \overline{S \pm r}$,

$$ \mathcal{B}'_{\overline{S \pm r \pm r}} \subseteq \mathcal{U}(\mathcal{A}'_{\overline{S \pm r}}) = \mathcal{U}(\mathcal{A}_{S \pm r}). \qquad (61) $$

Combining both inclusions gives the result.<fnref n="13" /> $\square$

## C Further notions on partitions

Proving this paper's main results requires the introduction of further notions, pinning down how the structure of a partition interacts with the structure of the graph that indexes it. The

<footnote n="13">¹³Note that this proof makes no use of the fact that the partitions are 1D; thus Proposition 2.1 applies to QCAs over any graph structure, as long as the overall $C^*$ algebra is a finite-dimensional factor — and, for $r$ not an integer, as long as it has been made clear what is meant by $\mathcal{U}$ being of radius $r$.</footnote>
first notion, correlation length, yields information about the partition's interaction with the graph's metric structure (i.e. distances on it), while the second, connectedness, ensures a kind of consistency with its topological structure (i.e. adjacency relationships). The third, strong connectedness, is a strengthening of connectedness in the case of a 1D graph.

Beyond their specific use for the technical proofs of this paper, we expect these natural notions to be relevant for other future applications of partitions.

### C.1 Correlations between subalgebras and correlation length

An important aspect of partitions, compared to the standard case of factorisations, is that there might be correlations between parts — i.e., they might contain redundant information. Mathematically, correlations between two commuting sub-C* algebras $\mathcal{F}$ and $\mathcal{G}$ are witnessed by the fact that a projector $\pi \in \mathcal{F}$ and a projector $\mu \in \mathcal{G}$ satisfy $\pi\mu = \pi$ — i.e., the information that we are in the sector designated by $\mu$ is redundant with the information that we are in the sector designated by $\pi$. This is equivalent to the statement that $\pi\bar{\mu} = 0$, where $\bar{\mu} := 1 - \mu \in \mathcal{G}$. We thus see that the presence of correlations is equivalently witnessed by the existence of null products between the projectors of $\mathcal{F}$ and those of $\mathcal{G}$.

Therefore, a convenient structural way to pin down the absence of correlations between two algebras is to link it to the injectiveness of the product $\cdot_{\mathcal{F},\mathcal{G}}$ between elements of one and elements of the other.

**Definition C.1** (Uncorrelated subalgebras). Let $\mathcal{F}, \mathcal{G}$ be commuting sub-C* algebras of the C*-algebra $\omega$. We say that they are uncorrelated (denoted as $\mathcal{F} \perp \mathcal{G}$) if the homomorphism of C*-algebras

$$ \begin{gathered} \cdot_{\mathcal{F},\mathcal{G}} : \mathcal{F} \otimes \mathcal{G} \to \omega \\ f \otimes g \mapsto fg = gf \end{gathered} \qquad (62) $$

is injective. Note that it then is a bijection to $\mathcal{F} \vee \mathcal{G}$.

In fact, as proven by the following proposition, correlations between two algebras come down to correlations between their centres. The latter

correspond to sectorial correlations: being in a certain sector (i.e. a certain block) of $\mathcal{F}$ is correlated with being in a certain sector of $\mathcal{G}$.

**Proposition C.1.** Suppose $\mathcal{F}$ and $\mathcal{G}$ commute; then

$$ \mathcal{F} \perp \mathcal{G} \iff \mathcal{Z}(\mathcal{F}) \perp \mathcal{Z}(\mathcal{G}). \qquad (63) $$

*Proof.* First, let us prove that $\mathcal{Z}(\mathcal{F} \otimes \mathcal{G}) = \mathcal{Z}(\mathcal{F}) \otimes \mathcal{Z}(\mathcal{G})$. By Theorem 2.2 and the fact that factor algebras are isomorphic to algebras of operators, we have $\mathcal{F} \cong \bigoplus_k \mathrm{Lin}(\mathcal{H}_F^k)$ and $\mathcal{G} \cong \bigoplus_l \mathrm{Lin}(\mathcal{H}_G^l)$, so $\mathcal{F} \otimes \mathcal{G} \cong (\bigoplus_k \mathrm{Lin}(\mathcal{H}_F^k)) \otimes (\bigoplus_l \mathrm{Lin}(\mathcal{H}_G^l)) \cong \bigoplus_{kl} \mathrm{Lin}(\mathcal{H}_F^k \otimes \mathcal{H}_G^l)$. Thus, $\mathcal{Z}(\mathcal{F} \otimes \mathcal{G})$'s atomic projectors are the $\pi^k \otimes \mu^l$'s, where the first are the atomic projectors of $\mathcal{Z}(\mathcal{F})$ and the second are the atomic projectors of $\mathcal{Z}(\mathcal{G})$. Yet the $\pi^k \otimes \mu^l$'s are also the atomic projectors of $\mathcal{Z}(\mathcal{F}) \otimes \mathcal{Z}(\mathcal{G})$, since they satisfy all the conditions (pairwise orthogonal, summing to the identity, and spanning $\mathcal{Z}(\mathcal{F}) \otimes \mathcal{Z}(\mathcal{G})$). Thus, $\mathcal{Z}(\mathcal{F} \otimes \mathcal{G}) = \mathcal{Z}(\mathcal{F}) \otimes \mathcal{Z}(\mathcal{G})$.

As $\cdot_{\mathcal{F},\mathcal{G}}$ is a homomorphism of finite-dimensional C*-algebras, we have by Proposition B.1 a $\mu \in \mathcal{Z}(\mathcal{F} \otimes \mathcal{G}) = \mathcal{Z}(\mathcal{F}) \otimes \mathcal{Z}(\mathcal{G})$ such that $\ker(\cdot_{\mathcal{F},\mathcal{G}}) = \mu(\mathcal{F} \otimes \mathcal{G})$. $\mu$ then satisfies $0 = \cdot_{\mathcal{F},\mathcal{G}}(\mu) = \cdot_{\mathcal{Z}(\mathcal{F}),\mathcal{Z}(\mathcal{G})}(\mu)$. Thus, $\cdot_{\mathcal{F},\mathcal{G}}$'s kernel being non null implies that $\cdot_{\mathcal{Z}(\mathcal{F}),\mathcal{Z}(\mathcal{G})}$'s kernel is non-null as well, while the reverse implication is direct. Since a linear map is injective if and only if its kernel is null, this yields the desired result. $\square$

We can use this to define a notion of correlation length in a partition, ensuring that the algebras of sufficiently far enough parts of the graph are always uncorrelated.

**Definition C.2** (Correlation length). Let $l \ge 0$. A partition $(A_S)_{S \subseteq \Gamma} \vdash \omega$ has correlation length at most $l$ if

$$ \forall S \subseteq \Gamma, \quad A_S \perp A_{\overline{S \pm l}}. \qquad (64) $$

Note that a partition of a factor has correlation length $l = 0$ if and only if it is a factorisation, since one then has, for any $S$, $\mathcal{Z}_S \perp \mathcal{Z}_{\bar{S}} = \mathcal{Z}_S$, which is only possible for $\mathcal{Z}_S$ trivial.

The following lemma yields an important consequence of the non-correlation of algebras.

**Lemma C.1.** Let $\mathcal{F}, \mathcal{G} \subseteq \omega$ be C* algebras, with $\mathcal{F} \perp \mathcal{G}$. Then for any C*-algebras $A_1, A_2 \subseteq \mathcal{F}$ and $A_3, A_4 \subseteq \mathcal{G}$,
$$ (\mathcal{A}_1 \vee \mathcal{A}_3) \cap (\mathcal{A}_2 \vee \mathcal{A}_4) \\ = (\mathcal{A}_1 \cap \mathcal{A}_2) \vee (\mathcal{A}_3 \cap \mathcal{A}_4), \qquad (65) $$

$$ (\mathcal{A}_1 \vee \mathcal{A}_3) \cap \mathcal{A}_2 = \mathcal{A}_1 \cap \mathcal{A}_2. \qquad (66) $$

*Proof.* We first look at (65). Let us start with the reverse inclusion: it is in fact always true and does not require the existence of $\mathcal{F}$ and $\mathcal{G}$. Indeed, an element of (65)'s RHS, being in the span of elements of $\mathcal{A}_1 \cap \mathcal{A}_2$ and $\mathcal{A}_3 \cap \mathcal{A}_4$, is in particular in the span of elements of $\mathcal{A}_1$ and $\mathcal{A}_3$, as well as in that of elements of $\mathcal{A}_2$ and $\mathcal{A}_4$.

The interesting inclusion here is therefore the direct one, to which we now turn. We fix a basis $\{g^1, \dots, g^{n_1}\}$ of $\mathcal{A}_3 \cap \mathcal{A}_4$, then complete it into a basis $\{g^1, \dots, g^{n_2}\}$ of $\mathcal{A}_3$ on the one hand, and into a basis $\{g^1, \dots, g^{n_1}, g^{n_2+1}, \dots, g^{n_3}\}$ of $\mathcal{A}_4$ on the other hand, yielding a basis $\{g^1, \dots, g^{n_3}\}$ of $\mathcal{A}_3 \uplus \mathcal{A}_4$, which we finally complete into a basis $\{g^1, \dots, g^{n_4}\}$ of $\mathcal{G}$. Taking an element of (65)'s LHS, it can be written as $f = \sum_i a_1^i g^i = \sum_i a_2^i g^i$, where the $a_1^i$'s are in $\mathcal{A}_1$ and are null for $i > n_2$ and the $a_2^i$'s are in $\mathcal{A}_2$ and are null for $n_1 < i \le n_2$ and for $i > n_4$. By injectiveness of $\cdot_{\mathcal{F}\otimes\mathcal{G}}$, this implies that $\sum_i a_1^i \otimes g^i = \sum_i a_2^i \otimes g^i$, which, since the $g^i$'s form a basis of $\mathcal{G}$, implies that $a_1^i = a_2^i$ for every $i$. Thus $f = \sum_i a_1^i g^i$ where the $a_1^i$'s are in $\mathcal{A}_1 \cap \mathcal{A}_2$ and are null for $i > n_1$, so $f \in (\mathcal{A}_1 \cap \mathcal{A}_2) \vee (\mathcal{A}_3 \cap \mathcal{A}_4)$.

Turning to (66), it is simply obtained by setting $\mathcal{A}_4$ to be the trivial algebra $\mathcal{I} := \{\lambda 1 | \lambda \in \mathbb{C}\}$ in (65), so that $\mathcal{A}_3 \cap \mathcal{A}_4 = \mathcal{I}$. $\square$

## C.2 Connected partitions over a graph

As shown by Ref. [21]' Proposition 4.3, in a partition $(\mathcal{A}_S)_{S \subseteq X} \vdash \omega$, the centres of joint algebras are determined by those of individual algebras. When $\omega$ is a factor, the rule that connects them can be intuitively expressed in the following way: a projector of $\bigvee_{x \in S} \mathcal{Z}_x$ is still present in $\mathcal{Z}_S$ if and only if it is also in $\mathcal{A}_{\bar{S}}$ (and therefore in $\mathcal{Z}_{\bar{S}}$, since $\mathcal{A}_S$ and $\mathcal{A}_{\bar{S}}$ commute). Otherwise, it is not in $\mathcal{Z}_S$, i.e. $\mathcal{A}_S$ contains elements that do not commute with it. In other words, some part of a centre gets 'resolved' in $\mathcal{A}_S$ (i.e. disappears from the centre $\mathcal{Z}_S$, so that elements of $\mathcal{A}_S$ are allowed not to commute with it) whenever $S$ is large enough to encompass 'all copies' of it, i.e. whenever there are no copies of it in $\mathcal{A}_{\bar{S}}$.

A consequence of this is that if, for disjoint parts $S$ and $T$, $\mathcal{Z}_S \cap \mathcal{Z}_T \subseteq \mathcal{Z}_{S \cup T}$, then there is no part of $\mathcal{Z}_S \vee \mathcal{Z}_T$ that their joint algebra $\mathcal{A}_{S \cup T}$ can 'resolve', so $\mathcal{Z}_{S \cup T} = \mathcal{Z}_S \vee \mathcal{Z}_T$ — which, by Proposition B.6, is equivalent to $\mathcal{A}_{S \cup T} = \mathcal{A}_S \vee \mathcal{A}_T$.

This leads to a natural way of pinning down a notion of 'connectedness of sectorial correlations' in partitions over a graph. Intuitively, what we want to formalise is the following requirement: taking $S$ and $T$ distinct parts of the graph, if an information can be found in $\mathcal{A}_S$ as well as in $\mathcal{A}_T$, i.e. if the corresponding projector is in $\mathcal{A}_S \cap \mathcal{A}_T$ (and therefore in $\mathcal{Z}_S \cap \mathcal{Z}_T$ as well, since $\mathcal{A}_S$ and $\mathcal{A}_T$ commute), then it can also be found along some connected path between $S$ and $T$. Broadly speaking, the idea is that sectorial correlations should be disposed in a connected way and not 'jump' from one node to a non-contiguous one.

Using the previous considerations, our idea for a definition is the following: if $S$ and $T$ are not adjacent in the graph, then necessarily any information they share is also in the complement of their union, $\overline{S \cup T}$; so, by the reasoning two paragraphs above, $\mathcal{A}_{S \cup T} = \mathcal{A}_S \vee \mathcal{A}_T$.

**Definition C.3 (Connected partition).** Let $\Gamma$ be a finite graph, and $\omega$ a factor $C^*$-algebra. A partition $(\mathcal{A}_S)_{S \subseteq \Gamma} \vdash \omega$ is connected (or more precisely, has connected sectorial correlations) if

$$ \begin{array}{l} \forall S, T \subseteq \Gamma, \quad S \text{ and } T \text{ are not adjacent} \\ \qquad \implies \mathcal{A}_{S \cup T} = \mathcal{A}_S \vee \mathcal{A}_T. \end{array} \qquad (67) $$

*Note that, by Proposition B.6, (67) could equivalently be written as*

$$ \begin{array}{l} \forall S, T \subseteq \Gamma, \quad S \text{ and } T \text{ are not adjacent} \\ \qquad \implies \mathcal{Z}_{S \cup T} = \mathcal{Z}_S \vee \mathcal{Z}_T. \end{array} \qquad (68) $$

We see that a connected partition is one in which the algebra corresponding to a disconnected set is the algebraic span of the algebras of its connected components. This will be useful to us later, since it allows us to define the partition by only specifying the algebras corresponding to connected components (i.e., in the 1D case, to intervals).

Let us also discuss a connection with boundaries. For $\Gamma$ a graph and $S \subseteq \Gamma$, we denote $S^{\text{adj}}$ to be the set containing $S$ and its adjacent points,
$\bar{S} := \Gamma \setminus S$, and $S^{\text{bound}} := S^{\text{adj}} \cap \bar{S}$. In other words, $S^{\text{bound}}$ is the outer boundary of $S$ (its inner boundary is given by $\bar{S}^{\text{bound}}$).

**Theorem C.1.** Let $\Gamma$ be a finite graph, and $\omega$ a factor $C^*$-algebra. A partition $(\mathcal{A}_S)_{S \subseteq \Gamma} \vdash \omega$ is connected if and only if

$$ \forall S \subseteq \Gamma, \quad \mathcal{Z}_S = \mathcal{A}_{S^{\text{bound}}} \cap \mathcal{A}_{\bar{S}^{\text{bound}}}. \quad (69) $$

This yields an equivalent characterisation of connected partitions, which will be very useful to us: that any $\mathcal{A}_S$'s centre lies entirely at $S$'s boundary, or more precisely, is the intersection of its inner boundary's algebra with its outer boundary's algebra. Note that (69) could equivalently be written with $\mathcal{Z}$'s in place of $\mathcal{A}$'s in the RHS: indeed $\mathcal{A}_{S^{\text{bound}}}$ and $\mathcal{A}_{\bar{S}^{\text{bound}}}$ commute since $S^{\text{bound}}$ and $\bar{S}^{\text{bound}}$ are disjoint.

*Proof.* Suppose $(\mathcal{A}_S)_{S \subseteq \Gamma}$ is connected. Then for $S \subseteq \Gamma$, we have $\mathcal{Z}_S \subseteq \mathcal{Z}_S \lor \mathcal{Z}_{\bar{S} \cap \bar{S}^{\text{bound}}} = \mathcal{Z}_{S \cup (\bar{S} \cap \bar{S}^{\text{bound}})} = \mathcal{Z}_{\bar{S}^{\text{bound}}}$, where in the middle equality we used (68). Furthermore, as $\omega$ is factor, we have, by Proposition B.4, $\mathcal{Z}_T = \mathcal{Z}_{\bar{T}} \forall T$, and thus here $\mathcal{Z}_{\bar{S}^{\text{bound}}} = \mathcal{Z}_{S^{\text{bound}}}$, so $\mathcal{Z}_S \subseteq \mathcal{Z}_{S^{\text{bound}}}$. Symmetrically, $\mathcal{Z}_S = \mathcal{Z}_{\bar{S}} \subseteq \mathcal{Z}_{\bar{S}^{\text{bound}}}$. This yields $\mathcal{Z}_S \subseteq \mathcal{Z}_{S^{\text{bound}}} \cap \mathcal{Z}_{\bar{S}^{\text{bound}}}$, and the reverse inclusion is direct.

Conversely, suppose that $(\mathcal{A}_S)_{S \subseteq \Gamma}$ satisfies (69). Then, fixing $S, T \subseteq \Gamma$ non-adjacent, we have $\mathcal{Z}_S \subseteq \mathcal{A}_{S^{\text{bound}}} \subseteq \mathcal{A}_{\bar{S} \cup T}$. As, moreover, $\mathcal{Z}_S \subseteq \mathcal{A}_{S \cup T} = \mathcal{A}_{\bar{S} \cup T}$ by Proposition B.4, we get $\mathcal{Z}_S \subseteq \mathcal{Z}_{\bar{S} \cup T} = \mathcal{Z}_{S \cup T}$. As we can symmetrically get the same for $\mathcal{Z}_T$, we find $\mathcal{Z}_S \lor \mathcal{Z}_T \subseteq \mathcal{Z}_{S \cup T}$, with the reverse inclusion coming from (10). $\square$

### C.3 Strong connectedness in 1D

Let us take a closer look at connectedness in 1D. As Definition C.3 makes clear, when a partition is connected, what is really important in it is the data of the algebras corresponding to its connected subsets; indeed the algebra of a non-connected subset simply is the span of its connected components' algebras. In 1D, connected proper subsets are evidently intervals, which we will denote as $[m, n]$ (note that due to cyclicity, such intervals might go 'over the edge' of $\mathbb{Z}_N$). Theorem C.1 tells us that in a connected 1D partition, the centre of any interval algebra is:

$$ \mathcal{Z}_{[m,n]} = (\mathcal{Z}_m \lor \mathcal{Z}_n) \cap (\mathcal{Z}_{m-1} \lor \mathcal{Z}_{n+1}). \quad (70) $$

The following definition makes this formula simpler.

**Definition C.4** (Strongly connected 1D partitions). A connected 1D partition $(\mathcal{A}_S)_{S \subseteq \Gamma_N} \vdash \omega$ is strongly connected if, for any $m, n \in \Gamma_N$,

$$ \mathcal{Z}_{[m,n]} = (\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \lor (\mathcal{Z}_n \cap \mathcal{Z}_{n+1}). \quad (71) $$

The 'strong' qualification stems from the fact that (71) is a more refined specification than (70). Indeed, the former's RHS is always included in the latter's RHS, by the general rule $(A \cap B) \lor (C \cap D) \subseteq (A \lor C) \cap (B \lor D)$ (see the first paragraph of the proof of Lemma C.1). Furthermore, in both equations, the important aspect is the direct inclusion, since it is easy to prove that the reverse one is always true. Therefore, (71) implies (70).¹⁴

The difference between connectedness and strong connectedness is a subtle but important one. The following theorem mitigates it: if a 1D graph is sufficiently large with respect to a connected partition's correlation length (which will always be the case for the partitions we'll consider), then the partition is necessarily strongly connected.¹⁵

**Theorem C.2.** Let $N \ge 3l + 2$. Any connected partition of a factor $C^*$-algebra over $\Gamma_N$ with correlation length at most $l$ is strongly connected.

¹⁴An example of a connected yet not strongly connected 1D partition is given by taking $N = 3$,

$$ \omega := \text{Lin}(\mathbb{C}^4), \quad \mathcal{A}_0 := \left\{ \begin{pmatrix} \alpha & 0 & 0 & 0 \\ 0 & \alpha & 0 & 0 \\ 0 & 0 & \beta & 0 \\ 0 & 0 & 0 & \beta \end{pmatrix} \right\}, \quad \mathcal{A}_1 := \left\{ \begin{pmatrix} \alpha & 0 & 0 & 0 \\ 0 & \beta & 0 & 0 \\ 0 & 0 & \alpha & 0 \\ 0 & 0 & 0 & \beta \end{pmatrix} \right\} \quad \text{and} \quad \mathcal{A}_2 := \left\{ \begin{pmatrix} \alpha & 0 & 0 & 0 \\ 0 & \beta & 0 & 0 \\ 0 & 0 & \beta & 0 \\ 0 & 0 & 0 & \alpha \end{pmatrix} \right\} $$

(with the algebras for pairs specified by $\mathcal{A}_S = \mathcal{A}'_{\bar{S}}$). This partition is trivially connected, since there exist no non-adjacent subsets of $\Gamma_3$. Yet, the strong connectedness condition would lead to $\mathcal{Z}_0$ being trivial, which is not the case (here, $\mathcal{Z}_0 = \mathcal{A}_0$).

¹⁵Note that this theorem is not needed for the proof of this paper's main theorem, since the partitions involved in the causal decomposition of a 1D QCA will directly be strongly connected by construction.
Proof. Let us take an interval of $\Gamma_N$. Without loss of generality, we can take it to be $\llbracket 0, n \rrbracket$. We can also suppose without loss of generality that $n \le \frac{N}{2} - 1$; indeed, since we have $\mathcal{Z}_{\llbracket 0, n \rrbracket} = \mathcal{Z}_{\llbracket 0, n \rrbracket}$, in the case $n > \frac{N}{2} - 1$ we can equivalently consider the complementary interval $\overline{\llbracket 0, n \rrbracket} = \llbracket n + 1, -1 \rrbracket$, which satisfies $-1 - (n + 1) \stackrel{\text{mod } N}{=} N - n - 2 < N - \frac{N}{2} - 1 \le \frac{N}{2} - 1$.

Suppose first that $n \ge l$. Let us then first prove that

$$n, n + 1 \in \llbracket l, -l - 2 \rrbracket. \qquad (72)$$

First, note that $n \le \frac{N}{2} - 1$ yields $N \ge 2n + 2$. If $n \ge l + 1$, we get $-l - 2 \stackrel{\text{mod } N}{=} N - l - 2 \ge 2n - l \ge n + 1$; while for $n = l$, we can use $N \ge 3l + 2 \ge 2l + 3$, giving in that case as well $-l - 2 \stackrel{\text{mod } N}{=} N - l - 2 \ge l + 1 = n + 1$.

By Proposition B.5 applied to $(A_{-1}, A_{\llbracket 0, n \rrbracket}) \vdash A_{\llbracket -1, n \rrbracket}$,

$$ \begin{aligned} \mathcal{Z}_{\llbracket 0, n \rrbracket} &\subseteq \mathcal{Z}_{-1} \vee \mathcal{Z}_{\llbracket -1, n \rrbracket} \\ &\stackrel{(70)}{\subseteq} \mathcal{Z}_{-1} \vee \mathcal{Z}_{-1} \vee \mathcal{Z}_n \\ &= \mathcal{Z}_{-1} \vee \mathcal{Z}_n. \end{aligned} \qquad (73) $$

Furthermore, by (70), $\mathcal{Z}_{\llbracket 0, n \rrbracket} \subseteq \mathcal{Z}_{-1} \vee \mathcal{Z}_{n+1}$. We therefore find

$$ \begin{aligned} \mathcal{Z}_{\llbracket 0, n \rrbracket} &\subseteq (\mathcal{Z}_{-1} \vee \mathcal{Z}_n) \cap (\mathcal{Z}_{-1} \vee \mathcal{Z}_{n+1}) \\ &= \mathcal{Z}_{-1} \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1}), \end{aligned} \qquad (74) $$

where we were able to use Lemma C.1 since by correlation length, $\mathcal{Z}_{-1} \subseteq A_{-1} \perp A_{\llbracket -1-l, -1+l \rrbracket} = A_{\llbracket l, -l-2 \rrbracket}$, of which $\mathcal{Z}_n$ and $\mathcal{Z}_{n+1}$ are subsets by (72). Symmetrically, we also have $\mathcal{Z}_{\llbracket 0, n \rrbracket} \subseteq \mathcal{Z}_0 \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})$. Combining the two, we find

$$ \begin{aligned} \mathcal{Z}_{\llbracket 0, n \rrbracket} &\subseteq (\mathcal{Z}_{-1} \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})) \\ &\quad \cap (\mathcal{Z}_0 \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})) \\ &= (\mathcal{Z}_{-1} \cap \mathcal{Z}_0) \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1}), \end{aligned} \qquad (75) $$

where we used Lemma C.1 with $\mathcal{F} := A_{\{0,1\}}$ and $\mathcal{G} = A_n$, which are uncorrelated by correlation length since $d(\{-1, 0\}, n) > l$. Since the reverse inclusion is direct, this proves (71).

We turn to the case $n \le l - 1$. Note then that the interval $\llbracket n + l + 2, -1 \rrbracket$ has length greater than $l$, since $-1 - (n + l + 2) \stackrel{\text{mod } N}{=} N - 1 - (n + l + 2) \le l - 1$.

2) $N - n - l - 3 \ge N - 2l - 2 \ge l$, where we used $n \le l - 1$ then $N \ge 3l + 2$. Therefore the previous reasoning applies to it, so $\mathcal{Z}_{\llbracket n+l+2, -1 \rrbracket} = (\mathcal{Z}_{n+l+1} \cap \mathcal{Z}_{n+l+2}) \vee (\mathcal{Z}_{-1} \cap \mathcal{Z}_0)$. The same applies to $\mathcal{Z}_{\llbracket n+1, n+l+1 \rrbracket}$. This allows us to compute

$$ \begin{aligned} \mathcal{Z}_{\llbracket 0, n \rrbracket} &= \mathcal{Z}_{\llbracket n+1, -1 \rrbracket} \\ &\subseteq \mathcal{Z}_{\llbracket n+1, n+l+1 \rrbracket} \vee \mathcal{Z}_{\llbracket n+l+2, -1 \rrbracket} \\ &\subseteq (\mathcal{Z}_n \cap \mathcal{Z}_{n+1}) \vee (\mathcal{Z}_{n+l+1} \cap \mathcal{Z}_{n+l+2}) \\ &\quad \vee (\mathcal{Z}_{-1} \cap \mathcal{Z}_0), \end{aligned} \qquad (76) $$

where we first used the formula $\mathcal{Z}_S = \mathcal{Z}_{\tilde{S}}$, then (10) for the partition $(A_{\llbracket n+1, n+l+1 \rrbracket}, A_{\llbracket n+l+2, -1 \rrbracket}) \vdash A_{\llbracket n+1, -1 \rrbracket}$. Since $\mathcal{Z}_{\llbracket 0, n \rrbracket} \subseteq A_{\llbracket 0, n \rrbracket}$, one can add an intersection with the latter in (76)'s RHS, then use (66) in Lemma C.1 with $A_1 := (\mathcal{Z}_{-1} \cap \mathcal{Z}_0) \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})$, $\mathcal{F} = A_2 := A_{\llbracket 0, n \rrbracket}$, $A_3 = \mathcal{Z}_{n+l+1} \cap \mathcal{Z}_{n+l+2}$ and $\mathcal{G} = A_{n+l+1}$; indeed, $\mathcal{F}$ and $\mathcal{G}$ are then uncorrelated by correlation length. This yields $\mathcal{Z}_{\llbracket 0, n \rrbracket} \subseteq (\mathcal{Z}_{-1} \cap \mathcal{Z}_0) \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})$, where the reverse inclusion is direct. $\square$

# D Proof of Theorem 3.1

## D.1 A few lemmas

The lemmas in this subsection will come in handy. They allow us to soundly manipulate projections of subalgebras by orthogonal projectors commuting with them (i.e. $\pi \mathcal{F}$'s, where $\pi \in \mathcal{F}'$ is an orthogonal projector), which will be ubiquitous in our proof.

The first lemma aims to understand whether, for a $\pi \in \mathcal{F}'$, $\pi \mathcal{F}$ is the same thing as the intersection of $\mathcal{F}$ with the image of the homomorphism $f \mapsto \pi f = f\pi$.<fnref n="16" /> The result only holds if $\pi$ is in $\mathcal{F}$, and consequently in $\mathcal{Z}(\mathcal{F})$.

**Lemma D.1.** Let $\omega$ and $\mathcal{F} \subseteq \omega$ be $C^*$ algebras. We take $\pi$ an orthogonal projector of $\omega$, commuting with the elements of $\mathcal{F}$. We denote $\hat{\pi} : f \mapsto \pi f = f\pi$, acting on $\{\pi\}$. Then

$$ \pi \in \mathcal{F} \iff \pi \mathcal{F} = \mathcal{F} \cap \text{Im}(\hat{\pi}). \qquad (77) $$

Note that, since $\pi \in \mathcal{F}'$ by assumption, assuming $\pi \in \mathcal{F}$ is the same as assuming $\pi \in \mathcal{Z}(\mathcal{F})$.

<footnote n="16">¹⁶In general this is not the case, e.g. if $\pi$ projects a first qubit on $|0\rangle$, whereas $\mathcal{F}$ is the algebra of operators local to a second qubit.</footnote>
Proof. For the direct implication, supposing the LHS, we have that for all $f' \in \mathcal{F}$, $\pi f' \in \mathcal{F}$ since $\pi$ is in $\mathcal{F}$ and by closure of $\mathcal{F}$. So, for the direct inclusion, an element of $\pi \mathcal{F}$, which can be written as $f = \pi f'$, is both in $\mathcal{F}$ and in $\mathrm{Im}(\hat{\pi})$. Conversely, an element $f \in \mathcal{F} \cap \mathrm{Im}(\hat{\pi})$ if of the form $f = \pi f' = \pi \pi f' = \pi f$, so it is in $\pi \mathcal{F}$.

For the reverse implication, the RHS yields in particular that $\pi \mathcal{F} \subseteq \mathcal{F}$, so $\pi = \pi \mathcal{1} \in \pi \mathcal{F}$ is in $\mathcal{F}$. $\square$

The rest of our lemmas are all directed at the goal of keeping some control over what happens when one combines intersections of subalgebras (i.e. $\mathcal{F}_1 \cap \mathcal{F}_2$'s) and projections of subalgebras (i.e. $\pi \mathcal{F}$'s). More precisely, we want to pin down the relationship between $\pi(\mathcal{F}_1 \cap \mathcal{F}_2)$ and $\pi \mathcal{F}_1 \cap \pi \mathcal{F}_2$. First, the former is included in the latter.

**Lemma D.2.** Let $\mathcal{F}_1, \mathcal{F}_2 \subseteq \omega$ be $C^*$ algebras, and $\pi$ an orthogonal projector in $\mathcal{F}'_1 \cap \mathcal{F}'_2$. Then

$$ \pi(\mathcal{F}_1 \cap \mathcal{F}_2) \subseteq \pi \mathcal{F}_1 \cap \pi \mathcal{F}_2. \quad (78) $$

Proof. An element of the LHS is of the form $\pi f$ with $f \in \mathcal{F}_1 \cap \mathcal{F}_2$, and is therefore in particular in $\pi \mathcal{F}_1$ and in $\pi \mathcal{F}_2$. $\square$

A crucial stake is our ability to go in the other direction, i.e. to factor a projection out of an intersection. Taking again $\pi$ a projection on a first qubit, $\mathcal{F}_1$ the algebra of the second, and $\mathcal{F}_2 = (cZ)\mathcal{F}_1(cZ)^\dagger$ shows this is not always possible.¹⁷ But the following allows us to figure out to which extent we can do it: if we find an intermediate algebra $\mathcal{G}$ such that $\mathcal{F}_1, \mathcal{F}_2 \subseteq \mathcal{G} \subseteq \{\pi\}'$, then we can ‘screen’ $\pi$’s effect by a projector $\mu \in \mathcal{Z}(\mathcal{G})$, and $\pi$ can then be factored out, as long as $\mu$ is applied to the two algebras instead of $\pi$.

**Lemma D.3.** Working in a $C^*$-algebra $\omega$, let $\mathcal{F}_1, \mathcal{F}_2 \subseteq \mathcal{G} \subseteq \omega$ be sub-$C^*$ algebras, and $\pi$ be an orthogonal projector in $\mathcal{G}'$. Then

$$ \begin{array}{l} \textsuperscript{17}In other words, taking $\omega = \mathrm{Lin}(\mathbb{C}^4)$, $\mathcal{F}_1 = \\ \left\{ \begin{pmatrix} \alpha & \beta & 0 & 0 \\ \gamma & \delta & 0 & 0 \\ 0 & 0 & \alpha & \beta \\ 0 & 0 & \gamma & \delta \end{pmatrix} \right\}$, $\mathcal{F}_2 = \\ \left\{ \begin{pmatrix} \alpha & \beta & 0 & 0 \\ \gamma & \delta & 0 & 0 \\ 0 & 0 & \alpha & -\beta \\ 0 & 0 & -\gamma & \delta \end{pmatrix} \right\} \\ \text{and } \pi = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ yields a strict inclusion in (78). \end{array} $$

$$ \pi \mathcal{F}_1 \cap \pi \mathcal{F}_2 = \pi(\mu \mathcal{F}_1 \cap \mu \mathcal{F}_2), \quad (79) $$

where $\mu$ is the minimal orthogonal projector in $\mathcal{Z}(\mathcal{G})$ such that $\pi \mu = \pi$, as defined in Corollary B.1.

Proof. We then have:

$$ \begin{align} \pi \mathcal{F}_1 \cap \pi \mathcal{F}_2 &= \pi \mu \mathcal{F}_1 \cap \pi \mu \mathcal{F}_2 \tag{80} \\ &= \pi(\mu \mathcal{F}_1 \cap \mu \mathcal{F}_2), \notag \end{align} $$

where the second line comes from (49c): $\hat{\pi}$ is injective on $\mu \mathcal{G}$, in which $\mu \mathcal{F}_1$ and $\mu \mathcal{F}_2$ are included, so that intersecting the images under $\hat{\pi}$ is the same as intersecting the preimages and then applying $\hat{\pi}$. $\square$

The following two lemmas provide analogous results for the case in which one of the algebras at hand is a commutant.

**Lemma D.4.** Let $\mathcal{F}_1, \mathcal{F}_2 \subseteq \omega$ be $C^*$ algebras, and $\pi$ an orthogonal projector in $\mathcal{F}'_1 \cap \mathcal{F}'_2$. Then

$$ \pi(\mathcal{F}'_1 \cap \mathcal{F}_2) \subseteq \mathcal{F}'_1 \cap \pi \mathcal{F}_2. \quad (81) $$

Proof. An element of the LHS is of the form $\pi f$ with $f \in \mathcal{F}'_1 \cap \mathcal{F}_2$; as $\pi \in \mathcal{F}'_1$, $\pi f \in \mathcal{F}'_1$, so $\pi f \in \mathcal{F}'_1 \cap \pi \mathcal{F}_2$. $\square$

**Lemma D.5.** Working in a $C^*$-algebra $\omega$, let $\mathcal{F}_1, \mathcal{F}_2 \subseteq \mathcal{G}$ be sub-$C^*$ algebras, and $\pi$ be an orthogonal projector in $\mathcal{G}'$. Then

$$ \mathcal{F}'_1 \cap \pi \mathcal{F}_2 = \pi(\mathcal{F}'_1 \cap \mu \mathcal{F}_2), \quad (82) $$

where $\mu$ is the minimal orthogonal projector in $\mathcal{Z}(\mathcal{G})$ such that $\pi \mu = \pi$ as defined in Corollary B.1.

Proof. Let us then take $\pi f_2 \in \mathcal{F}'_1 \cap \pi \mathcal{F}_2$. Then for any $f_1 \in \mathcal{F}_1$, $0 = [\pi f_2, f_1] = \pi[f_2, f_1] = \pi \mu[f_2, f_1]$, where in the first equality we used the fact that $\pi$ commutes with $f_1$ and $f_2$, and in the second we used $\pi = \pi \mu$. By (49c), $\hat{\pi}$ is injective on $\mu \mathcal{G} \ni \mu[f_2, f_1]$, so $\mu[f_2, f_1] = 0$. Using the fact that $\mu$ is in $\mathcal{Z}(\mathcal{G})$ and thus commutes with $f_1$ and $f_2$, we can turn this into $[\mu f_2, f_1] = 0$; as this holds for any $f_1$, we find $\mu f_2 \in \mathcal{F}'_1$. Therefore $\pi f_2 = \pi \mu f_2 \in \pi(\mathcal{F}'_1 \cap \mu \mathcal{F}_2)$, so $\mathcal{F}'_1 \cap \pi \mathcal{F}_2 \subseteq \pi(\mathcal{F}'_1 \cap \mu \mathcal{F}_2)$.

The reverse inclusion is obtained through $\pi(\mathcal{F}'_1 \cap \mu \mathcal{F}_2) \subseteq \mathcal{F}'_1 \cap \pi \mu \mathcal{F}_2 = \mathcal{F}'_1 \cap \pi \mathcal{F}_2$, where we used Lemma D.4 then $\pi \mu = \pi$. $\square$
Recall that in the context of a strongly connected 1D partition, what really matters are the ‘edge centres’ $\mathcal{Z}_k \cap \mathcal{Z}_{k+1}$: indeed we saw that the center of a $\mathcal{A}_{\llbracket m,n \rrbracket}$ eventually decomposes as $(\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})$. A typical situation we will be facing is that we want to apply the previous lemmas for $\mathcal{G} = \mathcal{A}_{\llbracket m,n \rrbracket}$ and $\pi \in \mathcal{Z}_k \cap \mathcal{Z}_{k+1}$, with $k \notin \llbracket m, n \rrbracket$. We then want to keep some control as to the whereabouts of the minimal $\mu \in \mathcal{G}$ such that $\pi\mu = \pi$, as defined in Corollary B.1.

The idea is that:

* if $k$ is sufficiently far to the right of $n$, then $\mu$ is ‘only on $m$’s side, i.e. in $\mathcal{Z}_{m-1} \cap \mathcal{Z}_m$;

* symmetrically, if $k+1$ is sufficiently far to the left of $m$, then $\mu$ is in $\mathcal{Z}_n \cap \mathcal{Z}_{n+1}$;

* if $k$ is sufficiently far from both, then $\mu = 1$.

**Lemma D.6.** 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 take $k, m, n \in \Gamma_N$ with $k, k+1 \notin \llbracket m, n \rrbracket$. We take an orthogonal projector $\pi \in \mathcal{Z}_k \cap \mathcal{Z}_{k+1}$ and take $\mu$ to be the minimal orthogonal projector in $\mathcal{Z}_{\llbracket m, n \rrbracket} = (\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})$ such that $\pi\mu = \pi$. Then (remember the subtractions are defined modulo $N$):

* if $m - (k+1) \ge l$, then $\mu \in \mathcal{Z}_n \cap \mathcal{Z}_{n+1}$;

* if $k - n \ge l$, then $\mu \in \mathcal{Z}_{m-1} \cap \mathcal{Z}_m$;

* if $d(\{k, k+1\}, \{m, n\}) \ge l$, then $\mu = 1$.

**Proof.** We can equivalently work with $\bar{\mu} := 1 - \mu$, the maximal projector in $\mathcal{Z}_{\llbracket m, n \rrbracket}$ such that $\pi\bar{\mu} = 0$. As a preliminary, we first note that if $d(\{k\}, \{m, n\}) > l$, then we have $\mu = 1$; indeed, $\mathcal{Z}_{\llbracket m, n \rrbracket} \subseteq \mathcal{A}_{\{k\} \pm l} \perp \mathcal{A}_{\{k\}} \supseteq \mathcal{Z}_k \cap \mathcal{Z}_{k+1}$ by correlation length, so $\pi\bar{\mu} = 0 \implies \bar{\mu} = 0$. Symmetrically, the same happens if $d(\{k+1\}, \{m, n\}) > l$. Therefore, we can restrict the proof to cases in which both of these distances are lesser or equal than $l$.

Let us start with the first item in the list. Suppose first that $m = k+l+1$; $d(\{k\}, \{m, n\}) \le l$, obtained by the previous restriction of the proof, then yields $d(k, n) = k - n \le l$. We thus have $l+1 \le m - (n+1) \le 2l$, so that $d(m, n+1) = \min(m-(n+1), N-(m-(n+1)) \ge \min(l+1, N-2l) \ge l+1$, where we used $N \ge 3l+1$. Thus,

$d(\{k, n+1\}, \{m\}) > l$, so $(\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1}) \subseteq \mathcal{A}_{\llbracket n+1,k \rrbracket} \perp \mathcal{A}_{\overline{\llbracket n+1,k \rrbracket} \pm l} \supseteq \mathcal{Z}_{m-1} \cap \mathcal{Z}_m$.

Let us decompose $\bar{\mu} \in \mathcal{Z}_{\llbracket m, n \rrbracket}$ along the atomic projectors of $\mathcal{Z}_{\llbracket m, n \rrbracket}$, which are themselves, by Lemma B.2, of the form $\pi_n^i \pi_m^j$ where the $\pi_n^i$'s are the atomic projectors of $\mathcal{Z}_n \cap \mathcal{Z}_{n+1}$ and the $\pi_m^j$'s are the atomic projectors of $\mathcal{Z}_{m-1} \cap \mathcal{Z}_m$. In particular, because $\bar{\mu}$ is a projector we have that $\bar{\mu} = \sum_{i,j} \alpha^{ij} \pi_n^i \pi_m^j$ with the $\alpha_{ij} \in \{0,1\}$. Factorising by the $\pi_n^i$'s, we can rewrite $\bar{\mu} = \sum_i \pi_n^i \bar{\mu}_m^i$, where the $\bar{\mu}_m^i$'s are orthogonal projectors of $\mathcal{Z}_{m-1} \cap \mathcal{Z}_m$.

Fixing an $i$ such that $\bar{\mu}_m^i \ne 0$, we have that $0 = \pi\bar{\mu}\pi_n^i = \pi \sum_i \pi_n^i \bar{\mu}_m^i \pi_n^i = \pi\pi_n^i \bar{\mu}_m^i$. Because $\pi\pi_n^i \in (\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})$ and $\bar{\mu}_m^i \in \mathcal{Z}_{m-1} \cap \mathcal{Z}_m$ which are uncorrelated, and since $\bar{\mu}_m^i \ne 0$ by assumption, this leads to $\pi\pi_n^i = 0$. Because $\bar{\mu}$ is maximal, this means that the $\bar{\mu}_m^i$'s are all equal to 1 and therefore $\bar{\mu} \in \mathcal{Z}_n \cap \mathcal{Z}_{n+1}$.

In the general case, $m \in \llbracket k+l+1, n \rrbracket$, so $\ker \hat{\pi} \cap \mathcal{Z}_{\llbracket m, n \rrbracket} = \ker \hat{\pi} \cap \mathcal{Z}_{\llbracket m, n \rrbracket} \cap \mathcal{A}_{\llbracket k+l+1, n \rrbracket}$. Yet the previous case showed that $\ker \hat{\pi} \cap \mathcal{A}_{\llbracket k+l+1, n \rrbracket} = \bar{\mu}\mathcal{A}_{\llbracket k+l+1, n \rrbracket}$, with $\bar{\mu} \in \mathcal{Z}_n \cap \mathcal{Z}_{n+1}$, so $\ker \hat{\pi} \cap \mathcal{A}_{\llbracket k+l+1, n \rrbracket} = \bar{\mu}\mathcal{A}_{\llbracket k+l+1, n \rrbracket} = \mathcal{A}_{\llbracket k+l+1, n \rrbracket} \cap \mathrm{Im}(\hat{\mu})$ by Lemma D.1. We thus find $\ker \hat{\pi} \cap \mathcal{Z}_{\llbracket m, n \rrbracket} = \mathcal{Z}_{\llbracket m, n \rrbracket} \cap \mathrm{Im}(\hat{\mu}) = \bar{\mu}\mathcal{Z}_{\llbracket m, n \rrbracket}$ by Lemma D.1, since $\bar{\mu} \in \mathcal{Z}_n \cap \mathcal{Z}_{n+1} \subseteq \mathcal{Z}_{\llbracket m, n \rrbracket}$.

The proof of the second item is symmetric. For the third item, the restriction introduced in the proof's first paragraph now reduces it to $m = k+l+1$, $n = k-l$. Applying the two first items then yields $\bar{\mu} \in (\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \cap (\mathcal{Z}_n \cap \mathcal{Z}_{n+1})$, and the latter is the trivial sub-C* algebra as $d(m, n+1) = \min(m-(n+1), N-(m-(n+1))) = \min(2l, N-2l) > l$ so that $\mathcal{Z}_{m-1} \cap \mathcal{Z}_m \subseteq \mathcal{A}_m \perp \mathcal{A}_{\overline{\{m\} \pm l}} \supseteq \mathcal{A}_{n+1} \supseteq \mathcal{Z}_n \cap \mathcal{Z}_{n+1}$. $\square$

The following two lemmas leverage the third item to yield results when one is looking at how intersection interplay, not with the projection by a given atomic projector of $\mathcal{Z}_k \cap \mathcal{Z}_{k+1}$, but with spans with the whole of the latter.

**Lemma D.7.** 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 take $k, m, n \in \Gamma_N$ with $k, k+1 \notin \llbracket m, n \rrbracket$. We take $\mathcal{F}_1, \mathcal{F}_2 \subseteq \mathcal{G} = \mathcal{A}_{\llbracket m, n \rrbracket}$. If $d(\{k, k+1\}, \{m, n\}) \ge l$, then

$$ ((\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{F}_1) \cap ((\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{F}_2) \qquad (83a) \\ = (\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee (\mathcal{F}_1 \cap \mathcal{F}_2); $$
$$((\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{F}_1) \cap \mathcal{F}_2 = \mathcal{F}_1 \cap \mathcal{F}_2. \quad (83b)$$

Proof. Since $\mathcal{Z}_k \cap \mathcal{Z}_{k+1}$ is the linear span of its atomic projectors, which we denote as $\pi^i$'s, any element of $(\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{A}_{[m,n]}$ is of the form $\sum_i \pi^i g^i$, where the $g^i$'s are in $\mathcal{A}_{[m,n]}$. Furthermore, let us prove that this way of writing these elements is unique. If we suppose $\sum_i \pi^i g^i_1 = \sum_i \pi^i g^i_2$, then for any $i$, multiplying by $\pi^i$ yields $\pi^i g^i_1 = \pi^i g^i_2$. Yet $\hat{\pi}^i$ is injective on $\mathcal{A}_{[m,n]}$, since Lemma D.6 (applied with our assumption $d(\{k, k+1\}, \{m, n\}) \ge l$), tells us that the intersection of its kernel with the latter is designated by a $\bar{\mu} = 0$; so $g^i_1 = g^i_2$ for every $i$.

Thus, taking an element $f$ of (83a)'s LHS, it is of the form $f = \sum_i \pi^i f^i_1 = \sum_i \pi^i f^i_2$, with the $f^i_1$'s in $\mathcal{F}_1$ and the $f^i_2$'s in $\mathcal{F}_2$; so for every $i$, $f^i_1 = f^i_2 \in \mathcal{F}_1 \cap \mathcal{F}_2$, so (83a)'s LHS is included in its RHS, while the reverse inclusion is direct.

Turning to (83b), taking an element $f$ of its LHS, $f$ is in $\mathcal{F}_2$ and can be written both as $f = \sum_i \pi^i f^i_1$, with the $f^i_1$'s in $\mathcal{F}_1$, and as $f = 1f = \sum_i \pi^i f$. Thus we have $f^i_1 = f$ for every $i$, so $f \in \mathcal{F}_1 \cap \mathcal{F}_2$. This shows that $((\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{F}_1) \cap \mathcal{F}_2 \subseteq \mathcal{F}_1 \cap \mathcal{F}_2$, and the reverse inclusion is immediate. □

**Lemma D.8.** 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 take $k, m, n \in \Gamma_N$ with $k, k+1 \notin [m, n]$. We take $\mathcal{F}_1, \mathcal{F}_2 \subseteq \mathcal{G} = \mathcal{A}_{[m,n]}$. If $d(\{k, k+1\}, \{m, n\}) \ge l$, then

$$((\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{F}_2) \cap \mathcal{F}'_1 = (\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee (\mathcal{F}_2 \cap \mathcal{F}'_1). \quad (84)$$

Proof. An element of the RHS is an algebraic combination of elements of $\mathcal{Z}_k \cap \mathcal{Z}_{k+1}$ and of $(\mathcal{F}_2 \cap \mathcal{F}'_1)$, so it is in particular an algebraic combination of elements of the former and of $\mathcal{F}_2$. It also commutes with $\mathcal{F}_1$ since $\mathcal{Z}_k \cap \mathcal{Z}_{k+1} \subseteq \mathcal{A}_{[n+1, m-1]} = \mathcal{A}'_{[n+1, m-1]} = \mathcal{A}'_{[m, n]} \subseteq \mathcal{F}'_1$, so we have the indirect inclusion in (84).

We turn to the direct inclusion, denoting (84)'s LHS as $\mathcal{D}$. Since $\mathcal{Z}_k \cap \mathcal{Z}_{k+1} \subseteq \mathcal{F}'_1$, $\mathcal{Z}_k \cap \mathcal{Z}_{k+1} \subseteq \mathcal{D}$. Moreover, $\mathcal{D} \subseteq (\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{F}_2 \subseteq (\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{A}_{[m,n]}$, and elements of $\mathcal{Z}_k \cap \mathcal{Z}_{k+1}$ commute with all elements of $\mathcal{Z}_k \cap \mathcal{Z}_{k+1}$ but also with all elements of $\mathcal{A}_{[m,n]}$ as seen in the previous paragraph since $k, k+1 \notin \mathcal{A}_{[m,n]}$; so $\mathcal{Z}_k \cap \mathcal{Z}_{k+1} \subseteq \mathcal{D}'$. Thus, $\mathcal{Z}_k \cap \mathcal{Z}_{k+1} \subseteq \mathcal{Z}(\mathcal{D})$.

We can therefore apply Lemma B.3; (52) yields $\mathcal{D} = \bigcup_{\pi \in \text{AtomProj}(\mathcal{Z}_k \cap \mathcal{Z}_{k+1})} \pi \mathcal{D}$. Let us fix a $\pi \in \text{AtomProj}(\mathcal{Z}_k \cap \mathcal{Z}_{k+1})$. By Lemma D.4,

$$\begin{align}
\pi \mathcal{D} &\subseteq \pi((\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \mathcal{F}_2) \cap \mathcal{F}'_1 \nonumber \\
&= (\pi(\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee \pi \mathcal{F}_2) \cap \mathcal{F}'_1 \nonumber \\
&= (\{\alpha \pi | \alpha \in \mathbb{C}\} \vee \pi \mathcal{F}_2) \cap \mathcal{F}'_1 \nonumber \\
&= \pi \mathcal{F}_2 \cap \mathcal{F}'_1, \tag{85}
\end{align}$$

where the second line comes from the fact that $\pi$ commutes with both terms in the bracket, and in the fourth we used $\{\alpha\pi | \alpha \in \mathbb{C}\} \subseteq \pi \mathcal{F}_2$. By Lemmas D.5 and D.6, the last line is equal to $\pi(\mathcal{F}_2 \cap \mathcal{F}'_1)$, so $\pi \mathcal{D} \subseteq \pi(\mathcal{F}_2 \cap \mathcal{F}'_1)$. Therefore, $\mathcal{D} \subseteq \bigcup_{\pi \in \text{AtomProj}(\mathcal{Z}_k \cap \mathcal{Z}_{k+1})} \pi(\mathcal{F}_2 \cap \mathcal{F}'_1) \subseteq (\mathcal{Z}_k \cap \mathcal{Z}_{k+1}) \vee (\mathcal{F}_2 \cap \mathcal{F}'_1)$. □

## D.2 Vertex-splitting

The crucial tool in the proof of Theorem 3.1 is the following refinement lemma. It tells us which data specification is sufficient to ‘split a vertex’ in a strongly connected 1D partition of a factor, i.e. to turn it into a partition over a graph in which a certain vertex (say, 0) has been replaced with two vertices, $-\frac{1}{4}$ and $\frac{1}{4}$. In other words, it tells us under which conditions we can do a ‘one-vertex fine-graining’ of a strongly connected 1D partition. See Figure 3 for reference.

**Lemma D.9 (Vertex-splitting).** Let $(\mathcal{A}_S)_{S \subseteq \Gamma_N} \vdash \omega$ be a connected 1D partition of a factor $\omega$, with correlation length at most $l$, where $N \ge 4l + 1$. We fix an element of $\Gamma_N$, which we take without loss of generality to be 0. If $N$ is odd, we write $p_-$ and $p_+$ := $p_- + 1$ as the two elements furthest away from 0; if $N$ is even, we write $p_-$ as the furthest element from 0 and $p_+ = p_- + 1$ (in other words, in both cases, $p_- := -\lfloor \frac{N-1}{2} \rfloor$ and $p_+ := \lfloor \frac{N-1}{2} \rfloor$). We also write $q_- := p_- + l$ and $q_+ := p_+ - l$.

Suppose there exists a dual pair of sub-$C^*$ algebras $\mathcal{A}^{\text{split}}_{[p_-, -\frac{1}{4}]}$, $\mathcal{A}^{\text{split}}_{[\frac{1}{4}; p_+]} := (\mathcal{A}^{\text{split}}_{[p_-, -\frac{1}{4}]})'$ satisfying:

$$\mathcal{A}_{[p_-, -1]} \subseteq \mathcal{A}^{\text{split}}_{[p_-, -\frac{1}{4}]} \subseteq \mathcal{A}_{[p_-, 0]}; \quad (86a)$$

$$\mathcal{A}^{\text{split}}_{[p_-, -\frac{1}{4}]} = \mathcal{A}_{[p_-, 0]} \cap (\mathcal{A}^{\text{split}}_{[\frac{1}{4}; p_+]} \cap \mathcal{A}_{[0, q_+]})'. \quad (86b)$$

Then the following rules define another strongly connected partition of $\omega$, also with cor-
relation length at most $l$, over $\Gamma^{\text{split}} := \Gamma_N \setminus \{0\} \cup \{-\frac{1}{4}, \frac{1}{4}\}$.<fnref n="18" />

1. $\forall S \subseteq \Gamma_N \setminus \{0\}, A_S^{\text{split}} := A_S$ and $A_{S \cup \{-\frac{1}{4}, \frac{1}{4}\}}^{\text{split}} := A_{S \cup \{0\}}$;

2. $\forall m \in [\llap{p_{-}}, -\frac{1}{4}]$, $A_{[m, -\frac{1}{4}]}^{\text{split}} := A_{[m, 0]} \cap A_{[p_{-}, -\frac{1}{4}]}^{\text{split}}$;

3. $\forall m \in [\llap{\frac{1}{4}}, p_{+}]$, $A_{[\llap{\frac{1}{4}}, m]}^{\text{split}} := A_{[0, m]} \cap A_{[\llap{\frac{1}{4}}, p_{+}]}^{\text{split}}$;

4. algebras for intervals of size larger than $\frac{N}{2}$ are defined via the rule $A_S^{\text{split}} := (A_S^{\text{split}})'$;

5. $A_S^{\text{split}}$ for $S$ not an interval is defined as the algebraic span of the algebras for each of $S$'s connected components.

Before we turn to the proof, let us make a few comments. First, note that since $N \ge 4l + 1 \ge 3l+2$, $(A_S)_{S \subseteq \Gamma_N}$ is necessarily strongly connected by Theorem C.2, i.e. all of its interval's algebras have centre $Z_{[m,n]} = (Z_{m-1} \cap Z_m) \cup (Z_n \cap Z_{n+1})$. (If $l=0$, then it is a factorisation and therefore strongly connected as well.)

On another note, the rules make clear that the important cases in defining the $A_S^{\text{split}}$'s are the ones where $S$ contains exactly one of the $\pm\frac{1}{4}$'s. Thus, to alleviate notations, we will denote the $A_S^{\text{split}}$'s as $A_S$'s; the presence of $\pm\frac{1}{4}$ in the specification of $S$ will suffice to understand that we are talking about algebras of the refined partitions. Note also that with the notations given by the rules, the assumption (86b) can be written more simply as

$$A_{[p_{-}, -\frac{1}{4}]} = A_{[p_{-}, 0]} \cap A'_{[\frac{1}{4}, q_{+}]} \quad (87)$$

Another thing to note is that $N \ge 4l + 1$ and the definitions of the $p_{\pm}$ and $q_{\pm}$ imply that $q_{-}$ is at a distance at least $l$ from 0 and from $p_{-}$, and symmetrically for $q_{+}$; this will be central in many arguments involving correlation length.

We also stress that the lemma's assumptions are completely symmetric under exchange of the 'left' and 'right' sides. Indeed, (86a) and (86b) are equivalent to

$$A_{[1, p_{+}]} \subseteq A_{[\frac{1}{4}, p_{+}]}^{\text{split}} \subseteq A_{[0, p_{+}]} \quad (88a)$$

$$ \begin{aligned} A_{[\frac{1}{4}, p_{+}]} &= A_{[0, p_{+}]} \cap \left( A_{[p_{-}, -\frac{1}{4}]}^{\text{split}} \cap A_{[q_{-}, 0]} \right)' \quad (88b) \\ &= A_{[0, p_{+}]} \cap A'_{[q_{-}, -\frac{1}{4}]} \end{aligned} $$

More specifically, taking commutants suffices to find that (86a) and (88a) are equivalent; and, under the assumption of the former, (86b) (or its simplified form (87)) is equivalent to (88b). Let us prove this latter equivalence; given the obvious symmetry, showing implication one way is sufficient. We assume (87) holds. First,

$$ \begin{align} A_{[q_{-}, p_{+}]} \cap A'_{[\frac{1}{4}, p_{+}]} & \tag{88a} \\ & \stackrel{(88a)}{=} A_{[q_{-}, p_{+}]} \cap A'_{[1, p_{+}]} \cap A'_{[\frac{1}{4}, p_{+}]} \nonumber \\ & \stackrel{(55)}{=} (Z_{[q_{-}, p_{+}]} \vee A_{[q_{-}, 0]}) \cap A'_{[\frac{1}{4}, p_{+}]} \nonumber \\ & = ((Z_{q_{-}-1} \cap Z_{q_{-}}) \vee (Z_{p_{-}} \cap Z_{p_{+}}) \nonumber \\ & \phantom{=} \quad \vee A_{[q_{-}, 0]}) \cap A'_{[\frac{1}{4}, p_{+}]} \tag{89} \\ & = ((Z_{p_{-}} \cap Z_{p_{+}}) \vee A_{[q_{-}, 0]}) \cap A'_{[\frac{1}{4}, p_{+}]} \nonumber \\ & \subseteq ((Z_{p_{-}} \cap Z_{p_{+}}) \vee A_{[q_{-}, 0]}) \cap A'_{[\frac{1}{4}, q_{+}]} \nonumber \\ & = (Z_{p_{-}} \cap Z_{p_{+}}) \vee (A_{[q_{-}, 0]} \cap A'_{[\frac{1}{4}, q_{+}]}) \nonumber \\ & \stackrel{(87)}{=} (Z_{p_{-}} \cap Z_{p_{+}}) \vee A_{[q_{-}, -\frac{1}{4}]} \nonumber \end{align} $$

As the reverse inclusion is clear, this is in fact an equality. In the penultimate line, we used Lemma D.8 with $\mathcal{G} = A_{[q_{-}, q_{+}]}$, taking advantage of the fact that $d(\{p_{-}, p_{+}\}, \{q_{-}, q_{+}\}) \ge l$.

This shows that the commutant of $A_{[\frac{1}{4}, p_{+}]}$ within $A_{[q_{-}, p_{+}]}$ is $(Z_{p_{-}} \cap Z_{p_{+}}) \vee A_{[q_{-}, -\frac{1}{4}]} = (Z_{p_{-}} \cap Z_{p_{+}}) \vee (Z_{q_{-}} \cap Z_{q_{-}-1}) \vee A_{[q_{-}, -\frac{1}{4}]} = Z_{[q_{-}, p_{+}]} \vee A_{[q_{-}, -\frac{1}{4}]}$, where the first equality comes from the fact that $(Z_{q_{-}} \cap Z_{q_{-}-1}) \subseteq A_{[q_{-}, -1]} \subseteq A_{[q_{-}, -\frac{1}{4}]}$. Yet, $A_{[q_{-}, -\frac{1}{4}]}$'s commutant within $A_{[q_{-}, p_{+}]}$ is equal to $Z_{[q_{-}, p_{+}]} \vee A_{[q_{-}, -\frac{1}{4}]}$'s commutant within the same, since the added first term commutes with all of $A_{[q_{-}, p_{+}]}$ by definition. Thus, $A_{[q_{-}, -\frac{1}{4}]}$'s commutant within $A_{[q_{-}, p_{+}]}$ is equal to $A_{[\frac{1}{4}, p_{+}]}$'s double commutant within the same, which by Proposition B.3 is equal to $Z_{[q_{-}, p_{+}]} \vee A_{[\frac{1}{4}, p_{+}]} = (Z_{q_{-}-1} \cap Z_{q_{-}}) \vee A_{[\frac{1}{4}, p_{+}]}$. We can use this to compute that (88b) holds:

<footnote n="18">¹⁸Γsplit's graph structure is the obvious 1D one, with the chain ..., -1, -1/4, 1/4, +1, ...</footnote>
<figure><img src="image_15.png" /><figcaption>Figure 3: Vertex-splitting. Lemma D.9 indicates what is a sufficient data specification to turn a partition over $\Gamma_N$ to one over $\Gamma^{\text{split}}$, in which the 0 vertex has been split into two vertices $\pm \frac{1}{4}$. The distances $d(0, q_-)$, $d(q_-, p_-)$, $d(0, q_+)$, $d(q_+, p_+)$ are all greater or equal to the correlation length $l$.</figcaption></figure>

$$ \begin{align} \mathcal{A}_{\llbracket 0, p_+ \rrbracket} \cap \mathcal{A}'_{\llbracket q_-, -\frac{1}{4} \rrbracket} & = \mathcal{A}_{\llbracket 0, p_+ \rrbracket} \cap \mathcal{A}_{\llbracket q_-, p_+ \rrbracket} \cap \mathcal{A}'_{\llbracket q_-, -\frac{1}{4} \rrbracket} \tag{90} \\ & = \mathcal{A}_{\llbracket 0, p_+ \rrbracket} \cap \left( (\mathcal{Z}_{q_-, 1} \cap \mathcal{Z}_{q_-}) \cup \mathcal{A}_{\llbracket \frac{1}{4}, p_+ \rrbracket} \right) \nonumber \\ & = \mathcal{A}_{\llbracket 0, p_+ \rrbracket} \cap \mathcal{A}_{\llbracket \frac{1}{4}, p_+ \rrbracket} \tag{88a} \\ & \stackrel{(88a)}{=} \mathcal{A}_{\llbracket \frac{1}{4}, p_+ \rrbracket}, \nonumber \end{align} $$

where, in the penultimate equality, $d(\{q_-, 1, q_-, \{0, p_+\}\}) \ge l$ allowed us to use Lemma D.7 with $\mathcal{G} = \mathcal{A}_{\llbracket 0, p_+ \rrbracket}$.

Finally, let us display a few formulas that stem from our assumptions and definitions, and will be useful to refer to. We fix $m \in \llbracket p_-, -\frac{1}{4} \rrbracket$ and $n \in \llbracket \frac{1}{4}, p_+ \rrbracket$.

$$ \mathcal{A}_{\llbracket m, -1 \rrbracket} \subseteq \mathcal{A}_{\llbracket m, -\frac{1}{4} \rrbracket} \subseteq \mathcal{A}_{\llbracket m, 0 \rrbracket}; \quad (91a) $$

$$ \mathcal{A}_{\llbracket 1, n \rrbracket} \subseteq \mathcal{A}_{\llbracket \frac{1}{4}, n \rrbracket} \subseteq \mathcal{A}_{\llbracket 0, n \rrbracket}; \quad (91b) $$

$$ \mathcal{A}_{\llbracket m, -\frac{1}{4} \rrbracket} = \mathcal{A}_{\llbracket m, 0 \rrbracket} \cap \mathcal{A}'_{\llbracket \frac{1}{4}, q_+ \rrbracket}; \quad (91c) $$

$$ \mathcal{A}_{\llbracket \frac{1}{4}, n \rrbracket} = \mathcal{A}_{\llbracket 0, m \rrbracket} \cap \mathcal{A}'_{\llbracket q_-, -\frac{1}{4} \rrbracket}. \quad (91d) $$

(91a) is simply the combination of $\mathcal{A}_{\llbracket m, -\frac{1}{4} \rrbracket}$'s definition with (86a), while (91c) is its combination with (87). The two other equations are symmetric.

More generally, we can infer from this the general rules for any $S, T \subseteq \Gamma^{\text{split}}$:

$$ S \subseteq T \quad \Longrightarrow \quad \mathcal{A}_S \subseteq \mathcal{A}_T; \quad (92a) $$

$$ S \cap T = \emptyset \quad \Longrightarrow \quad \mathcal{A}_T \subseteq \mathcal{A}'_S. \quad (92b) $$

Indeed, if both $S$ and $T$ contain either none or both of the $\pm \frac{1}{4}$, then these two properties come from the fact that the original partition is indeed a partition (using Proposition B.4 for the second). Otherwise, supposing they are intervals, enumerating the few possible cases and applying (91) proves the result. Finally when they are not, one can also infer that the formulas hold from the fact that they are algebraic spans of their connected components' algebras.

We now turn to Lemma D.9's proof.

*Proof.* We have to prove that for any disjoint $S, T \subseteq \Gamma^{\text{split}}$, $(\mathcal{A}_S^{\text{split}}, \mathcal{A}_T^{\text{split}}) \vdash \mathcal{A}_{S \cup T}^{\text{split}}$. If neither $S$ nor $T$ contains any of the $\pm \frac{1}{4}$, or if one of them contains both, then given bullet point 1 above, this is direct from the fact that $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ is a partition. Otherwise, the crucial cases are the ones where $S$ and $T$ are intervals, so we will start with them.

A first case to prove is: $\forall m \in \llbracket p_-, -\frac{1}{4} \rrbracket, \forall n \in \llbracket \frac{1}{4}, p_+ \rrbracket$, $(\mathcal{A}_{\llbracket m, -\frac{1}{4} \rrbracket}, \mathcal{A}_{\llbracket \frac{1}{4}, n \rrbracket}) \vdash \mathcal{A}_{\llbracket m, n \rrbracket}$. We will use Proposition B.7 with $\tilde{Z}_1 = \mathcal{Z}_{m-1} \cap \mathcal{Z}_m$ and $\tilde{Z}_2 = \mathcal{Z}_n \cap \mathcal{Z}_{n+1}$; indeed, using (92a) and (92b), $\tilde{Z}_1$ is then included in $\mathcal{A}_m \subseteq \mathcal{A}_{\llbracket m, -\frac{1}{4} \rrbracket}$, and is also included in $\mathcal{A}_{m-1} \subseteq \mathcal{A}'_{\llbracket m, 0 \rrbracket} \subseteq \mathcal{A}'_{\llbracket m, -\frac{1}{4} \rrbracket}$, so $\tilde{Z}_1 \subseteq \mathcal{Z}_{\llbracket m, -\frac{1}{4} \rrbracket}$, and symmetrically $\tilde{Z}_2 \subseteq \mathcal{Z}_{\llbracket \frac{1}{4}, n \rrbracket}$, and $\mathcal{Z}_{\llbracket m, n \rrbracket} = \tilde{Z}_1 \lor \tilde{Z}_2$ by strong connectedness. Proposition B.7 shows that it is sufficient for us to prove: $\forall \pi \in \text{AtomProj}(\mathcal{Z}_{m-1} \cap \mathcal{Z}_m), \pi \mathcal{A}'_{\llbracket m, -\frac{1}{4} \rrbracket} \cap \mathcal{A}_{\llbracket m, n \rrbracket} = \pi \mathcal{A}_{\llbracket \frac{1}{4}, n \rrbracket}$.

Before we start proving this equality, let us remark that the LHS can be rewritten as $\mathcal{A}'_{\llbracket m, -\frac{1}{4} \rrbracket} \cap \pi \mathcal{A}_{\llbracket m, n \rrbracket}$. Indeed, by hypothesis $\pi$ is an orthogonal.
onal projector and is an element of $\mathbb{Z}_{m-1} \cap \mathbb{Z}_m$ which is included both in $\mathbb{Z}_{[m, -\frac{1}{4}]}$ and in $\mathbb{Z}_{[m, n]}$. Applying Lemma D.1 for both algebras lets us rewrite $\pi A_{[m, -\frac{1}{4}]}' \cap A_{[m, n]} = A'_{[m, -\frac{1}{4}]} \cap \mathrm{Im}(\hat{\pi}) \cap A_{[m, n]} = A'_{[m, -\frac{1}{4}]} \cap \pi A_{[m, n]}$. The same move will occur again later for different algebras (for example, in (93), to show that $A'_{[m, -1]} \cap \pi A_{[m, n]} = \pi A_{[0, n]}$).

We thus fix $\pi \in \mathrm{AtomProj}(\mathbb{Z}_{m-1} \cap \mathbb{Z}_m)$. Let us first prove the reverse inclusion, which is the easier one: $\pi A_{[\frac{1}{4}, n]} \subseteq A'_{[m, -\frac{1}{4}]} \cap \pi A_{[m, n]}$. On the one hand, by (92a), $A_{[\frac{1}{4}, n]} \subseteq A_{[m, n]}$ so $\pi A_{[\frac{1}{4}, n]} \subseteq \pi A_{[m, n]}$. On the other hand, by their definition, we have $A_{[\frac{1}{4}, n]} \subseteq A_{[\frac{1}{4}, p_+]} = A'_{[p_-, -\frac{1}{4}]} \subseteq A'_{[m, -\frac{1}{4}]}$, and additionally $\pi \in \mathbb{Z}_{m-1} \cap \mathbb{Z}_m \subseteq \mathbb{Z}_{[m, 0]} \subseteq \mathbb{Z}'_{[m, 0]} \subseteq \mathbb{Z}'_{[m, -\frac{1}{4}]}$, so $\pi A_{[\frac{1}{4}, n]} \subseteq \mathbb{Z}'_{[m, -\frac{1}{4}]}$. Combining the two inclusions yields $\pi A_{[\frac{1}{4}, n]} \subseteq \mathbb{Z}'_{[m, -\frac{1}{4}]} \cap \pi A_{[m, n]}$.

We now turn to the direct inclusion, and compute

$$
\begin{align}
& A'_{[m, -\frac{1}{4}]} \cap \pi A_{[m, n]} \notag \\
& \stackrel{(92a)}{=} A'_{[m, -\frac{1}{4}]} \cap A'_{[m, -1]} \cap \pi A_{[m, n]} \notag \\
& \stackrel{(9)}{=} A'_{[m, -\frac{1}{4}]} \cap \pi A_{[0, n]} \notag \\
& \stackrel{(92a)}{=} A'_{[m, -\frac{1}{4}]} \cap \pi A_{[m, p_+]} \cap \pi A_{[0, n]} \notag \\
& \stackrel{(*)}{=} \pi A_{[\frac{1}{4}, p_+]} \cap \pi A_{[0, n]} \tag{93} \\
& = \pi \left( \mu A_{[\frac{1}{4}, p_+]} \cap \mu A_{[0, n]} \right) \notag \\
& \stackrel{(88b)}{=} \pi \left( \mu (A_{[0, p_+]} \cap A'_{[q_-, -\frac{1}{4}]}) \cap \mu A_{[0, n]} \right) \notag \\
& \subseteq \pi \left( \mu A_{[0, p_+]} \cap A'_{[q_-, -\frac{1}{4}]} \cap \mu A_{[0, n]} \right) \notag \\
& \stackrel{(92a)}{=} \pi \left( A'_{[q_-, -\frac{1}{4}]} \cap \mu A_{[0, n]} \right); \notag
\end{align}
$$

we leave the justification of the starred equality for later. The fifth equality comes from the application of Lemma D.3 with $\mathcal{G} = A_{[0, p_+]}$, yielding $\mu \in \mathbb{Z}_{[0, p_+]} = (\mathbb{Z}_0 \cap \mathbb{Z}_1) \cup (\mathbb{Z}_{p_+} \cap \mathbb{Z}_{p_-})$, and the penultimate line comes from Lemma D.4.

We now distinguish two cases. Suppose $m \in [q_-, -\frac{1}{4}]$; then $d(m-1, p_+) \ge l$, so by Lemma D.6, $\mu \in \mathbb{Z}_{-1} \cap \mathbb{Z}_0 \subseteq A_{[0, n]}$; therefore, by Lemma D.1, the last line of (93) becomes

$$
\begin{align}
& \pi \left( A'_{[q_-, -\frac{1}{4}]} \cap A_{[0, n]} \cap \mathrm{Im}(\hat{\mu}) \right) \notag \\
& \subseteq \pi \left( A'_{[q_-, -\frac{1}{4}]} \cap A_{[0, n]} \right) \tag{94} \\
& \stackrel{(91d)}{=} \pi A_{[\frac{1}{4}, n]} . \notag
\end{align}
$$

The other case is $m \in [p_-, q_-, 1]$; then $d(m, 0) \ge l$, so by Lemma D.6, $\mu \in (\mathbb{Z}_{p_+} \cap \mathbb{Z}_{p_-})$. Then by applying Lemma D.5 with $\mathcal{G} = A_{[q_-, n]}$, we get a $\nu \in \mathbb{Z}_{[q_-, n]} = (\mathbb{Z}_{q_-, 1} \cap \mathbb{Z}_{q_-}) \cup (\mathbb{Z}_n \cap \mathbb{Z}_{n+1})$ such that $\pi(A'_{[q_-, -\frac{1}{4}]} \cap \mu A_{[0, n]}) = \pi \mu(A'_{[q_-, -\frac{1}{4}]} \cap \nu A_{[0, n]}) = \pi(A'_{[q_-, -\frac{1}{4}]} \cap \nu A_{[0, n]}$); yet $d(p_-, q_-) \ge l$, so $\nu \in \mathbb{Z}_n \cap \mathbb{Z}_{n+1} \subseteq A_{[0, n]}$, and thus by Lemma D.1, $\nu A_{[0, n]} = A_{[0, n]} \cap \mathrm{Im}(\hat{\nu})$. Thus the last line of (93) becomes $\pi(A'_{[q_-, -\frac{1}{4}]} \cap A_{[0, n]} \cap \mathrm{Im}(\hat{\nu}))$, which by the same reasoning as in (94) is included in $\pi A_{[\frac{1}{4}, n]}$.

In both cases, we thus find $A'_{[m, -\frac{1}{4}]} \cap \pi A_{[m, n]} \subseteq \pi A_{[\frac{1}{4}, n]}$, the direct inclusion we set out to prove.

Let us return to the justification of the starred equality in (93). Note that for the case $m = p_-$, (93) is correct as one can move from the second to the fourth equality: since $\pi \in A'_{[p_-, -\frac{1}{4}]}$, we have by Lemma D.1 $A'_{[p_-, -\frac{1}{4}]} \cap \mathrm{Im}(\hat{\pi}) = \pi A'_{[p_-, -\frac{1}{4}]} = \pi A_{[\frac{1}{4}, p_+]}$. Therefore, the above reasoning shows that $(A_{[p_-, -\frac{1}{4}]}, A_{[\frac{1}{4}, n]}) \vdash A_{[p_-, n]}$. As our assumptions are completely symmetric with respect to a swap of the left and right directions, a symmetric reasoning would show that, for any $m \in [p_-, -\frac{1}{4}]$, $(A_{[m, -\frac{1}{4}]}, A_{[\frac{1}{4}, p_+]}) \vdash A_{[m, p_+]}$. Therefore, for $\pi \in \mathbb{Z}_{m-1} \cap \mathbb{Z}_m$, applying Proposition B.4 then gives $A'_{[m, -\frac{1}{4}]} \cap A_{[m, p_+]} = \mathbb{Z}_{[m, p_+]} \cup A_{[\frac{1}{4}, p_+]} = (\mathbb{Z}_{m-1} \cap \mathbb{Z}_m) \cup A_{[\frac{1}{4}, p_+]}$. This shows that, for $\pi \in \mathbb{Z}_{m-1} \cap \mathbb{Z}_m$,

$$
\begin{align}
& A'_{[m, -\frac{1}{4}]} \cap \pi A_{[m, p_+]} \cap \pi A_{[0, n]} \notag \\
& = ((\mathbb{Z}_{m-1} \cap \mathbb{Z}_m) \cup A_{[\frac{1}{4}, p_+]}) \notag \\
& \phantom{=} \cap \pi A_{[0, n]} \notag \\
& = ((\mathbb{Z}_{m-1} \cap \mathbb{Z}_m) \cup A_{[\frac{1}{4}, p_+]}) \tag{95} \\
& \phantom{=} \cap \mathrm{Im}(\hat{\pi}) \cap \pi A_{[0, n]} \notag \\
& = \pi ((\mathbb{Z}_{m-1} \cap \mathbb{Z}_m) \cup A_{[\frac{1}{4}, p_+]}) \notag \\
& \phantom{=} \cap \pi A_{[0, n]} \notag \\
& = \pi A_{[\frac{1}{4}, p_+]} \cap \pi A_{[0, n]}, \notag
\end{align}
$$
where we used Lemma D.1 in the third equality, and Lemma B.3 in the fourth one. This is the proof of the starred equality in (93), and concludes the proof that $(A_{\llbracket m, -\frac{1}{4} \rrbracket}, A_{\llbracket \frac{1}{4}, n \rrbracket}) \vdash A_{\llbracket m, n \rrbracket}$.

As an interlude, let us use this to compute $\mathcal{Z}_{\llbracket m, -\frac{1}{4} \rrbracket}$ for a given $m \in \llbracket p-, -\frac{1}{4} \rrbracket$. We start with the case $m = q_-$. First, using (92a) then Proposition B.4, we have

$$ \begin{aligned} \mathcal{Z}_{\llbracket q-, -\frac{1}{4} \rrbracket} &\subseteq A'_{\llbracket q-, -1 \rrbracket} \cap A_{\llbracket q-, 0 \rrbracket} \\ &= \mathcal{Z}_{\llbracket q-, 0 \rrbracket} \vee A_0 \\ &= (\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee A_0. \end{aligned} \quad (96) $$

Furthermore, $\mathcal{Z}_{\llbracket q-, -\frac{1}{4} \rrbracket} \subseteq A_{\llbracket q-, -\frac{1}{4} \rrbracket} \subseteq A'_{\llbracket \frac{1}{4}, q_+ \rrbracket}$ so

$$ \begin{aligned} \mathcal{Z}_{\llbracket q-, -\frac{1}{4} \rrbracket} &\subseteq ((\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee A_0) \cap A'_{\llbracket \frac{1}{4}, q_+ \rrbracket} \\ &= (\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee (A_0 \cap A'_{\llbracket \frac{1}{4}, q_+ \rrbracket}) \quad (97) \\ &\stackrel{(91c)}{=} (\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee A_{-\frac{1}{4}}, \end{aligned} $$

where we used Lemma D.8 with $\mathcal{G} = A_{\llbracket 0, q_+ \rrbracket}$, since $d(\{q_-, 1, q-\}, \{0, q_+\}) \ge l$. By symmetry, we also have $\mathcal{Z}_{\llbracket \frac{1}{4}, q_+ \rrbracket} \subseteq A_{\frac{1}{4}} \vee (\mathcal{Z}_{q_+} \cap \mathcal{Z}_{q_+1})$.

Using Proposition B.5 applied to $(A_{\llbracket q-, -\frac{1}{4} \rrbracket}, A_{\llbracket \frac{1}{4}, q_+ \rrbracket}) \vdash A_{\llbracket q-, q_+ \rrbracket}$, we find

$$ \begin{aligned} \mathcal{Z}_{\llbracket q-, -\frac{1}{4} \rrbracket} &\subseteq \mathcal{Z}_{\llbracket q-, q_+ \rrbracket} \vee \mathcal{Z}_{\llbracket \frac{1}{4}, q_+ \rrbracket} \\ &\subseteq \mathcal{Z}_{\llbracket q-, q_+ \rrbracket} \vee (\mathcal{Z}_{q_+} \cap \mathcal{Z}_{q_+1}) \vee A_{\frac{1}{4}} \quad (98) \\ &= (\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee (\mathcal{Z}_{q_+} \cap \mathcal{Z}_{q_+1}) \vee A_{\frac{1}{4}}. \end{aligned} $$

Combining (97) and (98), we find

$$ \begin{aligned} \mathcal{Z}_{\llbracket q-, -\frac{1}{4} \rrbracket} &\subseteq ((\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee A_{-\frac{1}{4}}) \\ &\cap ((\mathcal{Z}_{q_+} \cap \mathcal{Z}_{q_+1}) \vee (\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \\ &\qquad \vee A_{\frac{1}{4}}) \\ &= ((\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee A_{-\frac{1}{4}}) \\ &\cap ((\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee A_{\frac{1}{4}}) \\ &= (\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee (A_{-\frac{1}{4}} \cap A_{\frac{1}{4}}) \\ &= (\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}), \end{aligned} \quad (99) $$

where we first used Lemma D.7 with $\mathcal{G} = A_{\llbracket q-, 0 \rrbracket}$, since $d(\{q_+, q_+ + 1\}, \{q-, 0\}) \ge l$; then Lemma D.7

with $\mathcal{G} = A_0$, since $d(\{q-, 1, q-\}, 0) \ge l$. The last equality comes from the fact that $A_{-\frac{1}{4}} \cap A_{\frac{1}{4}} = \mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}$, since the two sets being intersected commute by (92b).

Since the reverse inclusion is direct, we get $\mathcal{Z}_{\llbracket q-, -\frac{1}{4} \rrbracket} = (\mathcal{Z}_{q-1} \cap \mathcal{Z}_{q-}) \vee (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}})$, and symmetrically, $\mathcal{Z}_{\llbracket \frac{1}{4}, q_+ \rrbracket} = (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}) \vee (\mathcal{Z}_{q_+} \cap \mathcal{Z}_{q_+1})$.

Let us now turn to the general case, taking $m \in \llbracket p-, -\frac{1}{4} \rrbracket$. By Proposition B.5 and the fact that $(A_{\llbracket m, -\frac{1}{4} \rrbracket}, A_{\llbracket \frac{1}{4}, q_+ \rrbracket}) \vdash A_{\llbracket m, q_+ \rrbracket}$, we have $\mathcal{Z}_{\llbracket m, -\frac{1}{4} \rrbracket} \subseteq \mathcal{Z}_{\llbracket \frac{1}{4}, q_+ \rrbracket} \vee \mathcal{Z}_{\llbracket m, q_+ \rrbracket} = (\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \vee (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}) \vee (\mathcal{Z}_{q_+} \cap \mathcal{Z}_{q_+1})$. Since $\mathcal{Z}_{\llbracket m, -\frac{1}{4} \rrbracket} \subseteq A_{\llbracket m, 0 \rrbracket}$, we find furthermore

$$ \begin{aligned} \mathcal{Z}_{\llbracket m, -\frac{1}{4} \rrbracket} &\subseteq ((\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \vee (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}) \\ &\qquad \vee (\mathcal{Z}_{q_+} \cap \mathcal{Z}_{q_+1})) \cap A_{\llbracket m, 0 \rrbracket} \\ &= ((\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \vee (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}})) \cap A_{\llbracket m, 0 \rrbracket} \\ &\stackrel{(92a)}{=} (\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \vee (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}), \end{aligned} \quad (100) $$

where in the middle equality we used Lemma D.7 with $\mathcal{G} = A_{\llbracket m, 0 \rrbracket}$, since $d(\{q_+, q_+ + 1\}, \{m, 0\}) \ge l$. With the reverse inclusion being direct, we can conclude that

$$ \mathcal{Z}_{\llbracket m, -\frac{1}{4} \rrbracket} = (\mathcal{Z}_{m-1} \cap \mathcal{Z}_m) \vee (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}), \quad (101a) $$

$$ \mathcal{Z}_{\llbracket \frac{1}{4}, n \rrbracket} = (\mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}) \vee (\mathcal{Z}_n \cap \mathcal{Z}_{n+1}), \quad (101b) $$

where we added the symmetric case for $n \in \llbracket \frac{1}{4}, p_+ \rrbracket$.

This proves that the refined partition is a strongly connected one. It will also be useful for the next step in our proof, to which we now turn: showing that for any given $m \in \llbracket p-, -1 \rrbracket$ and $n \in \llbracket m, -\frac{1}{4} \rrbracket$, $(A_{\llbracket m, n-1 \rrbracket}, A_{\llbracket n, -\frac{1}{4} \rrbracket}) \vdash A_{\llbracket m, -\frac{1}{4} \rrbracket}$. Here again we will make use of Proposition B.7, with $\tilde{Z}_1 = \mathcal{Z}_{m-1} \cap \mathcal{Z}_m$ and $\tilde{Z}_2 = \mathcal{Z}_{-\frac{1}{4}} \cap \mathcal{Z}_{\frac{1}{4}}$. (101a) indeed yields that $\mathcal{Z}_{\llbracket m, -\frac{1}{4} \rrbracket} = \tilde{Z}_1 \vee \tilde{Z}_2$ and $\tilde{Z}_2 \subseteq \mathcal{Z}_{\llbracket n, -\frac{1}{4} \rrbracket}$, while $\tilde{Z}_1 \subseteq \mathcal{Z}_{\llbracket m, n-1 \rrbracket}$ comes from the strong connectedness of the initial partition.

The proposition shows that it is sufficient for us to prove $A'_{\llbracket m, n-1 \rrbracket} \cap \pi A_{\llbracket m, -\frac{1}{4} \rrbracket} = \pi A_{\llbracket n, -\frac{1}{4} \rrbracket}$ for $\pi \in \mathcal{Z}_{m-1} \cap \mathcal{Z}_m$. Since the reverse inclusion is easy to prove, we focus on the direct one, and compute
$$
\begin{align}
& \mathcal{A}'_{[[m,n-1]]} \cap \pi \mathcal{A}_{[[m,-\frac{1}{4}]}} \nonumber \\
& \qquad \subseteq \mathcal{A}'_{[[m,n-1]]} \cap \pi \mathcal{A}_{[[m,-\frac{1}{4}]}} \cap \mathcal{A}'_{[[\frac{1}{4},q_+]]} \nonumber \\
& \qquad \subseteq \mathcal{A}'_{[[m,n-1]]} \cap \pi \mathcal{A}_{[[m,0]]} \cap \mathcal{A}'_{[[\frac{1}{4},q_+]]} \tag{92a} \\
& \qquad \stackrel{(59)}{=} \pi \mathcal{A}_{[[n,0]]} \cap \mathcal{A}'_{[[\frac{1}{4},q_+]]} \nonumber \\
&= \pi(\mu \mathcal{A}_{[[n,0]]} \cap \mathcal{A}'_{[[\frac{1}{4},q_+]]}) \nonumber \\
&= \pi(\mathcal{A}_{[[n,0]]} \cap \operatorname{Im}(\hat{\mu}) \cap \mathcal{A}'_{[[\frac{1}{4},q_+]]}) \nonumber \\
&\subseteq \pi(\mathcal{A}_{[[n,0]]} \cap \mathcal{A}'_{[[\frac{1}{4},q_+]]}) \nonumber \\
&= \pi \mathcal{A}_{[[n,-\frac{1}{4}]]}, \nonumber
\end{align}
$$

where we made use of Lemma D.5 with $\mathcal{G} = \mathcal{A}_{[[n,q_+]]}$, and then took advantage of the fact that, since $d(m-1, q_+) \ge l$, Lemma D.6 yields $\mu \in \mathcal{Z}_{n-1} \cap \mathcal{Z}_n \subseteq \mathcal{A}_{[[n,0]]}$, so we can use Lemma D.1.

By symmetry, we also have, for all $m \in [[1, p_+]]$ and $n \in [[\frac{1}{4}, m]]$, $(\mathcal{A}_{[[\frac{1}{4}, n]]}, \mathcal{A}_{[[n+1, m]]}) \vdash \mathcal{A}_{[[\frac{1}{4}, m]]}$.

We now turn to the cases involving 'big' intervals, those of size larger than $\frac{N}{2}$, which were defined as commutants of the small intervals. Let us start by proving that, for $m \in [[2, p_+]]$, $n \in [[\frac{1}{4}, m-2]]$, $(\mathcal{A}_{[[m, -\frac{1}{4}]}, \mathcal{A}_{[[\frac{1}{4}, n]]}) \vdash \mathcal{A}_{[[m, n]]}$. Here again, using Proposition B.7, it is sufficient to prove that for any $\pi \in \mathcal{Z}_{m-1} \cap \mathcal{Z}_m$, $\mathcal{A}'_{[[m, -\frac{1}{4}]]} \cap \pi \mathcal{A}_{[[m, n]]} = \pi \mathcal{A}_{[[\frac{1}{4}, n]]}$. Since the reverse inclusion is easy to prove, we focus on the direct one. It is given by the computation

$$
\begin{align}
& \mathcal{A}'_{[[m,-\frac{1}{4}]]} \cap \pi \mathcal{A}_{[[m,n]]} \nonumber \\
&= \mathcal{A}_{[[\frac{1}{4},m-1]]} \cap \pi \mathcal{A}_{[[m,n]]} \nonumber \\
&= \mathcal{A}_{[[\frac{1}{4},p_+]]} \cap \mathcal{A}_{[[0,m-1]]} \cap \pi \mathcal{A}_{[[m,n]]} \nonumber \\
&= \mathcal{A}_{[[\frac{1}{4},p_+]]} \cap \mathcal{A}'_{[[m,-1]]} \cap \pi \mathcal{A}_{[[m,n]]} \nonumber \\
&= \mathcal{A}_{[[\frac{1}{4},p_+]]} \cap \pi \mathcal{A}_{[[0,n]]} \nonumber \\
&\subseteq \mathcal{A}'_{[[q_{-},-\frac{1}{4}]]} \cap \pi \mathcal{A}_{[[0,n]]} \tag{103} \\
&= \pi \left( \mathcal{A}'_{[[q_{-},-\frac{1}{4}]]} \cap \mu \mathcal{A}_{[[0,n]]} \right) \nonumber \\
&= \pi \left( \mathcal{A}'_{[[q_{-},-\frac{1}{4}]]} \cap \mathcal{A}_{[[0,n]]} \cap \operatorname{Im}(\hat{\mu}) \right) \nonumber \\
&\subseteq \pi \left( \mathcal{A}'_{[[q_{-},-\frac{1}{4}]]} \cap \mathcal{A}_{[[0,n]]} \right) \nonumber \\
&= \pi \mathcal{A}_{[[\frac{1}{4},n]]}, \nonumber
\end{align}
$$

where we used Lemma D.5 with $\mathcal{G} = \mathcal{A}_{[[q_{-},n]]}$, then Lemma D.1, since Lemma D.6 proves that $\mu \in \mathcal{Z}_n \cap \mathcal{Z}_{n+1} \subseteq \mathcal{A}_{[[0,n]]}$ as $d(m-1, q_{-}) \ge l$.

Finally, we have to prove that, for $m \in [[1, p_+]]$ and $n \in [[m+1, -\frac{1}{4}]]$, $(\mathcal{A}_{[[m, n-1]]}, \mathcal{A}_{[[n, -\frac{1}{4}]}) \vdash \mathcal{A}_{[[m, -\frac{1}{4}]]}$. Like before, we only have to prove that for $\pi \in \mathcal{Z}_{m-1} \cap \mathcal{Z}_m$, $\mathcal{A}'_{[[m, n-1]]} \cap \pi \mathcal{A}_{[[m, -\frac{1}{4}]}} \subseteq \pi \mathcal{A}_{[[n, -\frac{1}{4}]]}$. This is given simply by

$$
\begin{align}
& \mathcal{A}'_{[[m,n-1]]} \cap \pi \mathcal{A}_{[[m,-\frac{1}{4}]]} \nonumber \\
&= \pi \mathcal{A}'_{[[m,n-1]]} \cap \mathcal{A}_{[[m,-\frac{1}{4}]}} \nonumber \\
&= \pi \mathcal{A}_{[[n,m-1]]} \cap \mathcal{A}'_{[[\frac{1}{4},m-1]]} \tag{104} \\
&= \pi \mathcal{A}_{[[n,-\frac{1}{4}]}}, \nonumber
\end{align}
$$

where we used Lemma D.1, and where the last line comes from (102) or (103), depending on whether $n \in [[p_{-}, -\frac{1}{4}}]]$ or $n \in [[\frac{1}{4}, p_{+}}]]$. The symmetric case gives an analogous result.

This concludes the proof that $(\mathcal{A}_S, \mathcal{A}_T) \vdash \mathcal{A}_{S \sqcup T}$ when $S, T$, and $S \sqcup T$ are all intervals. Let us now turn to the case in which only $S \sqcup T$ is an interval. We write it as $[[m_0, n]]$;<fnref n="19" /> then, taking without loss of generality $m_0 \in S$ and writing $[[m_0, m_1]]$ as the connected component of $S$ containing $m_0$, we have $S = [[m_0, m_1]] \sqcup \tilde{S}$ such that $\tilde{S} \sqcup T = [[m_1 + 1, n]]$. Then, if we suppose that $(\mathcal{A}_{\tilde{S}}, \mathcal{A}_T) \vdash \mathcal{A}_{[[m_1+1, n]]}$, we get, for $\pi \in \text{AtomProj}(\mathcal{Z}_{[[m_0, n]]})$,

$$
\begin{align}
& \mathcal{A}'_S \cap \pi \mathcal{A}_{[[m_0, n]]} \nonumber \\
&= (\mathcal{A}_{\tilde{S}} \vee \mathcal{A}_{[[m_0, m_1]]})' \cap \pi \mathcal{A}_{[[m_0, n]]} \nonumber \\
&= \mathcal{A}'_{\tilde{S}} \cap \mathcal{A}'_{[[m_0, m_1]]} \cap \pi \mathcal{A}_{[[m_0, n]]} \nonumber \\
&= \mathcal{A}'_{\tilde{S}} \cap \pi \mathcal{A}_{[[m_1+1, n]]} \tag{105} \\
&= \pi (\mathcal{A}'_{\tilde{S}} \cap \mu \mathcal{A}_{[[m_1+1, n]]}) \nonumber \\
&= \pi \mu \mathcal{A}_T \nonumber \\
&= \pi \mathcal{A}_T, \nonumber
\end{align}
$$

where we used the fact that $(\mathcal{A}_{[[m_0, m_1]]}, \mathcal{A}_{[[m_1+1, n]]}) \vdash \mathcal{A}_{[[m_0, n]]}$, then Lemma D.5 with $\mathcal{G} = \mathcal{A}_{[[m_1+1, n]]}$, then our assumption that $(\mathcal{A}_{\tilde{S}}, \mathcal{A}_T) \vdash \mathcal{A}_{[[m_1+1, n]]}$. This gives an induction on the size of $S \sqcup T$, allowing us to

<footnote n="19">¹⁹The case in which $S \sqcup T$ is the whole graph works in the same way: under an analogous induction hypothesis, we then directly have (105) by taking $[[m_0, m_1]]$ to be an arbitrary connected component of $S$.</footnote>
conclude that $(\mathcal{A}_S, \mathcal{A}_T) \vdash \mathcal{A}_{S \cup T}$ always holds when $S \cup T$ is an interval.

Finally, let us consider the general case. We can then decompose $S \cup T = \bigsqcup_i (S_i \cup T_i)$ where the $S_i \cup T_i$'s are non-contiguous intervals, and therefore $\mathcal{A}_{S \cup T} = \bigvee_i \mathcal{A}_{S_i \cup T_i}$. This also implies $\mathcal{Z}_{S \cup T} = \bigvee_i \mathcal{Z}_{S_i \cup T_i}$, so, by Lemma B.2, taking an atomic projector $\pi$ of $\mathcal{A}_{S \cup T}$, it is a product $\prod_i \pi_i$ of atomic projectors of each of the $\mathcal{Z}_{S_i \cup T_i}$'s. We then compute

$$ \begin{aligned} \mathcal{A}'_S \cap \mathcal{A}_{S \cup T} &= \left( \bigvee_i \mathcal{A}_{S_i} \right)' \cap \pi \left( \bigvee_i \mathcal{A}_{S_i \cup T_i} \right) \\ &= \left( \bigcap_i \mathcal{A}'_{S_i} \right) \cap \left( \prod_i \pi_i \right) \left( \bigvee_i \mathcal{A}_{S_i \cup T_i} \right) \\ &= \left( \bigcap_i \mathcal{A}'_{S_i} \right) \cap \left( \bigvee_i \pi_i \mathcal{A}_{S_i \cup T_i} \right); \end{aligned} \quad (106) $$

the third line is justified by the fact that an element of $(\prod_i \pi_i)(\bigvee_i \mathcal{A}_{S_i \cup T_i})$ can be written as $(\prod_i \pi_i)(\sum_t \prod_i f_t^t) = \sum_t \prod_i (\pi_i f_t^t)$, so $(\prod_i \pi_i)(\bigvee_i \mathcal{A}_{S_i \cup T_i}) = \bigvee_i \pi_i \mathcal{A}_{S_i \cup T_i}$.

Since the $\pi_i$'s are by assumption atomic projectors of each of the $\mathcal{A}_{S_i \cup T_i}$'s, each of the $\pi_i \mathcal{A}_{S_i \cup T_i}$'s is a factor. Thus, the homomorphism of $C^*$-algebras from $\otimes_i \pi_i \mathcal{A}_{S_i \cup T_i}$ to $\omega$ given by the product of elements is on a factor domain; so by Proposition B.1, being non-null, it is injective.

Writing an element of (107) as $f = \sum_t f_0^t \tilde{f}^t$, where the $\tilde{f}^t$'s form a basis of $\bigvee_{i \neq 0} \pi_i \mathcal{A}_{S_i \cup T_i}$ and the $f_0^t$'s are in $\mathcal{A}_{S_0 \cup T_0}$, we have for any $g \in \mathcal{A}_{S_0}$: $0 = [g, f] = \sum_t [g, f_0^t] \tilde{f}^t$, which by injectiveness of the product homomorphism yields $\forall t$, $[g, f_0^t] = 0$, so (107)'s RHS is equal to $(\mathcal{A}'_{S_0} \cap \pi_0 \mathcal{A}_{S_0 \cup T_0}) \vee ((\bigcap_{i \neq 0} \mathcal{A}'_{S_i}) \cap (\bigvee_{i \neq 0} \pi_i \mathcal{A}_{S_i \cup T_i}))$. Iterating the procedure for every $i$ yields $\bigvee_i (\mathcal{A}'_{S_i} \cap \pi_i \mathcal{A}_{S_i \cup T_i})$, so we can finish the computation:

$$ \begin{aligned} \mathcal{A}'_S \cap \mathcal{A}_{S \cup T} &= \bigvee_i (\mathcal{A}'_{S_i} \cap \pi_i \mathcal{A}_{S_i \cup T_i}) \\ &= \bigvee_i \pi_i \mathcal{A}_{T_i} \\ &= \left( \prod_i \pi_i \right) \left( \bigvee_i \mathcal{A}_{T_i} \right) \\ &= \pi \mathcal{A} \bigsqcup_i T_i \\ &= \pi \mathcal{A}_T, \end{aligned} \quad (107) $$

where we used the fact that $(\mathcal{A}_{S_i}, \mathcal{A}_{T_i}) \vdash \mathcal{A}_{S_i \cup T_i}$ since $S_i \cup T_i$ is an interval, then the fact that $\mathcal{A}_{\bigsqcup_i T_i} = \bigvee_i \mathcal{A}_{T_i}$ since the $T_i$'s are non-contiguous. This concludes the proof. $\square$

## D.3 Fine-graining

We can leverage Lemma D.9 to obtain the fine-graining theorem that will serve as the induction step for a proof of Theorem 3.1.<fnref n="20" />

**Theorem D.1 (Fine-graining).** Let $N$ be a natural number, and $r$ either a natural number or half a natural number. In the first case, we take $\Gamma = \Gamma_N$, and in the other, $\Gamma = \Gamma_N^{+1/2}$. Suppose $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ and $(\mathcal{B}_S)_{S \subseteq \Gamma}$ are two strongly connected 1D partitions of a factor $\omega$, of correlation length at most $l_A$ and $l_B$ respectively, satisfying

$$ \forall S \subseteq \Gamma_N, \quad \mathcal{A}_S \subseteq \mathcal{B}_{S \pm r}. \quad (108) $$

If $N \ge 4r + 2(l_A + l_B) + 1$, then there exists a strongly connected 1D partition $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}} \vdash \omega$, with correlation length at most $l_A + \frac{1}{2}$, satisfying

$$ \forall S \subseteq \Gamma_N, \quad \mathcal{A}_S = \tilde{\mathcal{A}}_{S \pm \frac{1}{4}}; \quad (109a) $$

$$ \forall S \subseteq \Gamma_N^{\text{fine}}, \quad \tilde{\mathcal{A}}_S \subseteq \mathcal{B}_{S \pm (r - \frac{3}{4})}. \quad (109b) $$

**Proof.** The idea is to build $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$ from $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ by repeatedly using Lemma D.9 to split each of $\Gamma$'s vertices in two, one by one.<fnref n="21" /> As a first step, for instance, the split of the vertex 0 is obtained by taking

$$ \mathcal{A}_{\llbracket p, -\frac{1}{4} \rrbracket} := \mathcal{A}_{\llbracket p, 0 \rrbracket} \cap \mathcal{B}'_{\llbracket r, r+l_B \rrbracket}, \quad (110) $$

<footnote n="20">²⁰In the theorem's statement, we added the possibility for the $\mathcal{B}$-partition not to be a factorisation for the sake of generality; but for the proof of our main result, one only needs the case in which it is a factorisation, i.e. the case $l_B = 0$.</footnote>

<footnote n="21">²¹Note that, strictly speaking, Lemma D.9 was written as starting with a partition on $\Gamma_N$ (or on the isomorphic $\Gamma_N^{+1/2}$). However, its result also holds when applied to a partition in which some vertices have been split already, i.e. a partition on a graph intermediary between $\Gamma_N$ and $\Gamma_N^{\text{fine}}$, in which some of the $n \in \Gamma_N$ have been replaced with two vertices $n \pm \frac{1}{4}$, and some have not. We deliberately refrained from writing it in this way, as defining and manipulating such partly-split graphs would have brought to the proof much formal burden at the detriment of the clarity of the main arguments.</footnote>
and more generally, for the step in which the vertex $n$ is split, we similarly take

$$ \mathcal{A}_{\llbracket n+p_{-}, n-\frac{1}{4} \rrbracket} := \mathcal{A}_{\llbracket n+p_{-}, n \rrbracket} \cap \mathcal{B}'_{\llbracket n+r, n+r+l_B \rrbracket}. \quad (111) $$

Let us focus on the first step, as every step is completely analogous. We first prove that the assumptions of Lemma D.9 are then satisfied. We start with the assumption that $N \ge 4l_A + 1$. For any $S \subseteq \Gamma$, we have $\mathcal{A}_S \subseteq \mathcal{B}_{S \pm r} \perp \mathcal{B}_{S \pm (r+l_B)} \supset \mathcal{B}_{S \pm (2r+l_B) \pm r} \supset \mathcal{A}_{S \pm (2r+l_B)}$; thus $l_A \le l_B + 2r$. Together with our assumption that $N \ge 4r + 2(l_A + l_B) + 1$, this yields $N \ge 4l_A + 1$.

Turning to (86a), we have by definition $\mathcal{A}_{\llbracket p_{-}, -\frac{1}{4} \rrbracket} \subseteq \mathcal{A}_{\llbracket p_{-}, 0 \rrbracket}$. Furthermore, $\mathcal{A}_{\llbracket p_{-}, -1 \rrbracket} \subseteq \mathcal{B}_{\llbracket p_{+}-(r-1), r-1 \rrbracket} \subseteq \mathcal{B}_{\llbracket r+l_B+1, r-1 \rrbracket} = \mathcal{B}'_{\llbracket r, l_B \rrbracket}$ (the second inclusion comes from the fact that $p_{+} = \lfloor \frac{N-1}{2} \rfloor \ge 2r+(l_A+l_B)+1$ and thus $p_{+}-(r-1) \ge r+l_B+1$). Therefore $\mathcal{A}_{\llbracket p_{-}, -1 \rrbracket} \subseteq \mathcal{A}_{\llbracket p_{-}, -\frac{1}{4} \rrbracket}$.

As for (86b), it is proven by the fact that $\mathcal{B}_{\llbracket r, r+l_B \rrbracket} = \mathcal{B}'_{\llbracket r+l_B+1, r-1 \rrbracket} \subseteq \mathcal{A}'_{\llbracket 2r+l_B+1, -1 \rrbracket} = \mathcal{A}_{\llbracket 0, 2r+l_B \rrbracket} \subseteq \mathcal{A}_{\llbracket 0, q_+ \rrbracket}$ (the last inclusion comes once again from $p_{+} \ge 2r+(l_A+l_B)+1$, from which we derive $q_{+} := p_{+} - l_A \ge 2r+l_B$). As by definition we also have $\mathcal{B}_{\llbracket r, r+l_B \rrbracket} \subseteq \mathcal{A}'_{\llbracket p_{-}, -\frac{1}{4} \rrbracket} := \mathcal{A}_{\llbracket \frac{1}{4}, p_{+} \rrbracket}$, we find that $\mathcal{B}_{\llbracket r, r+l_B \rrbracket} \subseteq \mathcal{A}_{\llbracket 0, q_+ \rrbracket} \cap \mathcal{A}_{\llbracket \frac{1}{4}, p_{+} \rrbracket}$ and thus

$$ \begin{aligned} & \mathcal{A}_{\llbracket p_{-}, 0 \rrbracket} \cap (\mathcal{A}_{\llbracket \frac{1}{4}, p_{+} \rrbracket} \cap \mathcal{A}_{\llbracket 0, q_{+} \rrbracket})' \\ & \subseteq \mathcal{A}_{\llbracket p_{-}, 0 \rrbracket} \cap \mathcal{B}'_{\llbracket r, r+l_B \rrbracket} \quad (112) \\ & = \mathcal{A}_{\llbracket p_{-}, -\frac{1}{4} \rrbracket}, \end{aligned} $$

and the reverse inclusion is direct.

We can thus use Lemma D.9 with the specification (110) to get a new partition $\mathcal{A}^{\text{split}}$ over the graph in which vertex 0 has been replaced with two vertices $\pm \frac{1}{4}$. Like before, we will denote $\mathcal{A}_S^{\text{split}}$ as $\mathcal{A}_S$, since the presence of $\pm \frac{1}{4}$ in $S$ is sufficient to disambiguate one from the other. First, (91a) and (91b) tell us that for any $m$, $\mathcal{A}_{\llbracket m, -\frac{1}{4} \rrbracket} \subseteq \mathcal{A}_{\llbracket m, 0 \rrbracket}$ and $\mathcal{A}_{\llbracket \frac{1}{4}, m \rrbracket} \subseteq \mathcal{A}_{\llbracket 0, m \rrbracket}$; after splitting every vertex, this yields (109a). To eventually obtain (109b), we need to prove, for any $m$, the following inclusions:

$$ \mathcal{A}_{\llbracket m, -\frac{1}{4} \rrbracket} \subseteq \mathcal{B}_{\llbracket m-r, r-1 \rrbracket}; \quad (113a) $$

$$ \mathcal{A}_{\llbracket \frac{1}{4}, m \rrbracket} \subseteq \mathcal{B}_{\llbracket r-1, m+r \rrbracket}. \quad (113b) $$

We start with (113a), for an $m \in \llbracket p_{-}, -\frac{1}{4} \rrbracket$. Then from (108) and (110), we get$^{22}$

$$ \begin{aligned} \mathcal{A}_{\llbracket m, -\frac{1}{4} \rrbracket} &\subseteq \mathcal{B}_{\llbracket m-r, r \rrbracket} \cap \mathcal{B}'_{\llbracket r, r+l_B \rrbracket} \\ &= \mathcal{B}_{\llbracket m-r, r \rrbracket} \cap \mathcal{B}_{\llbracket m-r, r+l_B \rrbracket} \cap \mathcal{B}'_{\llbracket r, r+l_B \rrbracket} \\ &= \mathcal{B}_{\llbracket m-r, r \rrbracket} \\ &\cap \left( (\mathcal{Z}_{r+l_B}^B \cap \mathcal{Z}_{r+l_B+1}^B) \cup \mathcal{B}_{\llbracket m-r, r-1 \rrbracket} \right) \quad (114) \\ &= \mathcal{B}_{\llbracket m-r, r \rrbracket} \cap \mathcal{B}_{\llbracket m-r, r-1 \rrbracket} \\ &= \mathcal{B}_{\llbracket m-r, r-1 \rrbracket}, \end{aligned} $$

where the penultimate equality is given by Lemma D.7 with $\mathcal{G} = \mathcal{B}_{\llbracket m-r, r \rrbracket}$, using $d(\{r+l_B, r+l_B+1\}, \{m-r, r\}) \ge l_B$. For the proof of (113b) for $m \in \llbracket \frac{1}{4}, p_{+} \rrbracket$, note that $\mathcal{B}_{\llbracket -(r+l_B), -r \rrbracket} \subseteq \mathcal{A}_{\llbracket p_{-}, 0 \rrbracket} \cap \mathcal{B}'_{\llbracket r, r+l_B \rrbracket} = \mathcal{A}_{\llbracket p_{-}, -\frac{1}{4} \rrbracket}$. Thus a computation completely symmetric to the previous one yields the result.

Finally, we have the case of large intervals. First, let us prove (113a) for $m \in \llbracket 1, p_{+} \rrbracket$. The previous results show that $\mathcal{A}_{\llbracket p_{-}, -\frac{1}{4} \rrbracket} \subseteq \mathcal{B}_{\llbracket p_{-}, r, r-1 \rrbracket} \subseteq \mathcal{B}_{\llbracket m-r, r-1 \rrbracket}$, so dually, $\mathcal{B}_{\llbracket r, m-r-1 \rrbracket} \subseteq \mathcal{A}_{\llbracket \frac{1}{4}, p_{+} \rrbracket}$. Since we also have $\mathcal{B}_{\llbracket r, m-r-1 \rrbracket} \subseteq \mathcal{A}_{\llbracket 0, m-1 \rrbracket}$, we conclude that $\mathcal{B}_{\llbracket r, m-r-1 \rrbracket} \subseteq \mathcal{A}_{\llbracket \frac{1}{4}, m-1 \rrbracket}$, which is the dual of (113a). The proof of (113b) for $m \in \llbracket p_{-}, -1 \rrbracket$ is symmetric.

After splitting every node in the same way, we obtain a partition $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$ which can be specified in the following way: for $m, n \in \Gamma_N$ satisfying $m \in \llbracket n+p_{-}, n-1 \rrbracket$,

$$ \tilde{\mathcal{A}}_{\llbracket m-\frac{1}{4}, n+\frac{1}{4} \rrbracket} := \mathcal{A}_{\llbracket m, n \rrbracket}; \quad (115a) $$

$$ \tilde{\mathcal{A}}_{\llbracket m-\frac{1}{4}, n-\frac{1}{4} \rrbracket} := \mathcal{A}_{\llbracket m, n \rrbracket} \cap \mathcal{B}'_{\llbracket m+r, m+r+l_B \rrbracket}; \quad (115b) $$

$$ \tilde{\mathcal{A}}_{\llbracket m+\frac{1}{4}, n+\frac{1}{4} \rrbracket} := \mathcal{A}_{\llbracket m, n \rrbracket} \cap \left( \mathcal{A}_{\llbracket m+p_{-}, m \rrbracket} \cap \mathcal{B}'_{\llbracket m+r, m+r+l_B \rrbracket} \right)'; \quad (115c) $$

$$ \begin{aligned} \tilde{\mathcal{A}}_{\llbracket m+\frac{1}{4}, n-\frac{1}{4} \rrbracket} &:= \mathcal{A}_{\llbracket m, n \rrbracket} \cap \mathcal{B}'_{\llbracket n+r, n+r+l_B \rrbracket} \quad (115d) \\ &\cap \left( \mathcal{A}_{\llbracket m+p_{-}, m \rrbracket} \cap \mathcal{B}'_{\llbracket m+r, m+r+l_B \rrbracket} \right)', \end{aligned} $$

while the specification for larger intervals is given by taking commutants, and that for non-intervals is given by (67).

<footnote>$^{22}$We denote $\mathcal{Z}_S^B$ for the centre of $\mathcal{B}_S$, to clarify that we are talking about centres of the $\mathcal{B}$ algebras and not the $\mathcal{A}$ ones.</footnote>
Finally, let us check that the partition $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}^-$ obtained after splitting every node has correlation length at most $l_A + \frac{1}{2}$. This comes from the fact that for every $S \subseteq \Gamma_N^{\text{fine}}$, we have $\tilde{\mathcal{A}}_S \subseteq \mathcal{A}_{S \pm \frac{1}{4}} \perp \mathcal{A}_{S \pm (\frac{1}{4} + l_A)} \supseteq \mathcal{A}_{S \pm (\frac{1}{2} + l_A)} \pm \frac{1}{4} \supseteq \tilde{\mathcal{A}}_{S \pm (\frac{1}{2} + l_A)}$. Note that in the first inclusion there, we used (109a) in the other direction; we can do this because of the invertibility of QCAs with a given causality radius, proven in Proposition 2.1. $\square$

## D.4 Envoi

Finally, we can prove Theorem 3.1, in a slightly generalised form, in which the QCA does not necessarily have to be between factorisations. Besides the added generality, this makes the proof by induction smoother.

**Theorem D.2.** Let $\mathcal{U}$ be a (generalised) 1D QCA with causality radius $r$, from a strongly connected partition of a factor, with correlation length at most $l$, to one with correlation length at most $l'$, over $\Gamma_N$ or $\Gamma_N^{+1/2}$.

If $N > 4r + 2(l + l')$, then there exist $2r$ (generalised) 1D QCAs $\mathcal{U}_1, \dots, \mathcal{U}_{2r}$, with causality radius $\frac{1}{2}$, such that

$$ \mathcal{U} = \mathcal{U}_{2r} \dots \mathcal{U}_1. \quad (116) $$

Furthermore, each of the $\mathcal{U}_i$'s decomposes as

$$ \mathcal{U}_i = \mathcal{U}_i^{\text{coarse}} \mathcal{U}_i^{\text{fine}}, \quad (117) $$

where $\mathcal{U}_i^{\text{fine}}$ is a fine-graining and $\mathcal{U}_i^{\text{coarse}}$ is a coarse-graining.

**Proof.** The proof is by induction. Suppose that the theorem is true up to a certain $r - \frac{1}{2}$, and take $\mathcal{U}$ to be of radius $r$ (or, for the base case, suppose nothing and take $r = \frac{1}{2}$). We write $\mathcal{U}$'s input partition as $(\mathcal{A}_S)_{S \subseteq \Gamma_N} \vdash \omega$, and its output partition as $(\mathcal{B}_S)_{S \subseteq \Gamma}$ (with $\Gamma$ being $\Gamma_N$ or $\Gamma_N^{+1/2}$ depending on whether $r$ is an integer or half an integer). Then, as $\mathcal{U}$ is a C*-algebra isomorphism and thus preserves inclusions and commutants, (19) is equivalent to (108) with $\mathcal{B}_S$'s replaced with $\mathcal{U}^{-1}(\mathcal{B}_S)$'s. We can thus apply Theorem D.1, which yields a strongly connected partition $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}} \vdash \omega$ with correlation length at most $l + \frac{1}{2}$, satisfying (109a) and (109b).

Let us then define $(\tilde{\mathcal{A}}_S^{\text{coarse}})_{S \subseteq \Gamma_N^{+1/2}} \vdash \omega$ to be the partition over $\Gamma_N^{+1/2}$ obtained by coarse-graining $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$ through

$$ \forall S \subseteq \Gamma_N^{+1/2}, \quad \tilde{\mathcal{A}}_S^{\text{coarse}} := \tilde{\mathcal{A}}_{S \pm \frac{1}{4}}. \quad (118) $$

From the properties of $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$, it is then clear that $(\tilde{\mathcal{A}}_S^{\text{coarse}})_{S \subseteq \Gamma_N^{+1/2}}$ is a strongly connected partition, with correlation length at most $l + 1$, and that (109b) yields

$$ \forall S \subseteq \Gamma_N^{+1/2}, \quad \tilde{\mathcal{A}}_S^{\text{coarse}} \subseteq \mathcal{U}^{-1}(\mathcal{B}_{S \pm (r - \frac{1}{2})}). \quad (119) $$

This means that $\mathcal{U}$, when seen as mapping from the partition $(\tilde{\mathcal{A}}_S^{\text{coarse}})_{S \subseteq \Gamma_N^{+1/2}}$ to the partition $(\mathcal{B}_S)_{S \subseteq \Gamma}$, has radius $r - \frac{1}{2}$. Furthermore, $N > 4r + 2(l + l') = 4(r - \frac{1}{2}) + 2((l + 1) + l')$.

If we are in the base case, we find $\forall S \subseteq \Gamma_N^{+1/2}$, $\tilde{\mathcal{A}}_S^{\text{coarse}} = \mathcal{U}^{-1}(\mathcal{B}_S)$; so, defining $\mathcal{U}_1^{\text{fine}} := \mathcal{I}$ (the identity automorphism on $\omega^{\text{in}}$) and $\mathcal{U}_1^{\text{coarse}} := \mathcal{U}$, and seeing the first one as mapping from $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ to $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$, and the second as mapping from the latter to $(\mathcal{B}_S)_{S \subseteq \Gamma}$, we find that the first is a fine-graining and the second a coarse-graining.

Otherwise, by our induction hypothesis, we can thus decompose $\mathcal{U}$ (seen as mapping from $(\tilde{\mathcal{A}}_S^{\text{coarse}})_{S \subseteq \Gamma_N^{+1/2}}$ to $(\mathcal{B}_S)_{S \subseteq \Gamma}$) as $\mathcal{U} = \mathcal{U}_{2r} \dots \mathcal{U}_2$. Furthermore, we can define $\mathcal{U}_1^{\text{fine}} = \mathcal{U}_1^{\text{coarse}} = \mathcal{I}$; seeing the first one as going from the partition $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ to $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$, and the second as going from the latter to the partition $(\tilde{\mathcal{A}}_S^{\text{coarse}})_{S \subseteq \Gamma_N^{+1/2}}$, the first one is then a fine-graining and the second a coarse-graining,<fnref n="23" /> and we then have

$$ \mathcal{U} = \mathcal{U}_{2r} \dots \mathcal{U}_2 \mathcal{U}_1. \quad (120) \quad \square $$

Theorem 3.1 is then simply the restriction of Theorem D.2 to the case of QCAs between factorisations, for which we have $l = l' = 0$.

<footnote n="23">²³It might feel awkward that the automorphisms in our decomposition are almost all identities, but this is in fact the crux of our proof strategy: focusing on the interplay between different partitions of the same algebra and on how, under the right conditions, it allows one to fine-grain these partitions. In this ‘passive’ picture, the ‘active’ components — the QCAs, yielding isomorphisms from one C*-algebra to another — play nothing more than a book-keeping role.</footnote>
## E Proof of Theorem 3.2

Our proof by induction of Theorem D.2 in Section D.4 shows that each $U_1^{\text{fine}}$ or $U_1^{\text{coarse}}$ in this Theorem can be taken to be the identity automorphism on $\omega^{\text{in}}$ (but typed as mapping between different partitions of it), except for $U_{2r}^{\text{coarse}}$ which is taken to be equal to $\mathcal{U}$. All we have to prove is thus that $\text{Sh}^{\text{in}}$ is a shift automorphism (i.e. satisfies (34a)) with respect to each of the intermediate partitions.

To achieve this, let us prove that in Theorem D.1, any automorphism $\text{Sh}$ of $\omega$ that is a shift with respect to both the $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ and the $(\mathcal{B}_S)_{S \subseteq \Gamma}$ partitions is also a shift with respect to the $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$ partition. Indeed, the equations (115) (together with the specification via commutants and spans of unions for large intervals and for non-intervals) show that for any given $S \subseteq \Gamma_N^{\text{fine}}$, $\tilde{\mathcal{A}}_S = F_S[(\mathcal{A}_T)_{T \subseteq \Gamma_N}, (\mathcal{B}_{T'})_{T' \subseteq \Gamma}]$'s, where the $F_S$ functions satisfy $F_S[(\mathcal{A}_{T^{\text{Sh}}})_{T \subseteq \Gamma_N}, (\mathcal{B}_{(T')^{\text{Sh}}})_{T' \subseteq \Gamma}] = F_{S^{\text{Sh}}}((\mathcal{A}_T)_{T \subseteq \Gamma_N}, (\mathcal{B}_{T'})_{T' \subseteq \Gamma})$, and, crucially, only involve commutants, intersections, and algebraic spans of union.

Therefore, as any $C^*$ automorphism distributes over these three operations, we find that if $\text{Sh}$ is a shift for both the $\mathcal{A}$ and $\mathcal{B}$ partitions, then for an arbitrary $S \subseteq \Gamma_N^{\text{fine}}$,

$$ \begin{aligned} \text{Sh} \tilde{\mathcal{A}}_S &= \text{Sh} (F_S [(\mathcal{A}_T)_{T \subseteq \Gamma_N}, (\mathcal{B}_{T'})_{T' \subseteq \Gamma}]) \\ &= F_S [(\text{Sh} \mathcal{A}_T)_{T \subseteq \Gamma_N}, (\text{Sh} \mathcal{B}_{T'})_{T' \subseteq \Gamma}] \\ &= F_S [(\mathcal{A}_{T^{\text{Sh}}})_{T \subseteq \Gamma_N}, (\mathcal{B}_{(T')^{\text{Sh}}})_{T' \subseteq \Gamma}] \quad (121) \\ &= F_{S^{\text{Sh}}} [(\mathcal{A}_T)_{T \subseteq \Gamma_N}, (\mathcal{B}_{T'})_{T' \subseteq \Gamma}] \\ &= \tilde{\mathcal{A}}_{S^{\text{Sh}}}, \end{aligned} $$

so $\text{Sh}$ is a shift with respect to $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$.

Returning to our proof by induction of Theorem D.2 in Section D.4, we note that translation-invariance can be restated as the fact that $\text{Sh}^{\text{in}}$ is a shift with respect to both $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ and $(\mathcal{U}^{-1}(\mathcal{B}_S))_{S \subseteq \Gamma}$. Thus, by our previous result, it is also a shift with respect to $(\tilde{\mathcal{A}}_S)_{S \subseteq \Gamma_N^{\text{fine}}}$, and thus also with respect to $(\tilde{\mathcal{A}}_S^{\text{coarse}})_{S \subseteq \Gamma_N^{+1/2}}$, which is the former's coarse-graining.

In the base case, we thus find that both $U_1^{\text{fine}} := \mathcal{I}$ and $U_1^{\text{coarse}} := \mathcal{U}$ are translation-invariant, where the input and output partitions of $U_1^{\text{fine}}$ are equipped with the shift $\text{Sh}^{\text{in}}$; while in the case of

the induction step, $U_1^{\text{fine}} := \mathcal{I}$ and $U_1^{\text{coarse}} := \mathcal{I}$ are translation-invariant, where all partitions involved are equipped with the shift $\text{Sh}^{\text{in}}$.

## F Proofs for Section 4

### F.1 Some Lemmas

The following two lemmas tell us that in a strongly connected 1D partition over a sufficiently large graph, sectorial correlations — the fact that some sectors at different points are incompatible, witnessed by the fact that the corresponding projectors' product is null — always come down to nearest-neighbour correlations: if the product of some projectors is null, it must be because two of them, at neighbouring edges, have a null product.

Note how, to better manipulate the structure of strong connectedness, we move from talking about centres at a vertex — the $\mathcal{Z}_n$'s for $n \in \Gamma_N$ — to centres at an edge — the $\mathcal{Z}_{n-\frac{1}{2}} \cap \mathcal{Z}_{n+\frac{1}{2}}$'s for $n \in \Gamma_N^{+1/2}$.

**Lemma F.1.** Let $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ be a strongly connected 1D partition with correlation length at most $l$, where $N \ge 3l+1$. For every $n$ in $\Gamma_N^{+1/2}$, we take $\pi_n \in \text{AtomProj}(\mathcal{Z}_{n-\frac{1}{2}} \cap \mathcal{Z}_{n+\frac{1}{2}})$.

Then $\prod_{n \in \Gamma_N^{+1/2}} \pi_n = 0$ if and only if $\exists n \in \Gamma_N^{+1/2}$ such that $\pi_n \pi_{n+1} = 0$.

*Proof.* Let us suppose $\prod_{n \in \Gamma_N^{+1/2}} \pi_n = 0$ without any of the $\pi_n \pi_{n+1}$'s being null, and reach a contradiction. We denote $p_-$ and $p_+$ as 0's antipodes, as in the statement of Lemma D.9. Then $d(p_-, 1) > l$, so $\pi_1 \in \mathcal{Z}_1 \perp \mathcal{Z}_{p_-} \ni \pi_{p_- - \frac{1}{2}}$, so $\pi_{p_- - \frac{1}{2}} \pi_1 \ne 0$.

Yet $\pi_{p_- - \frac{1}{2}} \pi_1$ is an atomic projector of $\mathcal{Z}_{[p_-, 0]} = \mathcal{Z}_{[1, p_+]}$, so $\pi_{p_- - \frac{1}{2}} \pi_1 \mathcal{A}_{[p_-, 0]}$ and $\pi_{p_- - \frac{1}{2}} \pi_1 \mathcal{A}_{[1, p_+]}$ are commuting factors, so the composition of elements of one with elements of the other (defined as in (62)), not being uniformly null, is necessarily injective. Therefore $\prod_{n \in \Gamma_N^{+1/2}} \pi_n = 0$ implies that at least one of $\prod_{n \in [p_- - \frac{1}{2}, \frac{1}{2}]} \pi_n$ and $\prod_{n \in [\frac{1}{2}, p_- - \frac{1}{2}]} \pi_n$ is null.

Suppose without loss of generality that $\prod_{n \in [\frac{1}{2}, p_- - \frac{1}{2}]} \pi_n = 0$. Since $0 - p_- \ge l$, Lemma D.6's second item tells us that $\ker \hat{\pi}_1 \cap \mathcal{A}_{[2, p_+]} = \bar{\mu} \mathcal{A}_{[2, p_+]}$ with $\bar{\mu} = 1 - \mu \in \mathcal{Z}_1 \cap \mathcal{Z}_2$. As $\pi_3$ is an atomic projector of $\mathcal{Z}_1 \cap \mathcal{Z}_2$ satisfying $\pi_1 \pi_3 \ne 0$,
$\pi_{\frac{3}{2}} = \pi_{\frac{3}{2}}\mu$, so $\hat{\pi}_{\frac{1}{2}}$ is injective on $\mu A_{\llbracket 2, p_+ \rrbracket} \ni \prod_{n \in \llbracket \frac{3}{2}, p_- - \frac{1}{2} \rrbracket} \pi_n$, so $\prod_{n \in \llbracket \frac{3}{2}, p_- - \frac{1}{2} \rrbracket} \pi_n = 0$. This procedure can be repeated to erase terms from the product until we reach $\pi_{p_- - \frac{1}{2}} = 0$, which is a contradiction. $\square$

**Lemma F.2.** Let $(A_S)_{S \subseteq \Gamma_N}$ be a strongly connected 1D partition with correlation length at most $l$, where $N \ge 3l + 1$. We take $m, n \in \Gamma_N$, and for every $x \in [m - \frac{1}{2}, n + \frac{1}{2}]$, we take $\pi_x \in \mathrm{AtomProj}(\mathcal{Z}_{x-\frac{1}{2}} \cap \mathcal{Z}_{x+\frac{1}{2}})$.

Then $\prod_{x \in [m - \frac{1}{2}, n + \frac{1}{2}]} \pi_x = 0$ if and only if $\pi_{m - \frac{1}{2}} \pi_{n + \frac{1}{2}} = 0$ or $\exists x \in [m - \frac{1}{2}, n - \frac{1}{2}]$, $\pi_x \pi_{x+1} = 0$.

*Proof.* Suppose $\pi_{m-\frac{1}{2}}\pi_{n+\frac{1}{2}} \neq 0$; it is then an atomic projector of $\mathcal{Z}_{[n+1,m-1]}$, so $\pi_{m-\frac{1}{2}}\pi_{n+\frac{1}{2}}A_{[n+1,m-1]}$ is a non-zero factor algebra, and we have $(\pi_{m-\frac{1}{2}}\pi_{n+\frac{1}{2}}A_S)_{S \subseteq [n+1,m-1]} \vdash \pi_{m-\frac{1}{2}}\pi_{n+\frac{1}{2}}A_{[n+1,m-1]}$. The identity is thus in $\bigvee_{t \in [n+1,m-1]} \mathcal{Z}(\pi_{m-\frac{1}{2}}\pi_{n+\frac{1}{2}}A_t) = \bigvee_{t \in [n+1,m-1]} \pi_{m-\frac{1}{2}}\pi_{n+\frac{1}{2}}\mathcal{Z}_t = \bigvee_{x \in [n+\frac{1}{2},m-\frac{1}{2}]} \pi_{m-\frac{1}{2}}\pi_{n+\frac{1}{2}}(\mathcal{Z}_{x-\frac{1}{2}} \cap \mathcal{Z}_{x+\frac{1}{2}})$ (the first equality uses Lemma B.1); in particular the latter cannot be null, so we can pick $\pi_x \in \mathrm{AtomProj}(\mathcal{Z}_{x-\frac{1}{2}} \cap \mathcal{Z}_{x+\frac{1}{2}})$ for every $x \in [n + \frac{3}{2}, m - \frac{3}{2}]$ such that $\prod_{x \in [n+\frac{1}{2},m-\frac{1}{2}]} \pi_x \neq 0$. Composition with the latter is necessarily an injective homomorphism on the factor $\pi_{m-\frac{1}{2}}\pi_{n+\frac{1}{2}}A_{[m,n]} \ni \prod_{x \in [m-\frac{1}{2},n+\frac{1}{2}]} \pi_x$; so $\prod_{x \in \Gamma_N^{+1/2}} \pi_x \neq 0$. We can then apply Lemma F.1 to get the result. $\square$

## F.2 Full representability of the partitions

Here, we adopt the notations and results of Ref. [21], to which the reader can refer for more details. As shown by Ref. [21]'s Theorem 5.2, any partition admits a concrete representation as a set of (routed) operators over a tensor product of Hilbert spaces, in which its individual parts' algebras — i.e. the $A_x$'s corresponding to singletons — are well-localised, i.e. of the form

$$A_x = \left\{ \begin{array}{l} (f_{A_x} \otimes \mathbb{1}_{A'_{x}, x' \neq x}) \tilde{\Pi} \\ \mid f \in \mathrm{Lin}_{\delta}(\mathcal{H}_{A_x}) \end{array} \right\}; \qquad (122)$$

however, it might not be the same for the algebras corresponding to composite systems. We say that the partition is fully representable if the representation can furthermore be picked such that

$$A_S = \left\{ \begin{array}{l} (f_{A_n, n \in S} \otimes \mathbb{1}_{A_n, n \notin S}) \tilde{\Pi} \\ \mid f \in \mathrm{Lin}_{\eta_S} \left( \bigotimes_{n \in S} \mathcal{H}_{A_n} \right) \end{array} \right\}, \qquad (123)$$

Some partitions are not fully representable; the general situation is that, if the partition is in a represented form (with (122) holding), then the algebras for composite systems take the form

$$A_S = \left\{ \begin{array}{l} \phi_S (f_{A_n, n \in S} \otimes \mathbb{1}_{A_n, n \notin S}) \phi_S^\dagger \tilde{\Pi} \\ \mid f \in \mathrm{Lin}_{\eta_S} \left( \bigotimes_{n \in S} \mathcal{H}_{A_n} \right) \end{array} \right\} \qquad (124) \\ = \hat{\phi}_S (\tilde{A}_S),$$

where $\phi_S$ is a dephasing unitary, belonging to Unit$(\bigvee_{x \in X} \mathcal{Z}_x)$, and $\tilde{A}_S$ denotes the 'well-localised' form (123) of the algebra. When dealing with a non-fully representable partition, it is not possible to find a representation in which all of the $\phi_S$'s jointly disappear.

Fortunately, we have the following result.

**Theorem F.1.** All the partitions at hand in Theorem D.2 are fully representable.

Since the construction of Theorem D.2 is based on successive node-splittings using Lemma D.9, Theorem F.1 is a direct consequence of the following Lemma.

**Lemma F.3.** In Lemma D.9, if $(A_S)_{S \subseteq \Gamma_N}$ is fully representable, then $(A_S^{\mathrm{split}})_{S \subseteq \Gamma^{\mathrm{split}}}$ is fully representable.

Before we prove this lemma, let us display three useful results. In each of them, we consider the partition $(A_S)_{S \subseteq X} \vdash \omega$, in a represented form given by (124).

**Lemma F.4.** We fix $R \subseteq S \subseteq T \subseteq X$. If $A_R$ and $A_T$ are well-localised (i.e. of the form (123)), then
$$ \phi_S \in \mathrm{Unit} \left( \mathcal{Z}_R \vee \bigvee_{x \in T \setminus R} \mathcal{Z}_x \right). \quad (125) $$

*Proof.* Since $\mathcal{A}_T$ is well-localised, the restricted partition $(\mathcal{A}_X)_{T' \subseteq T} \vdash \mathcal{A}_T$ is itself in a represented form satisfying (122), so we can apply the formula (124) to it and obtain

$$ \phi_S \in \mathrm{Unit} \left( \bigvee_{x \in T} \mathcal{Z}_x \right). \quad (126) $$

Furthermore, since $\mathcal{A}_R$ is well-localised, the coarse-grained version of the partition in which we consider $\mathcal{A}_R$ as an individual part without substructure is in a represented form as well, and applying the above result to it yields (125). □

**Lemma F.5.** We fix $S \subseteq X$ and suppose $\omega$ factor. If $\mathcal{A}_S$ is well-localised, then so is $\mathcal{A}_{\bar{S}}$.

*Proof.* Since $\omega$ is a factor, $\mathcal{A}_S = \mathcal{A}'_{\bar{S}}$, and from their form it is clear that $\tilde{\mathcal{A}}_S = \tilde{\mathcal{A}}'_{\bar{S}}$, so $\mathcal{A}_S = \tilde{\mathcal{A}}_S$ is equivalent to $\mathcal{A}_{\bar{S}} = \tilde{\mathcal{A}}_{\bar{S}}$. □

**Lemma F.6.** We fix $S, T \subseteq X$ such that $\mathcal{A}_{S \cap T} = \mathcal{A}_S \cap \mathcal{A}_T$ and suppose $\omega$ factor. If $\mathcal{A}_S$ and $\mathcal{A}_T$ are well-localised, then so is $\mathcal{A}_{S \cap T}$.

*Proof.* By Lemma F.5, $\mathcal{A}'_S = \mathcal{A}_{\bar{S}}$ and $\mathcal{A}'_T = \mathcal{A}_{\bar{T}}$ are well-localised, and thus so is their span $\mathcal{A}'_S \vee \mathcal{A}'_T = (\mathcal{A}_S \cap \mathcal{A}_T)' = \mathcal{A}'_{S \cap T} = \mathcal{A}_{S \cap T}$; so by Lemma F.5 again, $\mathcal{A}_{S \cap T}$ is well-localised. □

We now prove Lemma F.3.

*Proof.* Up to an isomorphism, we can assume without loss of generality that $(\mathcal{A}_S^{\mathrm{split}})_{S \subseteq \Gamma^{\mathrm{split}}}$ is in its represented form (124). Since its coarse-graining $(\mathcal{A}_S)_{S \subseteq \Gamma_N}$ is fully representable, we can pick a representation in which, for every $S \subseteq \Gamma_N \setminus \{0\}$, $\mathcal{A}_S^{\mathrm{split}}$ and $\mathcal{A}_{S \cup \{-\frac{1}{4}, \frac{1}{4}\}}^{\mathrm{split}}$ are well-localised, i.e. $\phi_S$ and $\phi_{S \cup \{-\frac{1}{4}, \frac{1}{4}\}}$ are identities.

In particular, $\mathcal{A}_{[p_{-}, \frac{1}{4}]}^{\mathrm{split}}$ and $\mathcal{A}_{[p_{-}, -1]}^{\mathrm{split}}$ are then well-localised; so, by Lemma F.4 applied with $R = [p_{-}, -1] \subseteq S = [p_{-}, -\frac{1}{4}] \subseteq T = [p_{-}, \frac{1}{4}]$, we have

$$ \phi_{[p_{-}, -\frac{1}{4}]} \in \mathrm{Unit} \left( \mathcal{Z}_{[p_{-}, -1]} \vee \mathcal{Z}_{-\frac{1}{4}} \vee \mathcal{Z}_{\frac{1}{4}} \right) \quad (127) \\ \subseteq \mathrm{Unit} \left( (\mathcal{Z}_{p_{+}} \cap \mathcal{Z}_{p_{-}}) \vee \mathcal{A}_{0} \right), $$

where $\mathcal{A}_{[p_{-}, -\frac{1}{4}]} = \hat{\phi}_{[p_{-}, -\frac{1}{4}]}(\tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]})$, with $\tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]}$ the localised algebra of the form (123).

Let us prove that $\phi_{[p_{-}, -\frac{1}{4}]} \in \mathrm{Unit}(\mathcal{A}_0)$. Denoting $\mathcal{Z}_{p_{+}} \cap \mathcal{Z}_{p_{-}}$'s atomic projectors as $\pi^k$'s, we have $\phi_{[p_{-}, -\frac{1}{4}]} = \sum_k \pi^k \phi^k$, with the $\phi^k$'s in $\mathrm{Unit}(\mathcal{A}_0)$.

We then take $f \in \mathcal{A}_{[q_{-}, -\frac{1}{4}]}$; since it is in $\mathcal{A}_{[p_{-}, -\frac{1}{4}]}$, $f = \hat{\phi}_{[p_{-}, -\frac{1}{4}]}(\tilde{f})$ with $\tilde{f} \in \tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]}$. Furthermore,

$$ \begin{aligned} \tilde{f} &\in \hat{\phi}_{[p_{-}, -\frac{1}{4}]}^{\dagger}(\mathcal{A}_{[q_{-}, -\frac{1}{4}]}) \\ &\subseteq \hat{\phi}_{[p_{-}, -\frac{1}{4}]}^{\dagger}(\mathcal{A}_{[q_{-}, p_{+}]}) \\ &\subseteq \mathcal{A}_{[q_{-}, p_{+}]} \\ &= \mathcal{A}'_{[p_{-}, q_{-} - 1]} \\ &= \tilde{\mathcal{A}}'_{[p_{-}, q_{-} - 1]}; \end{aligned} \quad (128) $$

therefore $\tilde{f} \in \tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]} \cap \tilde{\mathcal{A}}'_{[p_{-}, q_{-} - 1]} = (\mathcal{Z}_{p_{+}} \cap \mathcal{Z}_{p_{-}}) \vee \tilde{\mathcal{A}}_{[q_{-}, -\frac{1}{4}]}$.

We can thus write it as $\tilde{f} = \sum_k \pi^k \tilde{f}^k$, with the $\tilde{f}^k$'s in $\mathcal{A}_{[q_{-}, -\frac{1}{4}]}$. This yields $f = \sum_k \pi^k \hat{\phi}^k(\tilde{f}^k)$. Picking an arbitrary $k_0$, we then have $\pi^{k_0} f = \pi^{k_0} \hat{\phi}^{k_0}(\tilde{f}^{k_0})$, where $f$ and $\hat{\phi}^{k_0}(\tilde{f}^{k_0})$ are in $\mathcal{A}_{[q_{-}, 0]}$. However, multiplication by $\pi^{k_0}$ is injective on the latter, since Lemma D.6 tells us that the $\mu$ corresponding to it has to be the identity by correlation length, so $f = \hat{\phi}^{k_0}(\tilde{f}^{k_0})$.

Therefore, $\mathcal{A}_{[q_{-}, -\frac{1}{4}]} \subseteq \hat{\phi}^{k_0}(\tilde{\mathcal{A}}_{[q_{-}, -\frac{1}{4}]}) \subseteq \hat{\phi}^{k_0}(\tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]})$ with $\hat{\phi}^{k_0} \in \mathrm{Unit}(\mathcal{A}_0)$. We also note that, since $\hat{\phi}^{k_0}$ commutes with all elements of $\mathcal{A}_{[p_{-}, -1]} = \tilde{\mathcal{A}}_{[p_{-}, -1]} \subseteq \tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]}$, we have $\mathcal{A}_{[p_{-}, -1]} = \hat{\phi}^{k_0}(\mathcal{A}_{[p_{-}, -1]}) \subseteq \hat{\phi}^{k_0}(\tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]})$. Thus,

$$ \begin{aligned} \mathcal{A}_{[p_{-}, -\frac{1}{4}]} &= \mathcal{A}'_{[\frac{1}{4}, p_{+}]} \\ &\stackrel{(88b)}{=} (\mathcal{A}_{[0, p_{+}]} \cap \mathcal{A}'_{[q_{-}, -\frac{1}{4}]})' \\ &= \mathcal{A}'_{[0, p_{+}]} \vee \mathcal{A}_{[q_{-}, -\frac{1}{4}]} \\ &= \mathcal{A}_{[p_{-}, -1]} \vee \mathcal{A}_{[q_{-}, -\frac{1}{4}]} \\ &\subseteq \hat{\phi}^{k_0}(\tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]}) \end{aligned} \quad (129) $$

which we can turn into an equality $\mathcal{A}_{[p_{-}, -\frac{1}{4}]} = \hat{\phi}^{k_0}(\tilde{\mathcal{A}}_{[p_{-}, -\frac{1}{4}]})$ since $\hat{\phi}^{k_0}$ is bijective. This shows that $\phi_{[p_{-}, -\frac{1}{4}]}$ is in $\mathrm{Unit}(\mathcal{A}_0)$.

We move to a new representation by conjugating $\omega$ by $\phi_{[p_{-}, -\frac{1}{4}]}^\dagger$. Because it is in $\mathrm{Unit}(\mathcal{A}_0)$, this does not modify the representation of the $\mathcal{A}_S$ and
$\mathcal{A}_{S \cup \{-\frac{1}{4}, +\frac{1}{4}\}}$'s for $S \subseteq \Gamma_N \setminus \{0\}$; indeed $\phi_{[[p_{-}, -\frac{1}{4}]]}^{\dagger}$ commutes past the first, and can be absorbed in the second as they include $\mathcal{A}_0$. Thus, this yields a representation in which these as well as $\mathcal{A}_{[[p_{-}, -\frac{1}{4}]]}$ are localised. Symmetrically, $\mathcal{A}_{[[\frac{1}{4}, p_{+}]]} = \mathcal{A}'_{[[p_{-}, -\frac{1}{4}]]}$ is localised as well by Lemma F.5. Furthermore, for $n \in [[p_{-}, -\frac{1}{4}]]$, $\mathcal{A}_{[[n, -\frac{1}{4}]]} = \mathcal{A}_{[[n, 0]]} \cap \mathcal{A}_{[[p_{-}, -\frac{1}{4}]]}$ is the intersection of two localised algebras and is thus localised by Lemma F.6, and the same goes symmetrically for the $\mathcal{A}_{[[\frac{1}{4}, n]]}$'s with $n \in [[\frac{1}{4}, p_{+}]]$. One can find that the same is true for larger intervals by taking commutants (using Lemma F.5), and finally for non-intervals by taking spans. $\square$

<figure><img src="image_16.png" /></figure>

## F.3 Representations of strongly connected 1D partitions

Representations of partitions — i.e. how a partitioned algebra can be seen as a space of operators over a suitably partitioned Hilbert space, in such a way that each of the partitions' subalgebras becomes the subalgebra of operators acting non-trivially only on the corresponding part of the Hilbert space — were studied in the general case in Ref. [21]. We will work directly with this paper's notations, and assume knowledge of the notions it introduces.

The representation of a partition involves the use of Boolean matrices, pinning down sectorial correlations and routes, which in the general case can be quite convoluted. What we will now show is that, in the specific case of strongly connected 1D partitions (on a graph of size strictly greater than three times their correlation length), one can present these routes in a more accessible way, that only involves Kronecker deltas between neighbouring sites, as introduced in the main text.

For the remainder of this section, we take $(\mathcal{A}_S)_{S \subseteq X} \vdash \omega$ to be a strongly connected 1D partition of a factor, with correlation length at most $l$, where $N \ge 3l + 1$.

Since for every $n \in X$, $\mathbb{Z}_n = (\mathbb{Z}_{n-1} \cap \mathbb{Z}_n) \vee (\mathbb{Z}_n \cap \mathbb{Z}_{n+1})$, the idea is to label the former algebra's set of atomic projectors with a pair of labels, corresponding to atomic projectors of the latter two algebras. We will thus label these latter atomic projectors in the following way:

$$ \begin{aligned} \forall n \in \Gamma_N^{+1/2}, \quad & \text{AtomProj}(\mathbb{Z}_{n-\frac{1}{2}} \cap \mathbb{Z}_{n+\frac{1}{2}}) \\ = & \{\pi_n^{k_n} \mid k_n \in K_n\}. \end{aligned} \quad (130) $$

For a given $n \in \Gamma_N$, $\mathbb{Z}_n$'s atomic projectors are then almost the $\pi_{n-\frac{1}{2}}^{k_n}$, $\pi_{n+\frac{1}{2}}^{k_n}$ where $k_n^L \in K_{n-\frac{1}{2}}$ and $k_n^R \in K_{n+\frac{1}{2}}$, with the difference that the latter might also, in addition, include one or several instances of the null projector 0. We thus need the following proposition.

**Proposition F.1 (Null-sector trick).** *The construction of a partitioned algebra's routes displayed in Ref. [21] also works if, for every $n$, we allow some of the $\pi_n^{k_n}$'s to be null, as long as the $\pi_n^{k_n}$'s are still pairwise orthogonal projectors spanning $\mathbb{Z}_n$.*

*Proof.* The proofs in Ref. [21] never use the fact that the $\pi_n^{k_n}$'s are non-null. $\square$

<figure><img src="image_17.png" /></figure>

We will thus make use of this generalisation, which we call the 'null-sector trick', since it entails that in our representations, some of the sectors $\mathcal{H}_{A_n}^{k_n}$ now might be Hilbert spaces of dimension 0. Note that a coefficient $\sigma_k$ of the Boolean matrix of sectorial correlations will then be zero if (although not only if) any of the $k_n$'s is in one of these 'impossible values'.

In this labelling, we have, for $n \in \Gamma_N$, $K_n = K_{n-\frac{1}{2}} \times K_{n+\frac{1}{2}}$ — i.e., $\mathbb{Z}_n$'s atomic projectors are identified by a pair of labels: a left label $k_n^L \in K_{n-\frac{1}{2}}$, and a right label $k_n^R \in K_{n+\frac{1}{2}}$, and one can write each of the sites' Hilbert spaces as

$$ \mathcal{H}_{A_n} := \bigoplus_{k_n^L \in K_{n-\frac{1}{2}}, k_n^R \in K_{n+\frac{1}{2}}} \mathcal{H}_A^{(k_n^L, k_n^R)}, \quad (131) $$

with some of these summands potentially of null dimension: this corresponds to the main text's (41).

As we advertised, the routes corresponding to any part's algebra will then be writable using only Kronecker deltas. Let us focus on the algebras corresponding to intervals, since the other cases can then be immediately inferred through the use of algebraic spans. For any $m, n \in \Gamma_N$, we define the partial equivalence relation $\Delta_{[[m, n]]}$ on $K_m \times K_{m+1} \times \dots \times K_n = K_{m-\frac{1}{2}} \times K_{m+\frac{1}{2}} \times$