We provide a systematic description of the automorphism groups of specially cocompact $\CAT$ cube complexes. We show that these groups are topologically finitely generated, present a method to explicitly obtain generating sets, and prove a dichotomy on their size. Furthermore, we show that, under some extra assumptions, the normal subgroup known as $\Aut^{+}$ is simple, non-discrete, and tdlc. In particular, we obtain a new class of simple, non-discrete, tdlc groups that are accessible to further study. Finally, we study the relative size of $\Aut^{+} < \Aut$, providing a sufficient condition for its closure to be finite index and presenting a common example where it is not even cocompact.

§ INTRODUCTION

J. Tits showed in [22] that the (inversion-free) automorphism group of a regular tree with valence at least three is locally compact and uncountable, and that the subgroup generated by edge-fixators is simple. Following up on these ideas, Haglund and Paulin studied several types of negatively curved complexes in [15], including some negatively curved buildings, certain Coxeter systems, and polyhedral complexes that admit a geometric action by a hyperbolic group. They showed that, similarly, the automorphism groups of these complexes are locally compact and uncountable. Furthermore, they generalised the group generated by edge-fixators to a group denoted $\Aut^{+}$ and showed that, under the right assumptions, it is simple and has finite index. For right-angled buildings, the study of the automorphism group has yielded similar results and more [7, 12, 3, 4, 5] and in the case of trees, [18] has expanded on several results regarding subgroups of the automorphism group.

In this paper, we consider the automorphism groups $\Aut(X)$ of $\CAT$ cube complexes that are universal coverings of compact, non-positively curved special cube complexes. Particular instances of such cube complexes appear already in the work of Haglund and Paulin. Since then, automorphisms of $\CAT$ cube complexes have been studied implicitly in the work on automorphisms of right-angled buildings mentioned above and in [17], where Lazarovich studied regular $\CAT$ cube complexes (that is, all vertices have the same link) and in particular provided a method to determine simplicity of certain subgroups of $\Aut(X)$.

We will show that, for these cube complexes, $\Aut(X)$ satisfies both analogous and distinct properties to the cases studied by Tits, Haglund, Paulin, and Caprace. Specifically, we will show that these automorphism groups are topologically finitely generated and we present a characterisation as to when they are uncountable. Furthermore, we will show that, under some mild assumptions, the subgroup $\Aut^{+}(X)$ (see Section <ref>) is a simple, non-discrete tdlc (totally disconnected and locally compact) group and we will study the size of $\Aut^{+}(X)$ in $\Aut(X)$.

An additional motivation for our work arises as follows: Let $S$ be a compact, non-positively curved special cube complex, $\Gamma$ its fundamental group and $X$ its universal covering. In this case, $X$ is a $\CAT$ cube complex and $\Gamma$ acts cubically, freely, properly, and cocompactly on $X$. Such group actions have attracted significant attention over the last two decades, leading to significant results in group theory and topology [19, 20, 24, 6, 2]. Since $\Gamma$ acts cubically, properly, and cocompactly and $X$ is locally finite, $\Gamma$ embeds as a uniform lattice in $\Aut(X)$. A systematic understanding of $\Aut(X)$ may thus allow us to study cubulable groups using the theory of lattices in locally compact groups.

§.§ The key tool: admissible edge labelings

Our description of $\Aut(X)$ relies on the choice of a good edge labeling on $X$. Let $\vec{\mathcal{E}}(X)$ denote the set of oriented edges in $X$ and let $\vec{\mathcal{E}}(X)_v$ denote the set of outgoing edges at a vertex $v$. Our labelings are going to be as follows:

Let $\Sigma_0$ be a finite set with a fixpoint free involution $\cdot^{-1} : \Sigma_0 \rightarrow \Sigma_0$. A map $\ell : \vec{\mathcal{E}}(X) \rightarrow \Sigma_0$ is called an admissible edge labeling, if the following hold:

(A1) If $e, e'$ are in the same parallel class (as oriented edges), then $\ell(e) = \ell(e')$. If $e, e'$ are the same geometric edge with opposite orientation, then $\ell(e') = \ell(e)^{-1}$.

(A2) The restriction $\ell \vert_{\vec{\mathcal{E}}(X)_v}$ is injective for every $v \in X^{(0)}$.

(A3) For all $v, w \in X^{(0)}$, $e_1, e_2 \in \vec{\mathcal{E}}(X)_v$, $e'_1, e'_2 \in \vec{\mathcal{E}}(X)_w$, we have the following: If $\ell(e_1) = \ell(e'_1)$ and $\ell(e_2) = \ell(e'_2)$, then $e_1$ and $e_2$ span a square if and only if $e'_1$ and $e'_2$ span a square.

An admissible edge labeling is called $\Gamma$-invariant, if for all $e \in \vec{\mathcal{E}}(X)$ and all $g \in \Gamma$, we have $\ell(g(e)) = \ell(e)$.

The Salvetti complex of a right-angled Artin group admits an admissible edge labeling which is invariant under the action of the given right-angled Artin group. The labeling is the one coming from the Cayley-graph of the standard presentation of the right-angled Artin group, where oriented edges are labeled by the standard generators and their inverses.

It turns out that the existence of $\Gamma$-invariant admissible edge labelings is closely related to the existence of special cocompact actions. This is the content of our first result, which characterises the existence of $\Gamma$-invariant admissible edge labelings.

Let $S$ be a compact, non-positively curved special cube complex, $\Gamma$ its fundamental group, and $X$ its universal covering. Then $X$ admits a $\Gamma$-invariant admissible edge labeling.

Conversely, let $X$ be a locally finite $\CAT$ cube complex and $\Gamma$ a group acting cubically and cocompactly on $X$ such that there exists a $\Gamma$-invariant admissible edge labeling. Then the quotient $\faktor{X}{\Gamma}$ is a compact, non-positively curved special cube complex, $\Gamma$ acts freely on $X$, and $X$ is the universal covering of $\faktor{X}{\Gamma}$.

Given an admissible edge labeling, we can also label oriented diagonals. Namely, we label an oriented diagonal by the collection of labels of the corresponding oriented edges (see figure <ref> or Construction <ref>). This expands our alphabet $\Sigma_0$ to an alphabet $\Sigma \subset 2^{\Sigma_0}$ and the involution $\cdot^{-1}$ extends in the natural way.

Extending an admissible edge labeling to oriented diagonals. The indicated diagonals are labeled by the elements $\{ s, t \}, \{ s, t^{-1} \} \in \Sigma$ respectively.

Given a vertex $v \in X^{(0)}$, we write $\Sigma_v \subset \Sigma$ for the set of labels that the outgoing edges and diagonals at $v$ are labeled by. We write $\Sigma_{0,v} := \Sigma_v \cap \Sigma_0$ for the labels of the outgoing edges at $v$.

Furthermore, if $s, t \in \Sigma_{0,v}$ are the labels of two edges $e, e' \in \vec{ \mathcal{E} }(X)_v$, then we say that $s$ and $t$ commute if and only if $e$ and $e'$ span a square in $X$. We write $[s,t] = 1$ if they commute and $[s,t] \neq 1$, if they do not. (See Definition <ref> and Lemma <ref> as to why this does not depend on the vertex $v$ and Proposition <ref> for a justification of this terminology and notation.)

Having labeled oriented edges and diagonals, we can use these labels to identify concatenations of oriented edges and diagonals (we call such concatenations cube paths) with words over the alphabet $\Sigma$ together with a starting vertex. In particular, if we restrict to the cube paths that start at a fixed vertex $o$, we obtain a language $\mathcal{L}_o^{cube}$ over the alphabet $\Sigma$. Given two words $v, w \in \mathcal{L}_o^{cube}$, we say they are equivalent if and only if their corresponding cube paths, starting at $o$, have the same endpoint and we write $v \equiv_o w$. Each equivalence class under this equivalence relation corresponds to the set of all cube paths from $o$ to a particular vertex in $X^{(0)}$. In particular, the empty word $\epsilon$ represents the vertex $o$. We use this to identify vertices in $X$ with $\equiv_o$-equivalence classes in $\mathcal{L}_o^{cube}$.

§.§ Topological generators and the size of $\text{Aut}(X)$

Let $S$, $X$, and $\Gamma$ be as in Theorem <ref>. We always consider $\Aut(X)$ equipped with the compact-open topology. By a series of standard arguments, local finiteness of $X$ implies that $\Aut(X)$ is second countable and locally compact. Furthermore, since every element of $\Aut(X)$ preserves vertices in $X$ and the set of vertices is discrete, $\Aut(X)$ is totally disconnected.

Let $\ell$ be a $\Gamma$-invariant admissible edge labeling on $X$. In order to state our main results, we need to establish a few more definitions.

Let $\gamma$ be a concatenation of edges and diagonals, starting at $o$, and let $v \in \mathcal{L}_o^{cube}$ be the word spelt out by $\gamma$. We say $v$ is reduced, if $\gamma$ does not cross any hyperplane more than once. We call $v$ reducible, if it is not reduced.

Every vertex $v \in X^{(0)}$ admits a (usually not unique) cube path from $o$ to $v$ whose word is reduced.

Let $v \in \mathcal{L}_o^{cube}$ represent a vertex and let $\sigma : \Sigma_v \rightarrow \Sigma$ be an injective map. We call $\sigma$ a label-morphism, if it satisfies the following:

* $\sigma(\Sigma_{0,v}) \subset \Sigma_0$.

* $\forall s, t \in \Sigma_{0,v} : [s,t] = 1 \Leftrightarrow [\sigma(s), \sigma(t)] = 1$.

* $\forall \alpha = \{ s_1, \dots, s_n \} \in \Sigma_v : \quad \sigma(\alpha) = \{ \sigma(s_1), \dots, \sigma(s_n) \}.$

Since label-morphisms are uniquely determined by their restriction to $\Sigma_{0,v}$, we frequently write $\sigma : \Sigma_{0,v} \rightarrow \Sigma_0$.

If $\alpha = \{ s_1, \dots, s_n \} \in \Sigma$, we write $\sigma(\alpha) = \alpha$ if $\sigma$ fixes the set $\alpha$ pointwise. We write $\sigma( \{ \alpha \} ) = \{ \alpha \}$, if $\sigma$ preserves the set $\alpha$, although it may not fix its elements pointwise.

Given a reduced word $v \in \mathcal{L}_o^{cube} \setminus \{ \epsilon \}$ and a label-morphism $\sigma : \Sigma_{0,v} \rightarrow \Sigma_0$, we say that $\sigma$ is compatible with $v$ if and only if for all $\alpha \in \Sigma_v$ such that $v\alpha$ is reducible and for all $\beta \in \Sigma_v$ such that $[\alpha, \beta ] = 1$, we have $\sigma(\alpha) = \alpha$ and $\sigma(\beta) = \beta$.

Let $o \in X^{(0)}$. We denote $G_o := \Stab_{\Aut(X)}(o)$. For any $g \in G_o$ and any $v \in X^{(0)}$, $g$ induces a bijection

\[ \pre{v}{\sigma(g)} : \Sigma_v \rightarrow \Sigma_{g(v)} \]

which sends the label of an outgoing edge $e$ at $v$ to the label of the edge $g(e) \in \vec{\mathcal{E}}(X)_{g(v)}$. Equivalently, $\pre{v}{\sigma(g)}$ is defined by the equation

\[ g(v\alpha) = g(v) \pre{v}{\sigma(g)}(\alpha) \quad \forall \alpha \in \Sigma_v. \]

The maps $(\pre{v}{\sigma(g)})_{v \in X^{(0)}}$ form what we will call the portrait of $g$. Portraits will be of significant importance throughout the paper.

Let $v \in \mathcal{L}_o^{cube}$ represent a vertex other than $o$. Let $\hat{h}_1, \dots, \hat{h}_n$ be the collection of hyperplanes that intersect an edge incident to $v$ and that separate $o$ from $v$. Let $h_1, \dots, h_n$ denote the corresponding halfspaces containing $v$.

Let $w \in \mathcal{L}_o^{cube}$ represent a vertex. We say $w$ is of Type 1 relative to $(o,v)$ if $w$ is not contained in $h_1 \cap \dots \cap h_n$. We say $w$ is of Type 2 relative to $(o,v)$ if $w$ is contained in $h_1 \cap \dots \cap h_n$ and contained in the carrier of $\hat{h}_i$ for some $i$. It is of Type 3 relative to $(o,v)$ otherwise. (See Definition <ref> for an equivalent, but formally more useful definition.)

Let $\sigma : \Sigma_{0,o} \rightarrow \Sigma_0$ be a label-morphism. We say that $\sigma$ appears at $o$ in $G_o$, if there exists some isometry $g \in G_o$ such that $\pre{\epsilon}{\sigma(g)} = \sigma$.

Let $v \in \mathcal{L}_o^{cube}$ represent a vertex other than $o$ and let $\sigma : \Sigma_{0,v} \rightarrow \Sigma_0$ be a label-morphism. We say that $\sigma$ appears at $v$ in $G_o$ if there exists some isometry $g \in G_o$ such that $\pre{v}{\sigma(g)} = \sigma$ and $g$ fixes all vertices of Type 1 and 2 relative to $(o,v)$.

Suppose $X$ is a regular tree, $\ell$ an admissible edge labeling on $X$, $o$ a vertex in $X$, and let $v \in \mathcal{L}_o^{cube}$ represent a vertex other than $o$. We define $X_v$ to be the subtree containing all the vertices whose unique geodesic to $o$ contains $v$.

In this situation, $v$ is of Type 2 relative to $(o,v)$ and the vertices of Type 1 relative to $(o,v)$ are exactly those that are not contained in $X_v$. If $\sigma : \Sigma_{0,v} \rightarrow \Sigma_0$ is a label-morphism, then it is compatible with $v$ if and only if it fixes the label of the only edge at $v$ that is not contained in $X_v$. Furthermore, the label-morphism $\sigma$ appears at $v$ in $G_o$ if and only if there exists an automorphism $g$ of $X$ that fixes all vertices that are not in $X_v$. (Readers familiar with rooted automorphisms of regular trees will quickly observe that every label-morphism that is compatible with $v$ also appears at $v$ in $G_o$.)

As can be seen from the example above, the last few definitions heuristically mean the following: A label-morphism appears at $v$ in $G_o$, if there exists an element in $G_o$ that performs the permutation $\sigma$ on the edges emanating from $v$ and fixes all vertices that are not `behind $v$' from the vantage point of $o$. A label-morphism $\sigma$ is compatible with $v$ if it satisfies the basic requirements needed in order to appear at $v$ in $G_o$. The reasons for these definitions will become clear throughout the paper.

Suppose that for every reduced $v \in \mathcal{L}_o^{cube} \setminus \{ \epsilon \}$, every $\sigma$ that is compatible with $v$ also appears at $v$ in $G_o$. Then there exists a finitely generated dense subgroup of $\Aut(X)$.

As we will see, the assumption stated in Theorem <ref> is satisfied if the action of $\Gamma$ on $X$ is transitive on vertices. If the action of $\Gamma$ is not vertex-transitive, but cocompact, we can still find a finite set $V \subset \mathcal{L}_o^{cube} \setminus \{ \epsilon \}$ such that we only need to require for all $v \in V$ that every $\sigma$ compatible with $v$ also appears at $v$ in $G_o$ (see Lemma <ref> and Theorem <ref>). It seems likely that, with the development of some additional tools to understand $\mathcal{L}_o^{cube}$ in the non-vertex-transitive case, this assumption can either be described more concretely or removed.

We can get a more explicit version of Theorem <ref> if we assume the action of $\Gamma$ to be transitive on vertices. In that case, every letter in $\Sigma_0$ appears at every vertex of $X$, that is $\Sigma_{0,v} = \Sigma_0$ for every vertex $v$, and our label-morphisms become bijective maps $\sigma : \Sigma_0 \rightarrow \Sigma_0$. In this case, we call these maps label-isomorphisms. We also highlight that, if the action of $\Gamma$ is vertex-transitive, then one can easily show that $\Gamma$ is a right-angled Artin group and $X$ its Salvetti complex (see Remark <ref> for details).

If $\Gamma$ acts vertex-transitively on $X$, then we can do the following: For any vertex $o \in X^{(0)}$, any reduced word $v \in \mathcal{L}_o^{cube}$, and any label-isomorphism $\sigma$ that is compatible with $v$, we can construct an element $A_{v, \sigma} \in G_o$ such that $\pre{v}{A_{v,\sigma}} = \sigma$ and $A_{v, \sigma}$ fixes all vertices in $X$ that are of Type 1 or 2 relative to $(o,v)$. These automorphisms allow us to build a topologically generating set in the sense of the following Theorem.

Suppose $\Gamma$ acts vertex-transitively on $X$. Let $S$ be a finite generating set of $\Gamma$ such that $S^{-1} = S$. The set

\[ S \cup \{ A_{o, \sigma} \vert \sigma \text{ a label-isomorphism} \} \cup \{ A_{\alpha, \sigma} \vert \alpha \in \Sigma, \sigma \text{ compatible with } \alpha \} \]

generates a dense subgroup of $\Aut(X)$.

We now turn to the size of the automorphism group. The size of $\Aut(X)$ depends significantly on the `fractally self-repeating nature' of $\mathcal{L}_o^{cube}$. Since $\Aut(X)$ is a second countable, locally compact group, it is non-discrete if and only if it is uncountable. (One direction of this equivalence uses Baire's category theorem, the other simply follows from second countability of $\Aut(X)$.) Thus, the size of $\Aut(X)$ can be characterised in terms of whether it is discrete or not.

Suppose there exists some $o \in X^{(0)}$ such that for all reduced $v \in \mathcal{L}_o^{cube} \setminus \{ \epsilon \}$, every $\sigma$ that is compatible with $v$ also appears at $v$ in $G_o$. Then exactly one of the following holds:

* $\Aut(X)$ is finitely generated. Furthermore, all vertex-stabilizers are finite.

* $\Aut(X)$ is non-discrete.

In particular, the vertex-stabilizers are either uncountable or finite.

In the vertex-transitive case, there is a characterisation when $\Aut(X)$ is non-discrete entirely in terms of $\Gamma$. This characterisation has been proven independently in [21]. Unfortunately, this work seems to be unpublished and is difficult to get a hand on, so we provide the result including our proof using the technology developed in this paper.

[cf. [21]]
Suppose the action $\Gamma \curvearrowright X$ is vertex-transitive. Then the following are equivalent:

* The automorphism-group $\Aut(X)$ is non-discrete.

* There exist some distinct $s, t \in \Sigma_0$ such that $s \neq t^{-1}$ and $[s, t] \neq 1$.

* $\Gamma$ is not free abelian.

§.§ The subgroup $\text{Aut}^{+}(X)$ and simplicity

Our remaining results require some additional assumptions. A $\CAT$ cube complex $X$ is called reducible, if it is isomorphic to a non-trivial product. It is called irreducible, if it is not reducible. We say $X$ is essential, if every halfspace $h$ in $X$ contains points that are arbitrarily far away from its bounding hyperplane $\hat{h}$. A group $G < \Aut(X)$ is called elementary, if its limit set $\Lambda(G)$ has at most two points. It is called non-elementary, if it is not elementary. We will require that $X$ is irreducible and essential and that $\Aut(X)$ is non-elementary. Note that, if $\Aut(X)$ is non-elementary, then $X$ cannot be isometric to the real line.

We define $\Aut^{+}(X)$ to be the subgroup generated by all elements in $\Aut(X)$ that fix some halfspace of $X$ pointwise. We emphasize that we do not take the closure and $\Aut^{+}(X)$ may not be closed in general. One easily checks that $\Aut^{+}(X)$ is a normal subgroup of $\Aut(X)$, because conjugation sends halfspace-fixators to halfspace-fixators. Thus, this group is bound to appear when studying questions of simplicity in the automorphism group.

Both the literature on automorphisms of right-angled buildings (e.g. [7, 13, 12, 3, 4]) and the literature on automorphisms of $\CAT$ cube complexes (e.g. [17]) have a group denoted $\Aut^{+}(X)$. However, the definitions of this group in these two bodies of literature are distinct and likely do not coincide outside of some special cases. Both notions of $\Aut^{+}(X)$ are generalisations of the definition given by Tits [22], who studied the case where $X$ is a tree. It is worth noting that $\Aut^{+}(X)$ enjoys similar properties in both cases, although some properties are significantly harder to prove in one context than the other. Throughout this paper, we work with the definition given above, which is taken from [17] and is commonly used for automorphisms of $\CAT$ cube complexes.

In <cit.>, Lazarovich provides a sufficient condition for $\Aut^{+}(X)$ to be simple. We can combine his work with our results on non-discreteness to prove the following theorem.

Suppose $X$ is irreducible and essential, suppose $\Aut(X)$ is non-elementary, and suppose that the action $\Gamma \curvearrowright X$ is vertex-transitive.

Suppose that for every pair $s, t \in \Sigma_0$ such that $[s,t] = 1$, there exists $r \in \Sigma_0 \setminus \{ s^{-1}, t^{-1} \}$ such that $[s,r] \neq 1$ and $[t,r] \neq 1$. Then $\Aut^{+}(X)$ is a non-discrete, simple, totally disconnected group.

Suppose $\Gamma$ and $X$ are as in the theorem above. Using Theorem <ref>, one easily checks that, because $\Gamma$ acts vertex-transitively on $X$, there exists a finite graph $L$ such that $\Gamma$ is the right-angled Artin group induced by $L$ and $X$ is its corresponding Salvetti complex (cf. Remark <ref>). One can now reformulate the condition on labels required in Theorem <ref> as follows: Suppose that for every edge $e$ in $L$ there exists a vertex that is not adjacent to either endpoint of $e$. Then $\Aut^{+}(X)$ is non-discrete, simple, and totally disconnected. (Note that $L$ cannot be a single isolated vertex, as $\Aut(X)$ is assumed to be non-elementary.)

One of the reasons why simple, non-discrete, tdlc groups are of interest is their relevance in developing a structure theory of tdlc groups (see [9, 10] for an in-depth discussion of current developments on this subject). Since $\Aut(X)$ is tdlc whenever $X$ is locally finite and its non-discreteness can be understood in terms of Theorems <ref> and <ref>, the groups we are studying are an interesting class of examples to consider in the context of such a structure theory. Theorem <ref> tells us that, under some extra conditions, $\Aut^{+}(X)$ is a simple, non-discrete, totally disconnected subgroup of $\Aut(X)$. However, it is not a-priori obvious that $\Aut^{+}(X)$ is locally compact. The following result adresses this.

Suppose that the action $\Gamma \curvearrowright X$ is vertex-transitive and suppose there exists some $s \in \Sigma_0$ such that for any two distinct $t, r \in \Sigma_0$ such that $[s,t] = [s,r] = 1$, we have either $t = r^{-1}$ or $[t,r] = 1$. Then $\Aut^{+}(X)$ is locally compact.

Let $L$ be a finite graph, $\Gamma$ its induced right-angled Artin group, and $X$ the corresponding Salvetti complex. The condition required from $\Sigma_0$ in the theorem above can again be formulated as a property of $L$. Namely, $\Aut^{+}(X)$ is locally compact if there exists a vertex $v$ in $L$ such that all vertices adjacent to $v$ are adjacent to each other. In particular, if $L$ contains an isolated vertex, or a vertex of degree one, then $\Aut^{+}(X)$ is locally compact.

While this provides a sufficient condition for $\Aut^{+}(X)$ to be locally compact, it seems likely that one can obtain local compactness results for more examples by further enhancing the methods developed in this paper.

With $\Aut^{+}(X) \triangleleft \Aut(X)$ being a simple subgroup, a naturally arising question is how big $\Aut^{+}(X)$ is inside $\Aut(X)$. Lazarovich showed in [17] that $\Aut^{+}(X)$ has finite index in $\Aut(X)$ in some specific cases. We obtain a sufficient condition for the closure of $\Aut^{+}(X)$ to have finite index in $\Aut(X)$.

Let $L$ be a graph. We say $L$ is flexibly vertex-transitive, if for all vertices $v, w$ in $L$, there exists a vertex $u$ and a graph-automorphism $\varphi \in \Aut(L)$ such that $\varphi$ fixes $u$ and all its neighbours and $\varphi(v) = w$.

See section <ref> for a discussion of flexible vertex-transitivity. For now, we simply point out that flexible vertex-transitivity is a very strong condition that is rarely satisfied.

Since we assume that $\Gamma$ acts vertex-transitively on $X$, the links at the vertices of $X$ are all canonically isomorphic. We denote by $\lk(X)$ the link at any vertex of $X$.

Suppose $\Gamma$ acts vertex-transitively on $X$ and suppose the $1$-skeleton of $\lk(X)$ is flexibly vertex-transitive. Then $\overline{\Aut^{+}(X)}$ has finite index in $\Aut(X)$.

This provides us with a sufficient condition for the closure of $\Aut^{+}(X)$ to be large in $\Aut(X)$. On the other hand, we can present a relatively basic example in which $\overline{ \Aut^{+}(X) }$ is non-trivial, even uncountable and non-discrete, yet not even cocompact in $\Aut(X)$. This answers a question by Haglund. Our example is as follows: Let $\Gamma_{CK}$ be the right-angled Artin group defined by

\[ \Gamma_{CK} := \langle a, b, c, d \vert [a,b] = [b,c] = [c,d] = 1 \rangle. \]

Let $X_{CK}$ be the Salvetti-complex of $\Gamma_{CK}$.

The quotient $\faktor{ \Aut(X_{CK}) }{\overline{ \Aut^{+}(X_{CK}) }}$ is not compact, while $\Aut^{+}(X_{CK})$ is simple, non-discrete, and tdlc.

We point out that $X_{CK}$ does not satisfy the condition on commuting pairs $s,t$ required in Theorem <ref>. However, this condition is only required to prove that $\Aut(X_{CK})$ acts faithfully on $X_{CK}$, which can be proven in a different way for this particular example. So the conclusion of Theorem <ref> still holds for $X_{CK}$ (see section <ref> or <cit.>). The condition required in Theorem <ref> is satisfied and thus its conclusion holds here.

§.§ About the proofs

Key to our proofs is our ability to describe the stabilizer subgroups $G_o$ and generate them by elements that enjoy useful properties. We do so by noticing that $G_o$ canonically embeds into the automorphism group of a particular rooted tree.

The language $\mathcal{L}_o^{cube}$ carries the structure of a prefix tree (because this language is regular, see Proposition <ref>). That is, we obtain a rooted tree $T_o^{cube}$ with root $\epsilon$, whose vertices are given by $\mathcal{L}_o^{cube}$ such that $v$, $w$ are connected by an edge if and only if $w = v \alpha$ for some $\alpha$ in $\Sigma$. (This is called the path language tree and it is a standard construction that can be done for any regular language.) Since any automorphism $g \in G_o$ fixes $o$ and sends cube paths to cube paths, it induces a rooted tree-automorphism on $T_o^{cube}$. This provides us with a canonical embedding $G_o < \Aut(T_o^{cube})$ (see Lemma <ref>).

Our key technical result is a description of $G_o$ as a subgroup of $\Aut(T_o^{cube})$. For this, we use that rooted tree-automorphisms can be described in terms of a family of bijections.

Let $\mathcal{L}$ be a regular language over the alphabet $\Sigma$. For all $v \in \mathcal{L}$, we denote

\[ \Sigma(v) := \{ \alpha \in \Sigma \vert v\alpha \in \mathcal{L} \}. \]

A portrait in a language $\mathcal{L}$ is a family of maps $(\pre{v}{\sigma})_{v \in \mathcal{L}}$ such that $\pre{v}{\sigma} : \Sigma(v) \rightarrow \Sigma$ is injective for every $v \in \mathcal{L}$. For any portrait, we define the map $\sigma : \mathcal{L} \rightarrow \Sigma^*$ by

\[ \sigma(\alpha_1 \dots \alpha_n) := \pre{\epsilon}{\sigma}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_{i-1}}{\sigma}(\alpha_i) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n). \]

Denote the path language tree corresponding to $\mathcal{L}$ by $T$. We say a portrait defines an automorphism on $T$ if $\sigma$ is the vertex-map of a rooted automorphism on $T$. (Since rooted automorphisms on $T$ are uniquely determined by what they do on vertices, $\sigma$ determines the entire automorphism.) We will discuss in section <ref> how to check whether a portrait defines an automorphism on $T$.

Given an element $g \in G_o$, its portrait $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$ defined earlier is exactly the portrait in $\mathcal{L}_o^{cube}$ that recovers $g$ as an element in $\Aut(T_o^{cube})$. Thus, this notion of a portrait on a tree is compatible with the notion of portrait we defined previously for elements $g \in G_o$.

A portrait in the language $\mathcal{L}_o^{cube}$ is called admissible if the following two conditions hold:

* For every pair $v, w \in \mathcal{L}_o^{cube}$ such that $v \equiv_o w$, we have $\pre{v}{\sigma} = \pre{w}{\sigma}$.

* For every $v \in \mathcal{L}_o^{cube}$, the restriction of $\pre{v}{\sigma}$ to $\Sigma_{0,v}$ is a label-morphism and $\pre{v}{\sigma}$ is the canonical extension of that label-morphism to $\Sigma_v$.

Suppose $(\sigma_v)_{v \in \mathcal{L}_o^{cube}}$ is an admissible portrait that defines an automorphism on $T_o^{cube}$. We define the following two properties that this family may satisfy:

(par) For all $v \in \mathcal{L}_o^{cube}$, for all non-isolated $s \in \Sigma_{0,v}$, and for all $\beta \in \Sigma_v$ such that $[\beta, s ]  = 1$, we have

\[ \pre{v}{\sigma(g)}(\beta) = \pre{vs}{\sigma(g)}(\beta). \]

(inv) For all $\alpha_1 \dots \alpha_n \in \mathcal{L}_o^{cube}$ and all $s \in \alpha_n$, we have

\[ \pre{\alpha_1 \dots \alpha_n}{\sigma(g)}(s^{-1}) = \pre{\alpha_1 \dots \tilde{\alpha}_n}{\sigma(g)}(s)^{-1}, \]

where $\tilde{\alpha}_n := \alpha_n \setminus \{ s \}$. (If $\tilde{\alpha}_n$ is empty, then $\alpha_1 \dots \tilde{\alpha}_n = \alpha_1 \dots \alpha_{n-1}$.)

Let $o \in X^{(0)}$ and let $(\pre{v}{\sigma})_{v \in \mathcal{L}_o^{cube}}$ be a portrait in $\mathcal{L}_o^{cube}$. There exists some $g \in G_o$ such that $(\pre{v}{\sigma})_{v \in \mathcal{L}_o^{cube}}$ is the portrait of $g$ if and only if $(\pre{v}{\sigma})_{v \in \mathcal{L}_o^{cube}}$ is admissible, defines an automorphism on $T_o^{cube}$, and satisfies $par$ and $inv$.

The  characterisation of stabilizer elements as portraits that are admissible, define an automorphism on $T_o^{cube}$, and satisfy $par$ and $inv$, allows us to work and reason with the local behaviour of automorphisms. In addition, it enables us to explicitly construct useful stabilizer elements and do computations with them. This is the key to most of our results.

The remainder of the paper is structured as follows. In section <ref>, we establish all the necessary knowledge on rooted tree-automorphisms, $\CAT$ cube complexes, and their edge labelings. In particular, we discuss properties of the language $\mathcal{L}_o^{cube}$ and define the language $\mathcal{L}_o$ induced by normal cube paths. In section 3, we discuss the stabilizer subgroups $G_o$ and prove Theorem <ref>. In section <ref>, we prove Theorem <ref>. In section <ref>, we produce generators of dense subgroups of the stabilizer subgroups $G_o$ and use them to produce finitely generated dense subgroups of $\Aut(X)$, proving Theorems <ref> and <ref>. In section <ref>, we prove Theorems <ref> and <ref>. In section <ref>, we move to the subgroup $\Aut^{+}(X)$, prove its simplicity, and study its size in $\Aut(X)$, which yields Theorems <ref>, <ref>, <ref>, and <ref>.

§.§ Acknowledgments.

The authors thank Pierre-Emmanuel Caprace, Frédéric Haglund, Nir Lazarovich, and Frédéric Paulin for their comments and suggestions. We also thank Federico Berlai and Michal Ferov for informing us about and helping us find [21]. Finally, we thank Claudio Llosa Isenrich for giving a talk at KIT that sparked this project and for pointing us towards several useful references. The second author was partially supported by the FWF grant 10.55776/ESP124.

§ ENCODINGS OF $\MATHRM{CAT(0)}$ CUBE COMPLEXES

Throughout this article we are going to use the following notation:

* $X$ is a locally finite $\CAT$ cube complex with vertex set $X^{(0)}$ and $o \in X^{(0)}$ is a basepoint.

* Automorphisms of $X$ are cubical isometries from $X$ to itself and we denote the group of automorphisms on $X$ by $\Aut(X)$.

* We denote by $\mathfrak h(X)$ the set of halfspaces in $X$ and by $\widehat{\mathfrak h}(X)$ the set of hyperplanes. Given $h \in \mathfrak h(X)$ we denote by $\widehat{h}$ the bounding hyperplane and by $h^c$ the opposite halfspace. Given a cube $C$ we denote by $\widehat{\mathfrak h}(C)$ the set of hyperplanes which intersect $C$. We say that $\widehat{h}_1, \widehat{h}_2 \in \widehat{\mathfrak h}(X)$ intersect transversally and write $\widehat{h}_1 \pitchfork \widehat{h}_2$ if they intersect and are not equal.

* We denote by $\vec{\mathcal D}(X)$ the set of oriented diagonals of cubes in $X$. If $v$ and $w$ are opposite vertices on a cube $C$, then we denote by $(v, C, w)$ the oriented diagonal of $C$ from $v$ to $w$ and refer to $C$ as its underlying cube. There is a fixpoint free involution on $\vec{\mathcal D}$ given by $(v,C, w)^{-1} := (w, C, v)$. We refer to the corresponding orbit $[v, C, w]$ as a diagonal and denote the set of diagonals by
$\mathcal D(X)$.

* $\vec{\mathcal E}(X) \subset \vec{\mathcal D}(X)$ denotes the subset of oriented edges (i.e. oriented diagonals of $1$-dimensional cubes)  and $\mathcal E(X) \subset \mathcal D(X)$ denotes the corresponding subset of geometric edges. Given an oriented edge $e = (v, C, w)$ we denote by $h(e)$ the unique halfspace containing $w$, but not $v$. We then denote by $\widehat{h}(e)$ the corresponding hyperplane. If $[e]$ is the corresponding geometric edge, then we set

\[\widehat{h}([e]) := \widehat{h}(e) = \widehat{h}(e^{-1}).\]

* Two oriented edges $e, e'$ are parallel, if $h(e) = h(e')$. We call the resulting equivalence classes oriented parallel classes. Oriented parallel classes are in 1–1 correspondence with $\mathfrak{h}(X)$ and – for our purposes – they are a useful generalisation of halfspaces to general cube complexes.

* If $S$ denotes a cube complex, we abuse notation and denote the set of oriented parallel classes by $\mathfrak{h}(S)$. Similarly, we write $h(e)$ for the oriented parallel class containing an oriented edge $e$ and $h^c$ for the oriented parallel class obtained by reversing the orientation of all elements of $h \in \mathfrak{h}(S)$.

* We denote by $\Gamma_{\mathcal D}(X)$ the graph with vertex set $X^{(0)}$ and oriented edge set $\vec{\mathcal D}(X)$ with source and target map given by $s(v,C,w) = v$ and $t(v,C, w) = w$. There is a natural immersion $|\Gamma_{\mathcal D}(X)|\looparrowright X$ obtained by sending each edge in $\vert \Gamma_{\mathcal{D}}(X) \vert$ to the corresponding diagonal in $X$.

* A path in $\Gamma_{\mathcal D}(X)$ is called a cube path in $X$; if all underlying cubes are $1$-dimensional, then it is called an edge path. We denote by $D(X)$ and $E(X)$ the collections of all cube paths and edge path respectively. We consider $\vec{\mathcal D}(X)$ and $\vec{\mathcal E}(X)$ as subsets of $D(X)$ by considering oriented diagonals as cube paths of length one.

* Using the immersion $|\Gamma_{\mathcal D}(X)|\looparrowright X$, cube paths correspond to curves in $X$ which move exclusively along edges and diagonals, and edge paths correspond to curves in the $1$-skeleton of $X$. It thus makes sense to say that a cube path intersects a hyperplane in $X$.

* We abbreviate a cube path $((v, C_1, v_1), \dots, (v_{n-1}, C_n, w)) \in D(X)$ by $(v, C_1, \dots, C_n, w)$ and refer to $(C_1, \dots, C_n)$ as the underlying sequence of cubes.
We then denote by $\widehat{\mathfrak h}(v, C_1, \dots, C_n, w)$ the set of hyperplanes which intersect at least one of the cubes $C_1, \dots, C_n$.

* If $v \in X^{(0)}$ is a vertex, then we denote by $D(X)_v$ the collection of all cube paths emanating from $v$. The subsets $E(X)_v$, $\vec{\mathcal D}(X)_v$, $\vec{\mathcal E}(X)_v$ are defined accordingly.

§.§ Labelings of $\mathrm{CAT(0)}$ cube complexes

Let $\Sigma_0$ be a finite set with a fixpoint free involution $s \mapsto s^{-1}$.

Let $S$ be a cube complex. A map $\ell: \mathfrak h(S) \to \Sigma_0$ is
an oriented parallel class labeling of $S$ if

\[
\ell(h^c) = \ell(h)^{-1} \quad \text{for all }h \in \mathfrak h(S).
\]

If $S$ is $\CAT$, we also call this a half-space labeling. Either way, the induced edge labeling is defined as

\[\ell: \vec{\mathcal{E}}(S) \to \Sigma_0, \quad \ell(e) := \ell(h(e)).\]

We will be interested in a special class of induced edge labelings. To define these we introduce the following notion:

We say that two geometric edges $[e], [e']$ span a square if they have precisely one endpoint in common and if
$\widehat{h}([e]) \pitchfork \widehat{h}([e'])$. We say that two oriented edges span a square if the underlying geometric edges span a square.

A map $\ell: \vec{\mathcal{E}}(S) \to \Sigma_0$ is called an admissible edge labeling if the following hold:

* $\ell$ is induced from an oriented parallel class labeling.

* $\ell\vert_{\vec{\mathcal{E}}(S)_v}$ is injective for every $v \in S^{(0)}$.

* For all $v,w \in S^{(0)}$, $e_1, e_2 \in \vec{\mathcal{E}}(S)_v$, and $e'_1, e'_2 \in \vec{\mathcal{E}}(S)_w$, the following holds: If $\ell(e_1) = \ell(e_1')$ and $\ell(e_2) = \ell(e_2')$, then $e_1$ and $e_2$ span a square if and only if $e_1'$ and $e_2'$ span a square.

We then refer to the pair $(S, \ell)$ as a $\Sigma_0$-labeled cube complex.

If a group $\Gamma$ acts by automorphisms on $S$, then an admissible edge labeling is called $\Gamma$-invariant if $\ell(g \cdot e) = \ell(e)$ for all $e \in \vec{\mathcal{E}}(S)$.

We want to give a sufficient condition for the existence of an admissible labeling. As we will see later, this sufficient condition is also necessary. To describe this, we need the notion of special cube complexes introduced by Haglund and Wise in [16]. We use the terminology from <cit.> regarding special cube complexes.

Let $X$ be the universal covering of a compact, non-positively curved special cube complex $S$ with fundamental group $\Gamma$. Then there exists a $\Gamma$-invariant admissible edge labeling on $X$.

Our strategy is to put a labeling on the oriented parallel classes in $S$ which lifts to an admissible labeling on $X$. As the lift of a labeling in $S$, this will be $\Gamma$-invariant.

Since $S$ is compact, we can choose a sufficiently large finite set $\Sigma_0$ and an involution $\cdot^{-1}$ on $\Sigma_0$ such that there exists an injective map $\ell : \mathfrak{h}(S) \rightarrow \Sigma_0$ satisfying $\ell(h^c) = \ell(h)^{-1}$ for all $h \in \mathfrak{h}(S)$. Since all hyperplanes in $S$ are two-sided (that is, every hyperplane induces exactly two oriented parallel classes), $\cdot^{-1}$ is fixpoint free. We denote the induced edge labeling by $\ell$ as well and the lift of $\ell$ to $X$ by $\tilde{\ell}$. By construction $\tilde{\ell}$ satisfies (A1), as $\ell$ lifts both as a labeling of oriented parallel classes and as an edge labeling and the two lifts are compatible with each other. We are left to show that $\tilde{\ell}$ satisfies (A2) and (A3).

We first show that, if $\ell$ satisfies (A2) and (A3), then so does $\tilde{\ell}$. For (A2) this is obvious, as it is a local property and $X$ is a covering of $S$. For (A3), suppose we have vertices $v, w \in X^{(0)}$, $e_1, e_2 \in \vec{\mathcal{E}}(X)_v$, and $e'_1, e'_2 \in \vec{\mathcal{E}}(X)_w$ such that $\tilde{\ell}(e_1) = \tilde{\ell}(e'_1)$ and $\tilde{\ell}(e_2) = \tilde{\ell}(e'_2)$. Denoting the projection map by $p : X \rightarrow S$, this implies that $\ell(p(e_1)) = \ell(p(e'_1))$ and $\ell(p(e_2)) = \ell(p(e'_2))$. Suppose that $e_1, e_2$ span a square, while $e'_1$ and $e'_2$ do not. Since $p$ is a covering map, this implies that $p(e_1), p(e_2)$ span a square, while $p(e'_1), p(e'_2)$ do not. We see that, if $\tilde{\ell}$ does not satisfy (A3), neither does $\ell$. Therefore, it is sufficient to show that $\ell$ satisfies (A2) and (A3) to conclude that $\tilde{\ell}$ is admissible.

We first show (A2). Let $v \in S^{(0)}$ and suppose there exist edges $e_1, e_2 \in \vec{\mathcal{E}}(X)_v$ such that $\ell(e_1) = \ell(e_2)$. Since $\ell$ is injective on oriented parallel classes, we conclude that $h(e_1) = h(e_2)$. This means that the hyperplane $\hat{h}(e_1)$ self-osculates which contradicts the assumption that $S$ is special. We conclude that $\ell$ satisfies (A2).

We are left to show (A3). Let $v, w \in S^{(0)}$, $e_1, e_2 \in \vec{\mathcal{E}}(X)_v$, and $e'_1, e'_2 \in \vec{\mathcal{E}}(X)_w$ such that $\ell(e_1) = \ell(e'_1)$ and $\ell(e_2) = \ell(e'_2)$. Since $\ell$ is injective on halfspaces by assumption, this implies that $e_1$ and $e'_1$ cross the same hyperplane in the same direction and analogously for $e_2$ and $e'_2$. Let $\hat{h} = \hat{h}(e_1)$ and $\hat{k} = \hat{h}(e_2)$ and suppose that $e_1, e_2$ span a square, while $e'_1$ and $e'_2$ do not. This means that $\hat{h}$ and $\hat{k}$ interosculate, which is ruled out by the fact that $S$ is special. We conclude that $\ell$ satisfies (A3). This proves that its lift $\tilde{\ell}$ is a $\Gamma$-invariant admissible edge labeling.

The Salvetti complex of a RAAG admits an admissible edge labeling, which is invariant under the given RAAG. In fact, this is exactly the class of examples for which $\Gamma$ acts vertex-transitively on $X$ (see Remark <ref>).

Most surface groups appear as fundamental groups of compact, non-positively curved special cube complexes [11]. Furthermore, the fundamental group of any closed, hyperbolic 3-manifold $M$ has a finite index subgroup that appears as the fundamental group of a compact, non-positively curved special cube complex [1]. Thus, the cubulations of these groups provide us with $\CAT$ cube complexes that have admissible edge-labelings which are invariant under the corresponding group action.

§.§ Cube paths in labeled $\mathrm{CAT(0)}$ cube complexes

From now on, $(X, \ell)$ denotes a $\Sigma_0$-labeled $\CAT$ cube complex. We refer to $\Sigma_0$ as the edge-label set of $(X, \ell)$.
[Labeling oriented diagonals]
We can identify oriented diagonals in $X$ with sets of oriented edges as follows: If $d = (v,C,w)$ is an oriented diagonal, then we define its oriented edge set as

\[
\vec{\mathcal E}_d := \{(v, e, v') \in \vec{\mathcal{E}}(X)_v \mid e \text{ is an edge of }C\},
\]

and obtain an embedding

\[
\vec{\mathcal D}(X) \to 2^{\vec{\mathcal E}(X)}, \quad d \mapsto\vec{\mathcal E}_d.
\]

It is thus natural to extend $\ell$ to oriented diagonals (see also figure <ref>) by setting

\[
\ell(v,C,w) := \ell(\vec{\mathcal E}_d) \in 2^{\Sigma_0}.
\]

Given a vertex $v$, we denote by

\[
\Sigma_v := \ell(\vec{\mathcal{D}}(X)_v) \quad \text{and} \quad \Sigma_{0,v} := \ell(\vec{\mathcal{E}}(X)_v)
\]

the sets of labels at $v$ and edge-labels at $v$ respectively. By (A2) these sets are in bijection with the outgoing diagonals and outgoing edges at $v$ respectively.

The following crucial definition will enable us to describe geometric properties and manipulations of cube paths in algebraic terms.

Let $s,t \in \Sigma_0$, $\alpha, \beta \in 2^{\Sigma_0}$ and $v \in X^{(0)}$.
We say that $s$ and $t$ commute at $v \in X^{(0)}$ if there exist edges $e, e' \in \vec{\mathcal E}(X)_v$ which span a square such that $\ell(e) = s$ and $\ell(e') = t$. We say that $s$ and $t$ commute if $s$ and $t$ commute at some vertex $v$ and we write $[s,t] = 1$. Similarly, we say that $\alpha$ and $\beta$ commute (at $v$) if $\alpha, \beta \in \Sigma_v$ and every $s \in \alpha$ commutes (at $v$) with every $t \in \beta$. We write $[\alpha, \beta] = 1$.

Note that, if edge-labels $s$ and $t$ commute, then by (A3) they commute at every vertex $v$ with $s, t \in \Sigma_{0, v}$. Similarly, if labels $\alpha$ and $\beta$ commute, then they commute at every vertex $v$ with $\alpha, \beta \in \Sigma_v$. We define the set

\[ \Sigma := \{ \alpha \in 2^{\Sigma_0} \setminus \{ \emptyset \} \vert \forall s, t \in \alpha : (s = t) \vee ( [s,t] = 1 ) \]

and see that the extended labeling is a map

\[ \ell : \vec{\mathcal{D}}(X) \rightarrow \Sigma. \]

Note that there may be elements of $\Sigma$ that do not lie in any $\Sigma_v$. This happens when there is a collection of edge-labels $s_1, \dots s_n$ that commute pairwise, but for which there exists no vertex $v$ such that they are all contained in $\Sigma_{0,v}$. These elements are not contained in the image of $\ell$, as they do not correspond to any diagonal in $X$.

Geometrically, commutation of labels is a property of the links of $X$: By definition, the vertices of $\lk(v)$ are given by the outgoing edges at $v$, which we can identify with $\Sigma_{0,v}$ via $\ell$. Under this identification, two elements $s,t \in \Sigma_{0,v}$ are connected by an edge in $\lk(v)$ if and only if the commute. Similarly, if $\alpha, \beta \in \Sigma_v$, then $\alpha$ and $\beta$ commute if and only if the vertices in $\lk(v)$ corresponding to the set $\alpha \cup \beta$ span a simplex in $\lk(v)$. Equivalently, $\alpha, \beta \in \Sigma_v$ commute if and only if the corresponding oriented diagonals at $v$ lie in (and hence span) a common cube of dimension $|\alpha| + |\beta|$. Using this interpretation of commutation in links, one easily checks that $\Sigma_v = \Sigma \cap 2^{\Sigma_{0,v}}$ because of the link-condition of $\CAT$ spaces.

[Trivial examples]  Since admissible edge labelings satisfy (A2), $s$ does not commute with itself. We claim that $s$ and $s^{-1}$ do not commute either. Indeed, suppose $[s, s^{-1}] = 1$. Then there exists some vertex $v$ with two outgoing edges $e, e'$, labeled $s$ and $s^{-1}$ respectively, that span a square (see figure <ref>). Let $w$ be the vertex at the other end of $e'$ and $e''$ the outgoing edge at $w$ crossing $\hat{h}(e)$. Condition (A1) implies that $\ell(e'') = s$ and $\ell(e'^{-1}) = s$, which contradicts (A2) at the vertex $w$. Therefore, $s$ and $s^{-1}$ cannot commute.

[Commutation and inverses]
If $[s,t] = 1$, then $[s, t^{-1}] = [s^{-1}, t] = [s^{-1}, t^{-1} ] = 1$. This follows from considering a vertex which has outgoing edges $e_1, e_2$ with labels $s$ and $t$ and looking at the outgoing edges at every corner of the square spanned by $e_1, e_2$ (see figure <ref>).

[Isolated labels]
A label $s \in \Sigma_{0,v}$ corresponds to an isolated vertex in $\lk(v)$ if and only if it does not commute with any $t \in \Sigma_{0,v}$. We say that
a label $s \in \Sigma_0$ is isolated, if it does not commute with any $t \in \Sigma_0$. Geometrically, this means that all $s$-colored edges are isolated points in the corresponding links.

The basic rules about commutation and inversion can be seen from these two squares.

We can now extend our labeling $\ell: \vec{\mathcal D}(X) \to \Sigma$ further to a labeling $\ell: D(X) \to \Sigma^*$ such that

\[\ell((v_1, C_1, v_2), \dots, (v_n, C_n, v_{n+1})) = \ell(v_1, C_1, v_2) \cdots \ell(v_n, C_n, v_{n+1}).\]

The language of cube paths of $(X, \ell, o)$ is defined as
\[\mathcal L^{\mathrm{cube}}_{o} := \ell(D(X)_o) \subset \Sigma^*.\]

We observe that $\ell$ induces a bijection $\ell: D(X)_o \to  \mathcal L^{\mathrm{cube}}_{o} $, i.e. every cube path emanating from $o$ is uniquely determined by the corresponding word over $\Sigma$.

Suppose a subgroup $\Gamma < \Aut(X)$ acts cocompactly on $X$ and let $\ell$ be a $\Gamma$-invariant admissible edge labeling. Then the language $\mathcal L^{\mathrm{cube}}_{o}$ is a regular language, i.e. recognized by a finite automaton.

We define an equivalence relation on $X^{(0)}$ as follows: If $v_1, v_2 \in X^{(0)}$ are two vertices, then $v_1$ and $v_2$ are equivalent, if $v_2 \in \Gamma v_1$. By $\Gamma$-invariance, this implies that $\Sigma_{v_1} = \Sigma_{v_2}$. We denote by $Q$ the set of equivalence classes of vertices with respect to this equivalence relation. Since $\Gamma$ acts cocompactly on $X$, $Q$ is finite.

Now let $\mathcal A$ be the following finite automaton: The underlying alphabet is $\Sigma$, the set of states is given by $Q$ and the equivalence class of $o$ is the unique initial state. (All states are final states.) The transition function $\delta$ is the partial function on $Q \times \Sigma$ defined as follows:

At a state $\Gamma v$, every $\alpha \in \Sigma_v$ is accepted (the set $\Sigma_v$ does not depend on the choice of representative, as $\Sigma_{gv} = \Sigma_v$ for all $g \in \Gamma$) and $\delta( \Gamma v, \alpha)$ is the equivalence class of the endpoint of the diagonal with label $\alpha$ emanating at $v$.

One easily checks that this automaton accepts all words in $\ell( D(X)_o ) = \mathcal{L}_o^{cube}$. Now suppose $\alpha_1 \dots \alpha_n = w \in \Sigma^*$ is accepted by $\mathcal{A}$. Starting at the vertex $o$, one can inductively construct a cube path that spells out the word $w$. Since $w$ is accepted by $\mathcal{A}$, we know that $\alpha_1 \in \Sigma_o$ and thus there exists a diagonal $D_1$ emanating from $o$ with label $\alpha_1$. If $v_1$ denotes the endpoint of $D_1$, the automaton is now in the state of $\Gamma v_1$ and accepts only elements of $\Sigma_{v_1}$. Since $\mathcal{A}$ accepts $w$, we conclude that $\alpha_2 \in \Sigma_{v_1}$ and we obtain a diagonal emanating at $v_1$ labeled with $\alpha_2$. Continuing inductively, we construct an element $\gamma \in D(X)_o$ such that $\ell(\gamma) = w$. We conclude that $\mathcal{A}$ accepts exactly the language $\mathcal{L}_o^{cube}$.

§.§ Equivalence of paths and words

We say that two cube paths $\gamma_1, \gamma_2 \in D(X)_o$ are equivalent and write $\gamma_1 \equiv_o \gamma_2$ if $\gamma_1$ and $\gamma_2$ have the same endpoints. If $o$ is clear from the context, we just write $\gamma_1 \equiv \gamma_2$. Since $\ell|_{D(X)_o}$ is injective, this induces an equivalence relation (also denoted $\equiv_o$ or $\equiv$) on the language $\mathcal{L}_o^{cube}$. Since $X$ is connected, the quotients $D(X)_o/\equiv$ and $ \mathcal L^{\mathrm{cube}}_{o} /\equiv$ can be identified with $X^{(0)}$. This identification allows for the following notation:

If $w \in \mathcal{L}_o^{cube}$, we can identify $w$ with a unique cube path $D(X)_o$. We may thus talk of words in $\mathcal{L}_o^{cube}$ crossing hyperplanes and having start and endpoints. We write $\Sigma_w$ for the labels of outgoing diagonals at the endpoint of $w$ and $\Sigma_{0,w}$ for the labels of the outgoing edges. Similarly, we write $\mathcal{L}_w^{cube}$ for the language of cube paths that start at the endpoint of $w$. We define $D(X)_w$, $E(X)_w$, $\vec{\mathcal{D}}(X)_w$, and $\vec{\mathcal{E}}(X)_w$ analogously.

The following equivalences hold in $ \mathcal L^{\mathrm{cube}}_{o} $:

* We have $w_1 ss^{-1} w_2 \equiv w_1 w_2$ for all $w_1 \in \mathcal{L}_o^{cube}$, $s \in \Sigma_{0,w_1}$, and $w_2 \in \mathcal{L}_{w_1}^{cube}$.

* We have $w_1 st w_2 \equiv w_1 ts w_2$ for all $s, t \in \Sigma_{0,w_1}$ with $[s,t] = 1$, $w_1 \in \mathcal{L}_o^{cube}$, $w_2 \in \mathcal{L}_{w_1 st}^{cube}$.

* We have $w_1 \alpha \beta w_2 \equiv w_1 \beta \alpha w_2$ for all $\alpha, \beta \in \Sigma_{w_1}$ with $[\alpha, \beta] = 1$, $w_1 \in \mathcal{L}_o^{cube}$, $w_2 \in \mathcal{L}_{w_1 \alpha \beta}^{cube}$.

(i): $w_1$ and $w_2$ are chosen so that $w_1 w_2 \in \mathcal{L}_o^{cube}$ and there exists a cube path $\gamma \in D(X)_o$ such that $\ell(\gamma) = w_1 w_2$. Since $s \in \Sigma_{0,w_1}$, there exists an edge $e \in \vec{\mathcal{E}}(X)_{w_1}$ labeled $s$. We can thus define $\gamma'$ to be the cube path where we insert the path $e * e^{-1}$ into $\gamma$ at $w_1$. By construction, $\gamma$ and $\gamma'$ have the same endpoints and thus

\[ w_1 s s^{-1} w_2 = \ell(\gamma') \equiv \ell(\gamma) = w_1 w_2. \]

(ii): $w_1$, $s$, $t$, and $w_2$ are chosen so that $w_1 st w_2 \in \mathcal{L}_o^{cube}$ and there exists a cube path $\gamma$ such that $\ell(\gamma) = w_1 st w_2$. Let $e * e'$ be the segment of $\gamma$ that spells out the letters $st$ between $w_1$ and $w_2$. Since $[s, t] = 1$, the edges $e^{-1}$ and $e'$ span a square $C$ at $w_1 s$. We define $\gamma'$ to be the path obtained from $\gamma$ by replacing the segment $e * e'$ with the other edge path around $C$ (see figure <ref>). By construction, $\gamma$ and $\gamma'$ are cube paths with the same end points and thus

\[ w_1 t s w_2 = \ell(\gamma') \equiv \ell(\gamma) = w_1 s t w_2. \]

The two words $w_1 st w_2$ and $w_1 ts w_2$ correspond to the two paths going around the indicated square in the two possible ways.

(iii) $w_1$, $\alpha$, $\beta$, and $w_2$ are chosen so that $w_1 \alpha \beta w_2 \in \mathcal{L}_o^{cube}$ and there exists a cube path $\gamma$ such that $\ell(\gamma) = w_1 \alpha \beta w_2$. Since $\alpha, \beta \in \Sigma_{w_1}$ and $[\alpha, \beta] = 1$, there are two cubes $C_{\alpha}, C_{\beta}$ at $w_1$ with diagonals labeled $\alpha$ and $\beta$ respectively such that these cubes span a cube $C_{\alpha, \beta}$ together. We replace the segment in $\gamma$ that spells out $\alpha \beta$ by the concatenation of diagonals $(w_1, \beta, w_1 \beta) * (w_1 \beta, \alpha, w_1 \beta \alpha)$. (This concatenation exists because of the existence of $C_{\alpha, \beta}$). We define $\gamma'$ to be the cube path obtained by replacing the segment that spells out $\alpha \beta$ by the segment $(w_1, \beta, w_1 \beta) * (w_1 \beta, \alpha, w_1 \beta \alpha)$. By construction, this is a cube-path with the same endpoints as $\gamma$ and thus

\[ w_1 \beta \alpha w_2 = \ell(\gamma') \equiv \ell(\gamma) = w_1 \alpha \beta w_2. \]

This proposition explains the terminology of commuting labels. The next Lemma generalises a standard result about edge paths in $\CAT$ cube complexes to cube paths and formulates it using the terminology of $\mathcal{L}_o^{cube}$. For convenience of the reader, we provide a proof using our terminology.

Let $\alpha_1 \dots \alpha_n \in \mathcal{L}_o$. A pair $\alpha_i, \alpha_j$ (with $i < j$) contains an innermost cancellation if and only if there exists a hyperplane that is crossed by $\alpha_i$ and $\alpha_j$ in opposite directions and the path-segments $\alpha_i \dots \alpha_{j-1}$ and $\alpha_{i+1} \dots \alpha_j$ both do not cross any hyperplane twice.

Any cube path $\gamma$ that crosses some hyperplane more than once contains an innermost cancellation. One may start with any hyperplane $\hat{h}$ that is crossed twice by $\gamma$. If it does not produce an innermost cancellation, there has to be another hyperplane $\hat{h}_{new}$ that crosses $\gamma$ twice on a segment of $\gamma$ between two places where $\gamma$ crosses $\hat{h}$. (One of the two crossings of $\hat{h}_{new}$ may happen at the same place where $\gamma$ crosses $\hat{h}$.) We replace $\hat{h}$ by $\hat{h}_{new}$. After finitely many iterations, this yields an innermost cancellation.

If $\alpha_i, \alpha_j$ contains an innermost cancellation, then there exists a labeling pair $s, s^{-1} \in \Sigma_0$ that corresponds to the hyperplane being crossed by $\alpha_i$ and $\alpha_j$. In particular, $s^{-1} \in \alpha_i$ and $s \in \alpha_j$ (or vice-versa).

Let $\alpha_1 \dots \alpha_n \in \mathcal{L}_o$ and suppose $\alpha_i, \alpha_j$ contains an innermost cancellation. Let $\hat{h}$ be a hyperplane crossed by $\alpha_i$ and $\alpha_j$ and let $s, s^{-1} \in \Sigma_0$ be the labels of the oriented edges crossing $\hat{h}$, such that $s^{-1} \in \alpha_i$ and $s \in \alpha_j$.

Then, for all $i \leq k \leq j-1$, $s \in \Sigma_{0, \alpha_1 \dots \alpha_k}$, and and for all $i+1 \leq k \leq j-1$, $[s, \alpha_k] = 1$.

In particular,

\[ \alpha_1 \dots \alpha_n \equiv \alpha_1 \dots \tilde{\alpha}_i \dots \tilde{\alpha}_j \dots \alpha_n, \]

where $\tilde{\alpha}_i = \alpha_i \setminus \{ s^{-1} \}$ and $\tilde{\alpha}_j = \alpha_j \setminus \{ s \}$.

Since $\hat{h}$ crosses every hyperplane crossed by the path from $w_i$ to $w_{j-1}$, we obtain the cubes (indicated as squares in the figure) that are spanned by an edge labeled $s$ and the diagonals of the path $\alpha_{i+1} \dots \alpha_{j-1}$. Thus $s$ commutes with every letter along this path.

The essence of the proof is captured in Figure <ref>. We provide a formal proof for the sake of completeness. To abbreviate notation, we denote $w_k := \alpha_1 \dots \alpha_k$ for all $1 \leq k \leq n$. Let $\hat{h}$, $s$, and $s^{-1}$ as stated in the Lemma. Let $\hat{h'}$ be a hyperplane crossed by the path-segment $\alpha_{i+1} \dots \alpha_{j-1}$. We claim that $\hat{h}$ and $\hat{h'}$ intersect transversally. Indeed, since $\alpha_i \dots \alpha_{j-1}$ and $\alpha_{i+1} \dots \alpha_j$ are both reduced, we know that $\hat{h'}$ separates the vertices $w_{i-1}$ and $w_i$ from the vertices $w_{j-1}$ and $w_j$. On the other hand, $\hat{h}$ separates $w_{i-1}$ and $w_j$ from $w_{i}$ and $w_{j-1}$. We conclude that these four vertices lie in four distinct connected components of $X \setminus ( \hat{h} \cup \hat{h'} )$. Therefore, $\hat{h}$ and $\hat{h'}$ intersect transversally.

Since $s^{-1} \in \alpha_i$, $s \in \alpha_i^{-1}$, and therefore $s \in \Sigma_{0, w_i}$. Furthermore, the edge in $\vec{\mathcal{E}}(X)_{w_i}$ labeled $s$ crosses the hyperplane $\hat{h}$ (because this is how we chose $\hat{h})$. Similarly, since $s \in \alpha_j$, we know that $s \in \Sigma_{0, w_{j-1}}$. We now prove inductively that, for all $i+1 \leq k \leq j-1$, $s \in \Sigma_{0, w_k}$ and $[s, \alpha_k] = 1$.

Suppose $i+1 \leq k \leq j-1$, suppose $s \in \Sigma_{0, w_{k-1}}$, and suppose the outgoing edge at $w_{k-1}$ with label $s$ crosses the hyperplane $\hat{h}$. (The induction start for $k = i+1$ is given by what we have shown before.) Since $k \leq j-1$, every hyperplane crossed by $\alpha_{k}$ is crossed by $\hat{h}$. Thus, the outgoing edges at $w_{k-1}$ labeled by $s$ and the elements of $\alpha_k$ span a cube, $[s, \alpha_k] = 1$, and we obtain an outgoing edge at $w_k$ that is labeled by $s$ and crosses the hyperplane $\hat{h}$. This finishes the induction.

To prove the remainder of the Lemma, we note that for any $\alpha = \beta \cup \{ s \} \in \Sigma_v$, we have $\alpha \equiv_v \beta s \equiv_v s \beta$. This, combined with finitely many applications of Proposition <ref> implies that

\begin{equation*}
\begin{split}
\alpha_1 \dots \alpha_i \dots \alpha_j \dots \alpha_n & \equiv \alpha_1 \dots \tilde{\alpha}_i s^{-1} \dots \alpha_k \dots s \tilde{\alpha}_j \dots \alpha_n\\
& \equiv \alpha_1 \dots \tilde{\alpha}_i s^{-1} s \dots \alpha_k \dots \tilde{\alpha}_j \dots \alpha_n\\
& \equiv \alpha_1 \dots \tilde{\alpha}_i \dots \tilde{\alpha}_j \dots \alpha_n.
\end{split}
\end{equation*}

§.§ The language of normal cube paths

A cube path $(v, C_1, \dots, C_n, w)$ is normal if it does not cross any hyperplane more than once and if for all  $i \in \{ 1, \dots, n-1 \}$ there exists no hyperplane $\widehat{h} \in \widehat{\mathfrak h}(C_{i+1})$ which intersects all hyperplanes in $\widehat{\mathfrak h}(C_i)$ transversally.

The indicated cube path is not a normal cube path. However, the reverse path from $v_2$ to $v_0$ is a normal cube path.

Niblo and Reeves' original definition of normal cube paths in <cit.> does not require that the path crosses every hyperplane at most once. Instead of this condition, they require that for all $1 \leq i \leq n-1$, $\widehat{\mathfrak{h}} (C_i) \cap \widehat{\mathfrak{h}}(C_{i+1}) = \emptyset$ and then they implicitly show that their definition is equivalent to the one given above.

The set of normal cube paths in $X$ will be denoted by $N(X) \subset D(X)$ and for $o \in X^{(0)}$ we define $N(X)_o := N(X) \cap D(X)_o$.
We define the language of normal cube paths at $o$ to be

\[
\mathcal L_o :=  \ell(N(X)_o) \subset  \mathcal L^{\mathrm{cube}}_{o}.
\]

Given a word $\alpha_1 \dots \alpha_n \in \Sigma^*$, one can easily verify from the definition, together with Remark <ref>, that

\[ \alpha_1 \dots \alpha_n \in \mathcal{L}_o \Leftrightarrow \forall 1 \leq i \leq n-1, \forall s \in \alpha_{i+1} : [\alpha_i, s] \neq 1 \text{ and } s^{-1} \notin \alpha_i. \]

Suppose there is a subgroup $\Gamma < \Aut(X)$ that acts cocompactly on $X$ and let $\ell$ be a $\Gamma$-invariant admissible edge-labeling. The language $\mathcal{L}_{o}$ of normal cube paths is regular, i.e. it is the language recognized by a finite automaton.

By [19], every cube path is $\equiv$-equivalent to a unique normal cube path. This shows that we can identify $N(X)_o$ with $X^{(0)}$ and hence obtain a bijective encoding $\mathcal L_{o} \to X^{(0)}$. Moreover, there are
normalization operations

\[
\mathrm{norm}: D(X)_o \to N(X)_o \quad \text{and} \quad \mathrm{norm}:  \mathcal L^{\mathrm{cube}}_{o}   \to \mathcal L_{o},
\]

which map every cube path (or word in the cube path language) to the unique $\equiv_o$-equivalent normal cube path (or word in the normal cube path language).
This allows us to define a partial multiplication on $\bigcup_{o \in X^{(0)}} \mathcal L_o$ as follows:

\[
\forall o \in X^{(0)}, v_1 \in \mathcal{L}_o, v_2 \in \mathcal{L}_{v_1} : v_1 \ast v_2 := \mathrm{norm}(v_1 v_2)
\]

This partial multiplication is associative and defines a groupoid.

To describe the partial multiplication explicitly, it suffices to compute $v \ast s$ for $s \in \Sigma_{0,v}$. For this we need the following observation:

Let $v = \alpha_1 \dots \alpha_n \in \mathcal{L}_o$ and $s \in \Sigma_{0,v}$. Suppose $vs$ crosses some hyperplane more than once. Then there exists some $k \in \{ 1, \dots, n \}$ such that $s^{-1} \in \alpha_k$ and $[s, \alpha_i] = 1$ for all $i > k$.

Since $v$ is a normal cube path, the multi-crossing in $vs$ cannot occur within $v$. We conclude that the only hyperplane that can be crossed more than once by $vs$ is the hyperplane that is crossed in the last step, when passing through the edge labeled by $s$. We denote this hyperplane by $\hat{h}(s)$. If $vs$ crosses $\hat{h}(s)$ more than once, then $v$ has to cross the hyperplane $\hat{h}(s)$. Since $v$ is a normal cube path, it can cross $\hat{h}(s)$ exactly once and in opposite direction from the crossing in the last edge of $vs$. We conclude that $s^{-1} \in \alpha_k$ for some $k$.

By construction, $\alpha_k, s$ is an innermost cancellation. By Lemma <ref>, we conclude that for every $i > k$ and every $t \in \alpha_i$, $[s^{-1}, t] = 1$ (see also figure <ref>). This proves the Lemma.

Let $v = \alpha_1 \dots \alpha_n \in \mathcal{L}_o$ and $s \in \Sigma_{0,v}$.

* If $[s, \alpha_n] \neq 1$ and $s^{-1} \notin \alpha_n$, then $v \ast s = vs$.

* If $s^{-1} \in \alpha_n$, then  $v \ast s = \alpha_1 \dots \tilde{\alpha}_n$, where $\tilde{\alpha}_n := \alpha_n \setminus \{ s^{-1} \}$.

* Suppose $[s, \alpha_n] = 1$ and let $k \in \{ 1, \dots, n \}$ be the least integer such that $[s, \alpha_i] = 1$ for all $i \geq k$. If $s^{-1} \in \alpha_{k-1}$, then $ v \ast s = \alpha_1 \dots \tilde{\alpha}_{k-1} \dots \alpha_n$, where $\tilde{\alpha}_{k-1} := \alpha_{k-1} \setminus \{ s^{-1} \}$.

* Suppose $[s, \alpha_n] = 1$ and let $k \in \{ 1, \dots, n \}$ be the least integer such that $[s, \alpha_i] = 1$ for all $i \geq k$. If $s^{-1} \notin \alpha_{k-1}$, then $v \ast s = \alpha_1 \dots \tilde{\alpha}_k \dots \alpha_n$, where $\tilde{\alpha}_k := \alpha_k \cup \{ s \}$.

Accordingly, the word length of $v \ast s$ over $\Sigma$ can be $n-1$, $n$, or $n+1$.

Let $\hat{h}(s)$ denote the hyperplane crossed by the path $s \in \mathcal{L}_v^{cube}$. In statement (1), we suppose that $[s, \alpha_n] \neq 1$ and $s^{-1} \notin \alpha_n$. This implies that $vs$ cannot cross any hyperplanes twice (otherwise there would be a contradiction to Lemma <ref>). Furthermore, $vs$ satisfies the second condition required of normal cube paths, because $\alpha_1 \dots \alpha_n$ does so and $[s, \alpha_n] \neq 1$. Thus, $vs$ is a normal cube path.

In statement (2), the path $vs$ crosses the hyperplane $\hat{h}(s)$ twice, once in the segment labeled $\alpha_n$ and once in the segment labeled $s$. Using canceling of labels, we see that

\[ \alpha_1 \dots \alpha_n s \equiv_o \alpha_1 \dots \alpha_{n-1} \tilde{\alpha}_n, \]

where $\tilde{\alpha}_n = \alpha_n \setminus \{ s^{-1} \}$. The path $\alpha_1 \dots \tilde{\alpha}_n$ crosses the same hyperplanes as $\alpha_1 \dots \alpha_n$, except for $\hat{h}(s)$. Furthermore, it crosses every hyperplane at most once and it is a normal cube path, as we still have for every $t \in \tilde{\alpha}_n \subset \alpha_n$ that $[t, \alpha_{n-1}] \neq 1$. Thus, it is the unique normal cube path equivalent to $vs$.

In statement (3), the path $vs$ crosses $\hat{h}(s)$ twice, once in the segment labeled $\alpha_{k-1}$ and once in the segment labeled $s$. Since $[s, \alpha_i] = 1$ for every $i \geq k$ and $s^{-1} \in \alpha_{k-1}$, we know that

\[ \alpha_1 \dots \alpha_n s \equiv \alpha_1 \dots \alpha_{k-1} s \alpha_k \dots \alpha_n \equiv \alpha_1 \dots \tilde{\alpha}_{k-1} \dots \alpha_n, \]

where $\tilde{\alpha}_{k-1} := \alpha_{k-1} \setminus \{ s^{-1} \}$. The path $\alpha_1 \dots \tilde{\alpha}_{k-1} \dots \alpha_n$ crosses the same hyperplanes as $v$, except for $\hat{h}(s)$, and it crosses them exactly once. In order to show that $\alpha_1 \dots \tilde{\alpha}_{k-1} \dots \alpha_n$ is still a normal cube path, we are left to show that for every $t \in \alpha_k$, $[t, \tilde{\alpha}_{k-1}] \neq 1$. Since we know that $[s, \alpha_k] = 1$, we have that $[s,t] = 1$. On the other hand, $[t, \alpha_{k-1}] \neq 1$, as we started out with the normal cube path $v$. Therefore, there has to be some element in $\tilde{\alpha}_{k-1}$ that does not commute with $t$. We conclude that $\alpha_1 \dots \tilde{\alpha}_{k-1} \dots \alpha_n$ is a normal cube path.

In statement (4), the path $vs$ cannot cross any hyperplane twice, as that would require $s^{-1} \in \alpha_{k-1}$ by Lemma <ref>. Putting $\tilde{\alpha}_k := \alpha_k \cup \{ s \}$, we obtain that the path $\alpha_1 \dots \tilde{\alpha}_k \dots \alpha_n$ does not cross any hyperplane more than once. We are left to show that for every $t \in \tilde{\alpha}_k$, $[t, \alpha_{k-1}] \neq 1$. For $t \in \alpha_k$, this is true because $v$ is a normal cube path. For $t = s$, this holds because statement (4) supposes that $[s, \alpha_{k-1}] \neq 1$. This proves the final statement and the Lemma.

§.§ Regular languages and rooted automorphisms of trees

For the remainder of the paper, it will be useful to understand the geometric structure intrinsically present in a regular language and how one can describe automorphisms of this geometric structure. Our main reference for this is [14]. Here, we present the relation between regular languages and self-similar trees and the description of automorphisms as portraits as developed in said reference to the extent that we need in this paper.

Let $\mathcal{L}$ be a regular language over an alphabet $\Sigma$, that is, $\mathcal{L}$ is precisely the language accepted by a finite automaton. This language has an associated path language tree $T$, which is a rooted tree, whose vertices are the elements of $\mathcal{L}$ (with the empty word $\epsilon$ as the root). Given two vertices $v, w \in \mathcal{L}$, there is an edge from $v$ to $w$ in $T$ if and only if there exists some $\alpha \in \Sigma$ such that $w = v\alpha$. We can label the edges of $T$ by elements of $\Sigma$ by labeling the edge from $v$ to $v\alpha$ by $\alpha$. A rooted tree is called self-similar if it is the path language tree of a regular language.

By Proposition <ref>, there is a self-similar tree associated to the language $\mathcal{L}_o^{cube}$. We denote this tree by $T_o^{cube}$. Similarly, Lemma <ref> implies that there is a self-similar tree associated to the language $\mathcal{L}_o$, which we denote by $T_o$. We write $V(T_o^{cube}) = \mathcal{L}_o^{cube}$ and $V(T_o) = \mathcal{L}_o$ for the vertices of these trees.

Let $\mathcal{L}$ be a regular language over the alphabet $\Sigma$ and $T$ the associated path language tree. For all $v \in \mathcal{L}$, we denote

\[ \Sigma(v) := \{ \alpha \in \Sigma \vert v\alpha \in \mathcal{L} \}. \]

A portrait in a language $\mathcal{L}$ is a family of maps $(\pre{v}{\sigma})_{v \in \mathcal{L}}$ such that $\pre{v}{\sigma} : \Sigma(v) \rightarrow \Sigma$ is injective for every $v \in \mathcal{L}$. For any portrait, we define the map $\sigma : \mathcal{L} \rightarrow \Sigma^*$ by

\[ \sigma(\alpha_1 \dots \alpha_n) := \pre{\epsilon}{\sigma}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_{i-1}}{\sigma}(\alpha_i) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n). \]

We say a portrait defines an automorphism on $T$ if $\sigma$ is the vertex-map of a rooted automorphism on $T$. (Since rooted automorphisms on $T$ are uniquely determined by what they do on vertices, $\sigma$ determines the entire automorphism.)

Given an element $\sigma \in \Aut(T)$, we define a map

\[ \pre{v}{\sigma} : \Sigma(v) \rightarrow \Sigma(\sigma(v)) \]

by the equation

\[ \forall \alpha \in \Sigma_v : \quad \sigma(v\alpha) = \sigma(v) \pre{v}{\sigma}(\alpha). \]

As was shown in <cit.>, the collection $(\pre{v}{\sigma})_{v \in V(T)}$ is a portrait and recovers the automorphism $\sigma$ via the formula given in Definition <ref>.

By <cit.>, a portrait defines an automorphism on $T$ if and only if $\sigma(\mathcal{L}) \subset \mathcal{L}$ and, for every $v \in \mathcal{L}$, $\pre{v}{\sigma}(\Sigma(v)) \subset \Sigma(\sigma(v))$. By <cit.>, this condition can be checked by verifying inductively over the length of $v$ that $\pre{v}{\sigma}(\Sigma(v)) \subset \Sigma(\sigma(v))$ for every $v \in \mathcal{L}$.

One easily checks from the equation in Remark <ref> that the composition of automorphisms on $T$ satisfies the following formula, when written in portraits:

\begin{equation} \label{eq:multiplication.formula.of.portraits}
\forall \sigma, \tau \in \Aut(T) : \pre{v}{(\sigma \circ \tau)} = \pre{\tau(v)}{\sigma} \circ \pre{v}{\tau}.
\end{equation}

For the language $\mathcal{L}_o^{cube}$, we have already introduced the special notation $\Sigma_v$ for the set $\Sigma(v)$ introduced in Definition <ref>. For the language $\mathcal{L}_o$, we write $\Sigma_v^{norm}$ for its set $\Sigma(v)$ and $\Sigma_{0,v}^{norm} := \Sigma_v^{norm} \cap \Sigma_0$.

This allows us to describe elements of $\Aut(T_o^{cube})$ and $\Aut(T_o)$ in terms of the corresponding portraits, which we will make use of throughout the rest of the paper.

§ STABILIZER ELEMENTS OF $\AUT(X)$ AS TREE AUTOMORPHISMS

Let $X$ be a locally finite $\CAT$ cube complex, $\Gamma$ a group acting cubically and cocompactly on $X$, and $\ell$ a $\Gamma$-invariant admissible edge labeling on $X$. We fix a vertex $o \in X^{(0)}$. Our goal in this section is to find a description of the stabilizer subgroup

\[ G_o := \{ g \in \Aut(X) \vert g(o) = o \}. \]

We start by embedding $G_o$ in the group of rooted automorphisms of the trees $T_o$ and $T_o^{cube}$ as follows: For $T_o$, we note that vertices of $T_o$ correspond to normal cube paths in $X$ that start at $o$ (see Remark <ref>). Since the action of $G_o$ on $X$ preserves cube paths that start at $o$, it induces an action on the vertices of $T_o$ that fixes the root $\epsilon$.

For $T_o^{cube}$, we observe that vertices of $T_o^{cube}$ correspond to elements in $D(X)_o$ (see Remark <ref>). Since the action of $G_o$ on $X$ preserves cube paths that start at $o$, it induces an action on the vertices of $T_o^{cube}$ that fixes the root $\epsilon$.

Let $o \in X^{(0)}$. The actions $G_o \curvearrowright V(T_o)$, $G_o \curvearrowright V(T_o^{cube})$ described above are actions by rooted automorphisms. In particular, $G_o$ canonically embeds into $\Aut(T_o)$ and $\Aut(T_o^{cube})$.

We present the proof for $T_o$. The proof for $T_o^{cube}$ is analogous. Let $g \in G_o$ and consider its action on $V(T_o)$. Clearly, this action fixes the root, as $g$ fixes the vertex $o$. To prove that $G_o$ acts by rooted automorphisms, we need to show that the action of $g$ on $T_o$ preserves edges.

Let $e$ be an edge from $v$ to $w$ in $T_o$, i.e. $v = \alpha_1 \dots \alpha_n$ and $w = \alpha_1 \dots \alpha_{n+1}$ are two normal cube paths in $X$. Since the action of $g$ on $X$ is a cubical isometry that fixes $o$, it sends normal cube paths starting at $o$ to normal cube paths (of the same length) that start at $o$. In particular, $g$ sends $\alpha_1 \dots \alpha_{n+1}$ to a normal cube path $\alpha'_1 \dots \alpha'_{n+1}$, which implies that $g(v) = \alpha'_1 \dots \alpha'_n$ and $g(w) = \alpha'_1 \dots \alpha'_{n+1}$. Therefore, $g(w) = g(v) \alpha'_{n+1}$ and there is an edge from $g(v)$ to $g(w)$ in $T_o$.

Applying the same argument to $g^{-1} \in G_o$ shows that, if there is an edge from $g(v)$ to $g(w)$ in $T_o$, then there is an edge from $v$ to $w$ in $T_o$. Thus, $g$ acts on $T_o$ as a rooted tree-automorphism.

Since the action of $g$ on both $X$ and $T_o$ is determined by the action on the respective vertices, which are canonically bijective, we immediately see that this map is injective and a homomorphism.

The Lemma above tells us that, for every vertex $o \in X^{(0)}$, we may think of $G_o$ as a subgroup of $\Aut(T_o)$ and $\Aut(T_o^{cube})$. We can thus use the theory developed in <cit.> to describe elements of $G_o$ in terms of a portrait. We will define the portrait of $g$ explicitly and then show that it is indeed a portrait in the sense of Definition <ref>.

Let $g \in G_o$ and $v \in V(T_o^{cube})$. We define

\[ \pre{v}{\sigma(g)} : \Sigma_{v} \rightarrow \Sigma \]

\[ s \mapsto \ell \circ g \circ \ell\vert_{\vec{\mathcal{D}}(X)_v}^{-1} (s). \]

Equivalently, the map $\pre{v}{\sigma(g)}$ is defined to send the label $s$ of a diagonal $d$ that is outgoing at $v$ to the label of the edge $g(d)$. Equivalently, $\pre{v}{\sigma(g)}$ is defined as the unique map satisfying the following equation:

\[ \forall \alpha \in \Sigma_v : g(v\alpha) = g(v) \pre{v}{\sigma(g)}(\alpha). \]

We define the portrait of $g$ to be the family $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$.

Following Definition <ref>, we obtain a map $\sigma(g) : \mathcal{L}_o^{cube} \rightarrow \mathcal{L}_o^{cube}$ defined by

\[ \sigma(g)(\alpha_1 \dots \alpha_n) := \pre{\epsilon}{\sigma(g)}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_i}{\sigma(g)}(\alpha_{i+1}) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma(g)}(\alpha_n). \]

One easily verifies that this map satisfies the formula

\[ \forall \alpha_1 \dots \alpha_n \in \mathcal{L}_o^{cube} : g(\alpha_1 \dots \alpha_n) = \sigma(g)(\alpha_1 \dots \alpha_n). \]

We recall that, by the definition of a labeling, the restrictions $\ell\vert_{\vec{\mathcal{D}}(X)_v}$ and $\ell\vert_{\vec{\mathcal{D}}(X)_{g(v)}}$ are injective and $\ell(\vec{\mathcal{D}}(X)_v ) = \Sigma_{v}$. Therefore, $\pre{v}{\sigma(g)}$ is injective, making the portrait of $g$ a portrait in $\mathcal{L}_o^{cube}$ in the sense of Definition <ref>.

§.§ Label-morphisms

Given $g \in G_o$, the portrait of $g$ satisfies much better properties than a general portrait in $\mathcal{L}_o^{cube}$. In this subsection, we introduce the necessary terminology to discuss these properties. To do so, we define a suitable notion of morphism on the set $\Sigma$.

A map $\sigma : \Sigma_{0,v} \rightarrow \Sigma_0$ is called a label-morphism if it is injective and satisfies the following condition:

\[ \forall s, t \in \Sigma_{0,v} :  [s,t] = 1 \Leftrightarrow [ \sigma(s), \sigma(t) ] = 1. \]

If $\Sigma_{0,v} = \Sigma_0$ for all $v$, then label-morphisms are bijective and we call them label-isomorphisms.

A label-morphism admits a canonical extension $\sigma : \Sigma_v \rightarrow \Sigma$ defined by

\[ \sigma( \{ s_1, \dots, s_n \} ) := \{ \sigma(s_1), \dots, \sigma(s_n) \}. \]

Since label-morphisms preserve commutation, the expression above sends elements of $\Sigma_v$ to elements of $\Sigma$. Furthermore, the size of a label does not change, since label-morphisms are injective. We always denote this canonical extension by the same symbol as the initial label-morphism.

If $\beta \in \Sigma$, then $\beta \subset \Sigma_0$ and thus $\beta$ can be thought of as an element or a subset of $\Sigma$. In order to distinguish between $\beta$ as an element and $\beta$ as a subset, we write $\{ \beta \}$ whenever we speak of $\beta$ as a subset of $\Sigma$, and $\beta$ when we speak of it as an element of $\Sigma$. Under this convention, we have that, for a label-morphism $\sigma$, the expression

\[ \sigma( \{ \beta \} ) = \{ \beta \} \]

means that $\sigma$ preserves the subset $\{ \beta \} \subset \Sigma_0$, while we write

\[ \sigma(\beta) = \beta :\Leftrightarrow \forall s \in \beta : \sigma(s) = s. \]

We are rarely going to use the expression $\sigma( \{ \beta \} )$, but $\sigma(\beta)$ and in particular $\sigma(\beta) = \beta$ will occur frequently.

We emphasize that, in general, a label-morphism does not satisfy the equation

\[ \sigma(s^{-1}) = \sigma(s)^{-1}. \]

An easy way to produce a counterexample is to work with a cube complex that has a link containing more than two isolated vertices. However, even non-isolated labels do not necessarily satisfy the above equation. (Consider for example the right-angled Artin group induced by a square and set $X$ to be the universal cover of its Salvetti complex. Study the automorphism-group of its link.)

A portrait $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ in the language $\mathcal{L}_o^{cube}$ is called admissible, if the following two conditions hold:

* For every pair $v, w \in \mathcal{L}_o^{cube}$ such that $v \equiv w$, we have $\pre{v}{\sigma} = \pre{w}{\sigma}$.

* For every $v \in \mathcal{L}_o^{cube}$, $\pre{v}{\sigma}\vert_{\Sigma_{0,v}} : \Sigma_{0,v} \rightarrow \Sigma_0$ is a label-morphism and $\pre{v}{\sigma}$ is its canonical extension.

We frequently identify the maps $\pre{v}{\sigma}, \pre{w}{\sigma}$ for equivalent words. Furthermore, since $\pre{v}{\sigma}$ is determined by its restriction to $\Sigma_{0,v}$, we view $(\pre{v}{\sigma})_{v \in V(T_o)}$ as a collection of maps $\pre{v}{\sigma} : \Sigma_{0,v} \rightarrow \Sigma_0$.

Let $(\pre{v}{\sigma})_{v \in \mathcal{L}_o^{cube}}$ be an admissible portrait that defines an automorphism on $T_o^{cube}$. We define the following two properties that this family may satisfy:

(par) For all $v \in \mathcal{L}_o^{cube}$, for all non-isolated $s \in \Sigma_{0,v}$, and for all $\beta \in \Sigma_v$ such that $[\beta, s ]  = 1$, we have

\[ \pre{v}{\sigma(g)}(\beta) = \pre{vs}{\sigma(g)}(\beta). \]

(inv) For all $\alpha_1 \dots \alpha_n \in \mathcal{L}_o^{cube}$ and all $s \in \alpha_n$, we have

\[ \pre{\alpha_1 \dots \alpha_n}{\sigma(g)}(s^{-1}) = \pre{\alpha_1 \dots \tilde{\alpha}_n}{\sigma(g)}(s)^{-1}, \]

where $\tilde{\alpha}_n := \alpha_n \setminus \{ s \}$. (If $\tilde{\alpha}_n$ is empty, then $\alpha_1 \dots \tilde{\alpha}_n = \alpha_1 \dots \alpha_{n-1}$.)

§.§ Proof of Theorem <ref>

The main result of this section is a characterisation of the portraits in $\mathcal{L}_o$ and $\mathcal{L}_o^{cube}$ that are obtained from elements in $G_o$.

Let $o \in X^{(0)}$. The following statements hold:

* Let $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ be a portrait in $\mathcal{L}_o^{cube}$. There exists some $g \in G_o$ such that $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ is the portrait of $g$ if and only if $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ is an admissible portrait that defines an automorphism on $T_o^{cube}$ and satisfies $par$ and $inv$.

* Every admissible portrait in $\mathcal{L}_o^{cube}$ that defines an automorphism on $T_o^{cube}$ and satisfies $par$ and $inv$ restricts to a portrait in $\mathcal{L}_o$ that defines an automorphism on $T_o$ and satisfies $par$ and $inv$.

By <cit.>, we have a sufficient condition as to when the map $\sigma$ defines an automorphism on $T_o^{cube}$. This condition is subject of the next Lemma.

Let $g \in G_o$ and $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$ its portrait. Then we have the following:

* $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$ is an admissible portrait.

* For every $v \in V(T_o^{cube})$, $\pre{v}{\sigma(g)}( \Sigma_{0,v} ) \subset \Sigma_{0, \sigma(g)(v)}$

In particular, the portrait of $g$ defines an automorphism on $\mathcal{L}_o^{cube}$ and the restriction

\[ \pre{v}{\sigma(g)} : \Sigma_{0,v} \rightarrow \Sigma_{0, \sigma(g)(v)} \]

is bijective.

Let $g \in G_o$. The first condition of admissibility is satisfied by definition of $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$.

Now let $s, t \in \Sigma_{0,v}$ such that $[s,t] = 1$ and let $e, e'$ be the outgoing edges at $v$ such that $\ell(e) = s$, $\ell(e') = t$. Since $[s,t] = 1$, $e$ and $e'$ span a square at $v$. Since $g$ is a cubical isometry, the edges $g(e)$ and $g(e')$ span a square at $g(v)$ and the diagonal induced by the pair $e, e'$ is sent to the diagonal induced by the pair $g(e), g(e')$. Therefore, $[\pre{v}{\sigma(g)}(s), \pre{v}{\sigma(g)}(t)] = [\ell(g(e)), \ell(g(e'))] = 1$ and $\pre{v}{\sigma(g)}( \{ s, t \} ) = \{ \pre{v}{\sigma(g)}(s), \pre{v}{\sigma(g)}(t) \}$. Thus, $\pre{v}{\sigma}$ is the extension of a label-morphism $\Sigma_{0,v} \rightarrow \Sigma_{0, \sigma(g)(v)}$ for every $v$.

Next, we show that $\pre{v}{\sigma}(\Sigma_{0,v}) \subset \Sigma_{0,\sigma(g)(v)}$. Let $s \in \Sigma_{0,v}$ be the label of an edge $e$ emanating from $v$. By Definition, $\pre{v}{\sigma(g)}(s)$ is the label of $g(e)$, which is an outgoing edge at $g(v)$. Thus $\pre{v}{\sigma(g)}(s) \in \Sigma_{0, g(v)}$. By Remark <ref>, we know that $g(v) = \sigma(g)(v)$ and thus $\pre{v}{\sigma(g)}$ sends $\Sigma_{0,v}$ to $\Sigma_{0,\sigma(g)(v)}$. The rest of the Lemma now follows from <cit.>.

Let $g \in G_o$. Then the family $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$ satisfies $par$ and $inv$.

Let $g \in G_o$ and $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$ its portrait. By Lemma <ref>, $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$ is admissible and defines an automorphism on $T_o^{cube}$. Thus it makes sense to speak of properties $par$ and $inv$ (cf. the assumptions on the portrait in Definition <ref>).

First, we show that $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$ satisfies property $par$. Let $v = \alpha_1 \dots \alpha_n \in \mathcal{L}_o^{cube}$, $s \in \Sigma_0$ be non-isolated, and $w \in \mathcal{L}_o^{cube}$ such that $w \equiv vs$. Let $\beta \in \Sigma_v$ such that $[s, \beta]=1$ and $t \in \beta$. It is sufficient to show that

\[ \pre{v}{\sigma}(t) = \pre{w}{\sigma}(t) \]

for all such $t$. Since $[s,t] = 1$ by assumption, there exists a square with vertices $v, w, vt, wt$. (Recall that we identify $\faktor{\mathcal{L}_o^{cube}}{\equiv_o}$ with $X^{(0)}$ and $D(X)_o$ with $\mathcal{L}_o^{cube}$; see Remark <ref> and Notation <ref>.) Since $g$ is a cubical isomorphism, it sends this square to a square that has $g(v)$ and $g(w)$ as two of its vertices and it sends the oriented edges $e := (v, vt), e' := (w, wt)$ to a pair of edges $g(e), g(e')$ such that $h(g(e)) = h(g(e'))$. This implies that $\ell(g(e)) = \ell(g(e'))$ and thus

\[ \pre{v}{\sigma(g)}(t) = \ell(g(e)) = \ell(g(e')) = \pre{w}{\sigma(g)}(t). \]

We conclude that $\pre{v}{\sigma(g)}(t) = \pre{w}{\sigma(g)}(t)$ for all $t \in \beta$, which proves property $par$.

Next, we prove property $inv$. Consider the vertex $\alpha_1 \dots \tilde{\alpha}_n \in V(T_o^{cube}) = \mathcal{L}_o^{cube}$, where $\tilde{\alpha}_n = \alpha_n \setminus \{ s \}$. There is an oriented edge $e$ from $\alpha_1 \dots \tilde{\alpha}_n$ to $\alpha_1 \dots \alpha_n$, which is labeled by $s$. Since $\ell$ is an admissible edge-labeling, the edge $e^{-1}$, obtained by reversing the orientation of $e$, satisfies $\ell(e^{-1}) = s^{-1}$. Since $g$ is a cubical isomorphism, it sends $e$ to $g(e)$, which is the oriented edge from $g(\alpha_1 \dots \tilde{\alpha}_n)$ to $g(\alpha_1 \dots \alpha_n)$. By definition of $\pre{v}{\sigma(g)}$, we have that $\pre{\alpha_1 \dots \alpha_n}{\sigma(g)}(s^{-1})$ is the labeling of the edge $g(e)^{-1}$, while $\pre{\alpha_1 \dots \tilde{\alpha}_n}{\sigma}(s)$ is the labeling of the edge $g(e)$. Since $\ell$ is an admissible edge-labeling, we obtain

\[ \pre{\alpha_1 \dots \alpha_n}{\sigma}(s^{-1}) = \ell( g(e)^{-1} ) = \ell( g(e) )^{-1} = \pre{\alpha_1 \dots \tilde{\alpha}_n}{\sigma}(s)^{-1}, \]

which proves property $inv$.

The purpose of portraits is to encode automorphisms of their underlying tree. This requires that portraits send children of a vertex $v$ to children of the image of $v$. It turns out that, in the case of $\CAT$ cube complexes, admissibility of the portrait, together with $par$ and $inv$ implies that good behaviour on the language of diagonal-paths implies good behaviour on the language of normal cube paths. This is the subject of the next two Lemmata.

Suppose $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ is an admissible portrait that satisfies properties $par$ and $inv$. Then for all $\alpha_1 \dots \alpha_n \in \mathcal{L}_o^{cube}$ and all $s \in \Sigma_0$ such that $[s, \alpha_n] = 1$, we have

\[ \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(s) = \pre{\alpha_1 \dots \alpha_n}{\sigma}(s), \]

\[ \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n)^{-1} = \pre{\alpha_1 \dots \alpha_n}{\sigma}(\alpha_n^{-1}) \]

The first equation is simply a repeated application of property $par$. The second equation follows, if we can show for every $t \in \alpha_n$ that

\[ \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(t)^{-1} = \pre{\alpha_1 \dots \alpha_n}{\sigma}(t^{-1}). \]

Setting $\tilde{\alpha}_n := \alpha_n \setminus \{ t \}$, application of $par$ and $inv$ shows that

\[ \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(t)^{-1} = \pre{\alpha_1 \dots \alpha_{n-1} \tilde{\alpha}_n}{\sigma}(t)^{-1} = \pre{\alpha_1 \dots, \alpha_n}{\sigma}(t^{-1}). \]

Let $( \pre{v}{\sigma} )_{v \in V(T_o^{cube})}$ be an admissible portrait that defines an automorphism on $T_o^{cube}$ and satisfies $par$ and $inv$. Then, for every $v \in \mathcal{L}_o$, $\pre{v}{\sigma}$ restricts to a bijection $\Sigma_{v}^{norm} \rightarrow \Sigma_{\sigma(v)}^{norm}$. In particular, the map

\[ \sigma^{norm} : \mathcal{L}_o \rightarrow \mathcal{L}_o \]

\[ \sigma^{norm}(\alpha_1 \dots \alpha_n) := \pre{\epsilon}{\sigma}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_{i-1}}{\sigma}(\alpha_i) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n) \]

is well-defined and defines an automorphism on $T_o$.

Let $v = \alpha_1 \dots \alpha_n \in \mathcal{L}_o = V(T_o)$. We denote $v_- := \alpha_1 \dots \alpha_{n-1}$ and $v_{end} := \alpha_n$. By <cit.>, we know that $\pre{v}{\sigma}$ is in fact a bijection $\Sigma_{0,v} \rightarrow \Sigma_{0, \sigma(v)}$. We prove the Lemma by showing that $\pre{v}{\sigma}$ restricts to a bijection between the complements of $\Sigma_v^{norm}$ and $\Sigma_{\sigma(v)}^{norm}$. We know from Remark <ref> that

\begin{equation*}
\begin{split}
\Sigma_v^{norm} & = \Sigma_v \setminus ( \{ \alpha \in \Sigma_v \vert   \alpha \cap v_{end}^{-1} \neq \emptyset \} \cup \{ \alpha \in \Sigma_v \vert \exists s \in \alpha : [s, v_{end}] = 1 \} ),\\
\Sigma_{\sigma(v)}^{norm} & = \Sigma_{\sigma(v)} \setminus ( \{ \alpha \in \Sigma_{\sigma(v)} \vert \alpha \cap \pre{v_-}{\sigma}(v_{end})^{-1} \neq \emptyset \}\\
& \qquad \qquad \qquad \cup \{ \alpha \in \Sigma_{\sigma(v)} \vert \exists s \in \alpha : [s, \pre{v_-}{\sigma}(v_{end})] = 1 \} ).
\end{split}
\end{equation*}

Let $\alpha \in \Sigma_v$ such that there exists $s \in \alpha \cap v_{end}^{-1}$. By Lemma <ref>, we have

\[ \pre{v}{\sigma}(v_{end}^{-1}) = \pre{v_-}{\sigma}(v_{end})^{-1} \]

and thus $\pre{v}{\sigma}(s) \in \pre{v}{\sigma}(\alpha \cap v_{end}^{-1}) = \pre{v}{\sigma}(\alpha) \cap \pre{v_-}{\sigma}(v_{end})^{-1}$. We conclude that $\alpha \cap v_{end}^{-1} \neq \emptyset$ if and only if $\pre{v}{\sigma}(\alpha) \cap \pre{v_-}{\sigma}(v_{end})^{-1} \neq \emptyset$.

Now suppose $\alpha \in \Sigma_v$ such that there exists $s \in \alpha$ satisfying $[s, v_{end}] = 1$. Since all $\pre{v}{\sigma}$ are label-isomorphisms, this holds if and only if

\[ [ \pre{v_-}{\sigma}(s) , \pre{v_-}{\sigma}(v_{end}) ] = 1. \]

Since $[s, v_{end}] = 1$, property par implies that $\pre{v_-}{\sigma}(s) = \pre{v}{\sigma}(s)$ and thus the commutation above is equivalent to

\[ [ \pre{v}{\sigma}(s), \pre{v_-}{\sigma}(v_{end}) ] = 1. \]

We conclude that there exists $s \in \alpha$ such that $[s, v_{end}] = 1$ if and only if there exists $s' \in \pre{v}{\sigma}(\alpha)$ such that $[s', \pre{v_-}{\sigma}(v_{end})]=1$ (with $s' = \pre{v}{\sigma}(s)$). Thus, $\pre{v}{\sigma}$ sends the complement of $\Sigma_v^{norm}$ bijectively to $\Sigma_{\sigma(v)}^{norm}$.

Lemma <ref> and Lemma <ref> tell us that the map that sends $g$ to its portrait $(\pre{v}{\sigma(g)})_{v \in V(T_o)}$ sends $G_o$ into the collection of admissible portraits that define an automorphism on $T_o^{cube}$ and that satisfy the properties $par$ and $inv$. To prove the first statement of Theorem <ref>, we are left to prove the converse. Before we can do so, we need to show the following preliminary result.

Let $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ be an admissible portrait that defines an automorphism on $T_o^{cube}$ and satisfies $par$ and $inv$. Then, for any two $v, w \in \mathcal{L}_o^{cube}$, $v \equiv_o w$ if and only if $\sigma(v) \equiv_o \sigma(w)$.

The proof of this proposition is built on the following two lemmata.

Let $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ be an admissible portrait that defines an automorphism on $T_o^{cube}$ and satisfies $par$. Then, for any $v \in \mathcal{L}_o^{cube}$ and any $\alpha = \{ s_1, \dots, s_n \} \in \Sigma_v$, we have

\[ \pre{v}{\sigma}(\alpha) \equiv_{\sigma(v)} \pre{v}{\sigma}(s_1) \dots \pre{v}{\sigma}(s_n) = \pre{v}{\sigma}(s_1) \dots \pre{vs_1\dots s_{n-1}}{\sigma}(s_n). \]

Since $\pre{v}{\sigma}$ is a label-morphism for every $v$, we know that

\[ \pre{v}{\sigma}(\alpha) = \{ \pre{v}{\sigma}(s_1), \dots, \pre{v}{\sigma}(s_n) \}. \]

Since any directed diagonal $\{ s_1, \dots, s_n \}$ has the same start and end point as the edge-path $s_1 \dots s_n$, we obtain the equivalence

\[ \pre{v}{\sigma}(\alpha) \equiv_{\sigma(v)} \pre{v}{\sigma}(s_1) \dots \pre{v}{\sigma}(s_n). \]

The second equation is a consequence of $par$.

The other result we need is a well-known fact about edge-paths in $\CAT$ cube complexes. For convenience of use, we formulate it using our labelings.

Consider two edge-paths in $X$ that both start at a vertex $o$ and end at a vertex $p$. Let $v, w \in \mathcal{L}_o^{cube}$ be the words corresponding to these edge-paths under the labeling $\ell$. Then $v$ can be transformed into $w$ by finitely many applications of the following two moves:

\[ u_1 u_2 \equiv_o u_1 s s^{-1} u_2 \]

\[ u_1 st u_2 \equiv_o u_1 ts u_2. \]

We introduce the following auxiliary notation for this proof. Given a word $u_1 u_2 \in \mathcal{L}_o^{cube}$, where $u_2 = \alpha_{n+1} \dots \alpha_{n+m}$, we write

\[ \pre{u_1}{\sigma}(u_2) := \pre{u_1}{\sigma}(\alpha_{n+1}) \dots \pre{u_1 \alpha_{n+1} \dots \alpha_{n+m-1}}{\sigma}(\alpha_{n+m}). \]

Admissibility of $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ immediately implies that, if $u_1 \equiv u'_1$, then $\pre{u_1}{\sigma}(u_2) = \pre{u'_1}{\sigma}(u_2)$.

Let $v \in \mathcal{L}_o^{cube}$. By Lemma <ref>, we can build an edge-path $v' \equiv_o v$ such that $\sigma(v) \equiv \sigma(v')$. It is thus sufficient to check that $v \equiv w$ if and only if $\sigma(v) \equiv \sigma(w)$ for edge-paths. By Lemma <ref>, we only need to check the cases, where $v$ and $w$ differ by one of the two moves above.

$``\Rightarrow"$: Suppose $v \equiv w$ and there are $u_1, s, u_2$ such that

\[ v = u_1 s s^{-1} u_2, \quad w = u_1 u_2. \]

Since $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ is admissible and $u_1 \equiv u_1 s s^{-1}$, we know that

\[ \pre{u_1}{\sigma}(u_2) = \pre{u_1 s s^{-1}}{\sigma}(u_2). \]

Furthermore, property $inv$ implies that

\[ \pre{u_1}{\sigma}(s) = \pre{u_1 s}{\sigma}(s^{-1})^{-1}. \]

We compute

\begin{equation*}
\begin{split}
\sigma(v) & = \sigma(u_1 s s^{-1} u_2)\\
& = \sigma(u_1) \pre{u_1}{\sigma}(s) \pre{u_1 s}{\sigma}(s^{-1}) \pre{u_1 s s^{-1}}{\sigma}(u_2)\\
& = \sigma(u_1) \pre{u_1}{\sigma}(u_2)\\
& = \sigma(w).
\end{split}
\end{equation*}

Thus $\sigma(v) \equiv \sigma(w)$.

Now suppose $v \equiv w$ and suppose there are $u_1, s, t, u_2$ such that $[s,t] = 1$ and

\[ v = u_1 s t u_2, \quad w = u_1 t s u_2. \]

Using admissibility again, we obtain

\[ \pre{u_1 st}{\sigma}(u_2) = \pre{u_1 ts}{\sigma}(u_2) \]

and, since $[s,t] = 1$, $[\pre{u_1}{\sigma}(s), \pre{u_1}{\sigma}(t)] = 1$. This, combined with $par$, implies that

\[ \pre{u_1}{\sigma}(s) = \pre{u_1 t}{\sigma}(s), \]

\[ \pre{u_1}{\sigma}(t) = \pre{u_1 s}{\sigma}(t). \]

We thus compute

\begin{equation*}
\begin{split}
\sigma(v) & = \sigma(u_1 s t u_2)\\
& = \sigma(u_1) \pre{u_1}{\sigma}(s) \pre{u_1 s}{\sigma}(t) \pre{u_1 s t}{\sigma}(u_2)\\
& \equiv \sigma(u_1) \pre{u_1}{\sigma}(t) \pre{u_1 t}{\sigma}(s) \pre{u_1 t s}{\sigma}(u_2)\\
& = \sigma(u_1 t s u_2)\\
& = \sigma(w).
\end{split}
\end{equation*}

$``\Leftarrow"$: We are left to show the converse, that is $\sigma(v) \equiv \sigma(w) \Rightarrow v \equiv w$. One could show this by arguing that the inverse of $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ in the group $\Aut(T_o^{cube})$ is also admissible and satisfies $par$ and $inv$. This requires dealing with the encoding of inverses in portraits, which becomes cumbersome in notation. Instead, we simply prove the converse directly.

Suppose $\sigma(v) \equiv \sigma(w)$ and there are $u'_1, s', u'_2$ such that

\[ \sigma(v) = u'_1 s' s'^{-1} u'_2, \quad \sigma(w) = u'_1 u'_2. \]

Since $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ defines an automorphism on $T_o^{cube}$, we obtain the following:

* There exists $u_1 \in \mathcal{L}_o^{cube}$ such that $\sigma(u_1) = u'_1$.

* There exists $s \in \Sigma_{0, u_1}$ such that $\pre{u_1}{\sigma}(s) = s'$.

* There exists $u_2 \in \mathcal{L}_{u_1}^{cube}$ such that $\pre{u_1}{\sigma}(u_2) = u'_2$. (The decomposition of $\sigma(w)$ into $u'_1$ and $u'_2$ induces a decomposition of the path $w$ into $u_1$ and a second path segment. This second path-segment is $u_2$.)

Since $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ is admissible, we know that

\[ \pre{u_1}{\sigma}(u_2) = \pre{u_1 s s^{-1}}{\sigma}(u_2). \]

By $inv$, we have that

\[ \pre{u_1}{\sigma}(s) = \pre{u_1 s}{\sigma}(s^{-1})^{-1}. \]

We thus compute

\begin{equation*}
\begin{split}
\sigma(u_1 s s^{-1} u_2) & = \sigma(u_1) \pre{u_1}{\sigma}(s) \pre{u_1 s}{\sigma}(s^{-1}) \pre{u_1 s s^{-1}}{\sigma}(u_2)\\
& = u'_1 s' s'^{-1} u'_2\\
& = \sigma(v)
\end{split}
\end{equation*}

Since $\sigma$ is bijective, this implies that $v = u_1 s s^{-1} u_2$, while $w = u_1 u_2$. Thus, $v \equiv w$.

Suppose now $\sigma(v) \equiv \sigma(w)$ and there are $u'_1, s', t', u'_2$ such that $[s', t'] = 1$ and

\[ \sigma(v) = u'_1 s' t' u'_2, \quad \sigma(w) = u'_1 t' s' u'_2. \]

Since $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ defines an automorphism on $T_o^{cube}$, the decomposition of the path $\sigma(v)$ into $u'_1$, $s'$, $t'$, $u'_2$ induces a decomposition of the path $v$ with the following properties:

* There exists $u_1 \in \mathcal{L}_o^{cube}$ such that $\sigma(u_1) = u'_1$.

* There exists $s \in \Sigma_{0,u_1}$ such that $\pre{u_1}{\sigma}(s) = s'$.

* There exists $t \in \Sigma_{0,u_1 s}$ such that $\pre{u_1 s}{\sigma}(t) = t'$.

* There exists $u_2 \in \mathcal{L}_{u_1 s t}^{cube}$ such that $\pre{u_1 s t}{\sigma}(u_2) = u'_2$.

Since $\pre{u_1}{\sigma}$ is a label-morphism and $[s',t'] = 1$, we conclude that

\[ [s, t] = [\pre{u_1}{\sigma}(s), \pre{u_1}{\sigma}(t)] = [s',t'] = 1. \]

In particular, $u_1 s t \equiv u_1 t s$. Admissibility now implies that

\[ \pre{u_1 t s}{\sigma}(u_2) = \pre{u_1 s t}{\sigma}(u_2) = u'_2. \]

By $par$, we obtain

\[ \pre{u_1}{\sigma}(t) = \pre{u_1 s}{\sigma}(t) = t', \]

\[ \pre{u_1 t}{\sigma}(s) = \pre{u_1}{\sigma}(s) = s'. \]

We conclude that

\[ \sigma(u_1 s t u_2) = u'_1 s' t' u'_2 = \sigma(v) \]

\[ \sigma(u_1 t s u_2) = \sigma(u_1) \pre{u_1}{\sigma}(t) \pre{u_1 t}{\sigma}(s) \pre{u_1 t s}{\sigma}(u_2) = u'_1 t' s' u'_2 = \sigma(w). \]

Since $\sigma$ is bijective, this implies that $v = u_1 s t u_2$ and $w = u_1 t s u_2$, while $[s,t] = 1$. It follows that $v \equiv w$.

This shows that $v \equiv w$ if and only if $\sigma(v) \equiv \sigma(w)$, which concludes the proof.

Let $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ be an admissible portrait that defines an automorphism on $T_o^{cube}$ and satisfies $par$ and $inv$. Then there exists a unique $g \in G_o$ such that $(\pre{v}{\sigma})_{v \in V(T_o)}$ is the portrait of $g$.

Let $( \pre{v}{\sigma} )_{v \in V(T_o^{cube})}$ be an admissible portrait satisfying $par$ and $inv$ that defines an automorphism on $T_o^{cube}$. We want to show that $\sigma$ is induced by a cubical isometry of $X$ that fixes $o$. We recall that any portrait that defines an automorphism on $T_o^{cube}$ has an associated map $\sigma : \mathcal{L}_o^{cube} \rightarrow \mathcal{L}_o^{cube}$ defined by

\[ \sigma(\alpha_1 \dots \alpha_n) = \pre{\epsilon}{\sigma}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_{i-1}}{\sigma}(\alpha_i) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n) \]

By Proposition <ref>, $\sigma$ defines a map on the vertices of $X^{(0)}$, that is, for any vertex $v \in X^{(0)}$, we set $\sigma(v)$ to be the equivalence class represented by $\sigma(\alpha_1 \dots \alpha_n)$ for any representative $\alpha_1 \dots \alpha_n \in \mathcal{L}_o^{cube}$ of $v$. We claim that this map extends to a cubical isometry on $X$. To prove this claim, we are left to show that two vertices $v, w \in X^{(0)}$ are connected by an edge if and only if $\sigma(v), \sigma(w)$ are.

$``\Rightarrow"$: Suppose there is an edge between $v$ and $w$ in $X$. Let $e$ denote the oriented edge from $v$ to $w$ and put $s := \ell(e)$. Identifying both $v$ and $w$ with elements in $\mathcal{L}_o^{cube}$ that represent these two vertices, we can thus write $vs \equiv w$. Let $\alpha_1 \dots \alpha_n$ be a representative of $v$. Then $\alpha_1 \dots \alpha_n s$ is a cube path from $o$ to $w$ and we conclude that

\[ \sigma(v) \equiv \pre{\epsilon}{\sigma}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_{i-1}}{\sigma}(\alpha_i) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n), \]

\[ \sigma(w) \equiv \sigma(vs) \equiv \pre{\epsilon}{\sigma}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_{i-1}}{\sigma}(\alpha_i) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n) \pre{v}{\sigma}(s). \]

Thus, $\sigma(v)$ and $\sigma(w)$ are connected by an edge in $X$ (whose label, when oriented from $\sigma(v)$ to $\sigma(w)$ is $\pre{v}{\sigma}(s)$).

$``\Leftarrow"$: Suppose there is an oriented edge $e'$ from $\sigma(v)$ to $\sigma(w)$ and put $s' := \ell(e')$. We obtain $\sigma(w) \equiv \sigma(v)s'$. Since $(\pre{v}{\sigma})_{v \in V(T_o^{cube})}$ defines an automorphism on $T_o^{cube}$, <cit.> implies that $\pre{v}{\sigma} : \Sigma_{0,v} \rightarrow \Sigma_{0,\sigma(v)}$ is bijective. Thus, there exists some $s \in \Sigma_0$ such that $\pre{v}{\sigma}(s) = s'$.

Let $\alpha_1 \dots \alpha_n \in \mathcal{L}_o^{cube}$ be a representative of $v$. Then

\[ \sigma(v) \equiv \pre{\epsilon}{\sigma}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_{i-1}}{\sigma}(\alpha_i) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n), \]

\[ \sigma(vs) \equiv \pre{\epsilon}{\sigma}(\alpha_1) \dots \pre{\alpha_1 \dots \alpha_{i-1}}{\sigma}(\alpha_i) \dots \pre{\alpha_1 \dots \alpha_{n-1}}{\sigma}(\alpha_n) \pre{v}{\sigma}(s) \equiv \sigma(v) s' \equiv \sigma(w). \]

We conclude that $\sigma(vs) \equiv \sigma(w)$. By Proposition <ref>, this implies that $vs \equiv w$. Therefore, $v$ and $w$ are connected by an edge.

We conclude that the action of $\sigma$ on $V(T_o) \cong X^{(0)}$ extends to an isometric action on $X$, which fixes the vertex $o$. Since the action of $g$ on the vertices is given by $\sigma$, the portrait of $g$ is exactly given by $(\pre{v}{\sigma})_{v \in V(T_o)}$. Uniqueness of $g$ follows from the fact that $g$ is uniquely determined by its action on the vertices, which is determined by $\sigma$. This proves the Lemma.

The first statement is the combined result of Lemma <ref>, <ref>, and <ref>. The second statement is the content of Lemma <ref>.

Having this characterisation of the image of $G_o \hookrightarrow \Aut(T_o)$ and the image of $G_o \hookrightarrow \Aut(T_o^{cube})$, we identify $G_o$ with its image under these inclusions from now on.

§ SPECIAL ACTIONS AND EDGE LABELINGS

In this section, we present a characterisation of compact special cube complexes in terms of actions on $\CAT$ cube complexes and edge labelings.

Suppose there exists a $\Gamma$-invariant admissible edge labeling on $X$. Then $\Gamma$ acts freely on $X$.

Let $\ell$ be a $\Gamma$-invariant admissible edge labeling on $X$. We start by showing that $\Gamma$ acts freely on $X^{(0)}$. Let $g \in \Gamma$ such that $g$ fixes a vertex $o \in X^{(0)}$. Then $g \in G_o$ and we can identify $g$ with its portrait $(\pre{v}{\sigma(g)})_{v \in V(T_o^{cube})}$. By definition of the portrait, $\pre{v}{\sigma(g)}(s)$ is the label of $g(e)$, where $e$ is the unique outgoing edge at $v$ labeled $s$. Since $\ell$ is $\Gamma$-invariant, $g$ has to preserve the labeling of oriented edges. Therefore, $\pre{v}{\sigma(g)} = \id$ for every $v \in \mathcal{L}_o$ and, by admissibility, for every $v \in \mathcal{L}_o^{cube}$. Since $g$ is uniquely characterised by its portrait, this implies that $g = \Id_X$. Thus, $\Gamma$ acts freely on $X^{(0)}$.

Now suppose $g$ does not fix any vertices, but fixes a point $p \in X$. Let $C(p)$ be the lowest-dimensional cube that contains $p$. Without loss of generality, we may assume that $p$ is chosen such that $C(p)$ has minimal dimension among all fixed points of $g$. Since $g$ fixes $p$, it has to preserve $C(p)$ and we can look at the restriction of $g$ to $C(p)$. This restriction has to send the edges of $C(p)$ to the edges of $C(p)$ while preserving the labeling $\ell$. Let $s_1^{\pm 1}, \dots, s_n^{\pm 1}$ be the labels appearing on the edges of $C(p)$. Since every vertex of $C(p)$ is uniquely determined by an orientation of all the half-spaces in $C(p)$, every vertex $v$ in $C(p)$ is uniquely determined by a choice $s_1^{\epsilon_1}, \dots s_n^{\epsilon_n}$, where $\epsilon_1, \dots \epsilon_n \in \{ \pm 1 \}$. Since $g$ has to preserve the labeling, it can only preserve $C(p)$ if it fixes all vertices of $C(p)$. This contradicts our assumption and we conclude that every $g \in \Gamma$ that has a fixed point also fixes a vertex. Since $\Gamma$ acts freely on $X^{(0)}$, this implies that $\Gamma$ acts freely on $X$.

Suppose $\Gamma < \Aut(X)$ acts cocompactly on $X$ and suppose there exists a $\Gamma$-invariant admissible edge-labeling on $X$. Then $\faktor{X}{\Gamma}$ is a non-positively curved special cube complex and $X$ is the universal covering of $\faktor{X}{\Gamma}$.

Let $\tilde{\ell}$ be a $\Gamma$-invariant admissible edge labeling on $X$. Consider the quotient $S := \faktor{X}{\Gamma}$. Since $\Gamma$ acts cubically on $X$, this quotient is a cube complex. Since $\tilde{\ell}$ is $\Gamma$-invariant, it descends to a labeling $\ell$ of the parallel classes of oriented edges in $S$.

In order to show that $S$ is special, we need to show that the four pathologies do not occur. Suppose there exists a hyperplane $\hat{h}$ in $S$ that self-intersects. This produces two oriented edges $e, e'$ in $S$ that span a square and have same label $s$. This square lifts to a square in $X$ spanned by two edges with the same label. This contradicts property (A2) in the definition of admissible edge labels and implies that $\tilde{\ell}$ is not admissible, a contradiction.

Suppose there exists a hyperplane $\hat{h}$ in $S$ that is not two-sided. This produces an edge $e$ in $S$ such that both of its orientations have the same label. This lifts to an edge in $X$ such that both of its orientations have the same label, which contradicts the fact that the map $\cdot^{-1} : \Sigma_0 \rightarrow \Sigma_0$ in the definition of an oriented parallel class labeling is assumed to be fixpoint free.

Suppose there exists a hyperplane $\hat{h}$ in $S$ that self-osculates. Then we find a vertex in $S$ with two outgoing edges $e, e'$ that have the same label. This lifts to a vertex in $X$ with two outgoing edges that have the same label, which contradicts (A2).

Suppose there exist two hyperplanes $\hat{h}, \hat{k}$ in $S$ that interosculate. We find two vertices $v$, $w$ with outgoing edges $e_1, e_2 \in \vec{\mathcal{E}}(S)_v$, $e'_1, e'_2 \in \vec{\mathcal{E}}(S)_w$ such that $e_1$ and $e_2$ span a square, $e'_1$ and $e'_2$ do not span a square, $\ell(e_1) = \ell(e'_1)$, and $\ell(e_2) = \ell(e'_2)$. This lifts to two vertices $\overline{v}$ $\overline{w}$ and edges $\overline{e_1}, \overline{e_2} \in \vec{\mathcal{E}}(X)_{\overline{v}}$, $\overline{e'_1}, \overline{e'_2} \in \vec{\mathcal{E}}(X)_{\overline{w}}$ such that $\overline{e_1}, \overline{e_2}$ span a square, $\overline{e'_1}, \overline{e'_2}$ do not span a square and $\tilde{\ell}(\overline{e_1}) = \tilde{\ell}(\overline{e'_1})$, $\tilde{\ell}(\overline{e_2}) = \tilde{\ell}(\overline{e'_2})$. This contradicts (A3).

We conclude that all four pathologies cannot occur. Thus, $S$ is a special cube complex. Non-positivity follows from the fact that $X$ is non-positively curved and that the projection $X \rightarrow S$ is a local isometry since $\Gamma$ acts freely (by Lemma <ref>) and cubically. Thus, $X \rightarrow S$ is a covering map and, since it is contractible, $X$ is the universal covering of $\faktor{X}{\Gamma}$.

Lemma <ref>, Lemma <ref>, and Theorem <ref> together imply Theorem <ref>.

It turns out that these findings allow us to show that the action of $\Gamma$ on $X$ can be expressed in a particularly nice way with respect to a $\Gamma$-invariant labeling $\ell$.

Let $v \in \mathcal{L}_o^{cube}$ such that $\mathcal{L}_o^{cube} = \mathcal{L}_v^{cube}$. This means that the concatenation of words

\[ \mathcal{L}_o^{cube} \rightarrow \mathcal{L}_o^{cube}, \quad w \mapsto vw \]

is well-defined. It induces a map on the vertices of $X$ that can be written as

\[ L_v : \mathcal{L}_o \rightarrow \mathcal{L}_o, \quad w \mapsto \mathrm{norm}(vw). \]

We call the map $L_v$ the (left-)translation by $v$.

One immediately sees that two vertices $w, w'$ are connected by an edge if and only if $L_v(w), L_v(w')$ are. Thus, $L_v$ extends to an isometry on the $1$-skeleton of $X$ and, since $X$ is $\CAT$, to an automorphism of $X$. Translations form a subgroup of $\Aut(X)$ that acts freely on $X^{(0)}$ and $L_{v^{-1}} = L_v^{-1}$. The following Lemma shows that there are many interesting instances where such a $\Gamma$ can be produced.

Let $S$ be a compact, non-positively curved special cube complex, $\Gamma$ its fundamental group, and $X$ its universal covering. Let $\ell$ be an admissible edge labeling on $S$. Then the action $\Gamma \curvearrowright X$ by decktransformation is an action by translations.

Choose $o \in X^{(0)}$ and let $\bar{o}$ be its projection in $S$. Viewing $\Gamma$ as the fundamental group with basepoint $\bar{o}$, every element of $\Gamma$ can be represented by an edge path $\bar{\gamma}$ in the $1$-skeleton of $S$ that starts and ends at $\bar{o}$. Let $\gamma$ be the unique lift of $\bar{\gamma}$ to that starts at $o$. The action of $[\bar{\gamma}] \in \pi_1(S)$ on $X$ is exactly the translation by the edge path $\gamma$. (Note that, for any $v \in X^{(0)}$ and any $g \in \pi_1(S)$, $\mathcal{L}_v^{cube} = \mathcal{L}_{g \cdot v}^{cube}$, since $g$ preserves the chosen edge labeling.) Thus, the fundamental group acts on $X$ by translations.

Suppose $\Gamma < \Aut(X)$ acts cocompactly on $X$ and suppose there exists a $\Gamma$-invariant admissible edge-labeling on $X$. Then $\Gamma$ acts by translations.

By Lemma <ref> and Theorem <ref>, the action of $\Gamma$ on $X$ is free and the quotient $\faktor{X}{\Gamma}$ is a special cube complex with $X$ as its universal covering. By Lemma <ref>, the action of $\Gamma$ is by translations.

§ TOPOLOGICALLY GENERATING SET

Throughout this section, we assume $X$ to be a locally finite $\CAT$ cube complex and $\Gamma < \Aut(X)$ a subgroup whose action on $X$ is cocompact. Suppose there exists a $\Gamma$-invariant admissible edge labeling $\ell : \vec{\mathcal{E}}(X) \rightarrow \Sigma_0$. By our results from section <ref>, we know that $\Gamma$ acts freely and by translations.

Let $o \in X^{(0)}$. Given two words $w_1, w_2 \in \mathcal{L}_o^{cube}$, we say that $w_1$ and $w_2$ commute if all letters in $w_1$ commute with all letters in $w_2$ and we write $[w_1, w_2] = 1$.

A word $w \in \mathcal{L}_o^{cube}$ is called reducible if its corresponding cube path with starting point $o$ crosses some hyperplane more than once. (Equivalently, there exists a letter $s \in \Sigma_0$ and a word $u$ such that $[s, u] = 1$ and $\alpha u\beta^{-1}$ is a subword in $w$, where $s \in \alpha$ and $s^{-1} \in \beta$.) A word is called reduced, if it is not reducible.

Note that every normal cube path induces a reduced word, that is the language $\mathcal{L}_o$ consists entirely of reduced words. Furthermore, every cube path that starts at $o$ is $\equiv_o$-equivalent to an edge path that does not cross any hyperplane more than once. These edge paths are exactly the shortest paths in the $1$-skeleton of $X$ from $o$ to the endpoint of the initial cube path. In the notation of $\mathcal{L}_o^{cube}$, this means that for every word $w \in \mathcal{L}_o^{cube}$ there exists a reduced word $w' \in \mathcal{L}_o^{cube} \cap \Sigma_0^*$ such that $w \equiv_o w'$.

Our goal in this section is to construct a family of elements in $\Aut(T_o^{cube})$ that generates a dense subgroup of $G_o = \Stab_{\Aut(X)}(o)$. We will first give an abstract argument that provides us with such a family and use this to prove Theorem <ref>. After that, we will explicitly construct a topologically generating family of $G_o$ under the assumption that $\Gamma$ acts vertex-transitively on $X$. This will yield a proof of Theorem <ref>.

§.§ Generating a dense subgroup of the stabilizer

Fix a basepoint $o \in X^{(0)}$. For every $v \in \mathcal{L}_o$, we will find a finite set of isometries in $G_o$ that addresses `everything that can happen at $v$'. We need to distinguish between the case $v = \epsilon$ and $v \neq \epsilon$.

[Generators at $o$]
Let $\sigma : \Sigma_{0,o} \rightarrow \Sigma_0$ be a label-morphism. We say that $\sigma$ appears at $o$ in $G_o$ if there exists some isometry $g \in G_o$ such that $\pre{\epsilon}{\sigma(g)} = \sigma$. For every label-morphism $\sigma$ that appears at $o$ in $G_o$, we choose one such isometry and denote it by $A_{\epsilon, \sigma}$.

To introduce a choice of generators at $v$ for $v \neq \epsilon$, we need some preparation. First, we sort all vertices $w \in \mathcal{L}_o$ into three distinct types relative to $o$ and $v$.

Let $v \in \mathcal{L}_o \setminus \{ \epsilon \}$ and $w \in \mathcal{L}_o$. Since every cube path is $\equiv$-equivalent to a reduced cube path, we can choose a, usually not unique, reduced word $w' \in \mathcal{L}_v^{cube}$ such that $w' \equiv_v v^{-1} w$. We obtain the equivalence

\[ w \equiv vv^{-1}w \equiv vw'. \]

We now distinguish between three types:

Type 1: $vw'$ is reducible.

Type 2: $vw'$ is reduced and there exists $\alpha \in \Sigma$ such that $v\alpha$ is reducible and $[\alpha, w'] = 1$.

Type 3: $vw'$ is reduced and for all $\alpha \in \Sigma$ such that $v\alpha$ is reducible, $[\alpha, w'] \neq 1$.

We say that $w$ is of Type 1, 2, or 3 relative to $(o,v)$ respectively.

We highlight the following points:

* The type of $w$ does not depend on the choice of $w'$. Since every choice of $w'$ crosses the same hyperplanes, reducibility of $vw'$ is independent of $w'$ and all choices of $w'$ commute with the same letters $\alpha \in \Sigma$.

* The vertex $v$ is of Type 2 relative to $(o,v)$. (Following Definition <ref>, the empty word commutes with everything.) The vertex $o$ is of Type 1 relative to $(o,v)$. In particular, swapping $o$ and $v$ in the pair $(o,v)$ changes the Type of vertices.

* If $w$ is of Type 3, then $d_{\ell^1}(o,w) > d_{\ell^1}(o,v)$, as $w \equiv vw'$, where $vw'$ is reduced and $w' \neq \epsilon$.

For every $v \in \mathcal{L}_o \setminus \{ \epsilon \}$, our goal is to find elements $A_{v, \sigma} \in G_o$ that fix all vertices of Type 1 and 2. To do so, we make use of the following definitions.

Given a vertex $v \in \mathcal{L}_o \setminus \{ \epsilon \}$ and a label-morphism $\sigma : \Sigma_{0,v} \rightarrow \Sigma_0$, we say that $\sigma$ is compatible with $v$ if and only if for all $\alpha \in \Sigma_v$ such that $v\alpha$ is reducible and for all $\beta \in \Sigma_v$ such that $[\alpha, \beta ] = 1$, we have $\sigma(\alpha) = \alpha$ and $\sigma(\beta) = \beta$.

[Generators at $v$]
Let $v \in \mathcal{L}_o \setminus \{ \epsilon \}$ and $\sigma : \Sigma_{0,v} \rightarrow \Sigma_0$ a label-morphism. We say that $\sigma$ appears at $v$ in $G_o$ if there exists some isometry $g \in G_o$ such that $\pre{v}{\sigma(g)} = \sigma$ and $g$ fixes all vertices of Type 1 and 2 relative to $(o,v)$. For every label-morphism $\sigma$ that appears at $v$ in $G_o$, we choose one such isometry and denote it by $A_{v, \sigma}$.

One easily checks that, if $\sigma$ appears at $v$ in $G_o$, then $\sigma$ has to be compatible with $v$ as otherwise $A_{v, \sigma}$ would not fix all vertices of Type 1 and 2.

Fix a choice of $A_{v, \sigma}$ for all $v \in \mathcal{L}_o$ and all $\sigma$ that appear at $v$ in $G_o$. With these choices, we obtain a family of automorphisms $A_{v, \sigma} \in G_o$.

Suppose that for every $v \in \mathcal{L}_o \setminus \{ \epsilon \}$, every $\sigma$ that is compatible with $v$ also appears at $v$ in $G_o$. Then the family

\[ \{ A_{v, \sigma} \vert v \in \mathcal{L}_o, \sigma \text{ appears at $v$ in $G_o$ } \} \]

generates a dense subgroup of $G_o$.

We will show in section <ref> that the assumption of Proposition <ref> is satisfied when the action $\Gamma \curvearrowright X$ is transitive on vertices.

Recall that we have a canonical identification of the sets $X^{(0)}$, $\faktor{ \mathcal{L}_o^{cube} }{\equiv_o}$, and $\mathcal{L}_o$ (see Remark <ref> and Notation <ref>).

§.§.§ Step 1: Preparing for induction.

Let $g \in G_o$. By Lemma <ref>, we obtain label-morphisms $\pre{v}{\sigma(g)} : \Sigma_{0,v} \rightarrow \Sigma_0$ for every $v \in X^{(0)}$. To prevent notation from getting cluttered in a moment, we define

\[ \sigma_{\epsilon} := \pre{\epsilon}{\sigma(g)}. \]

We will write $g$ as the limit of a composition of elements $A_{v, \sigma}$. We start by observing that for all $s \in \Sigma_0$,

\[ A_{\epsilon, \sigma_{\epsilon}}^{-1} \circ g(s) = \pre{\epsilon}{A_{\epsilon, \sigma_{\epsilon}}}^{-1} \circ \pre{\epsilon}{\sigma(g)}(s) = \sigma_{\epsilon}^{-1} \circ \sigma_{\epsilon}(s) = s. \]

We see that $A_{\epsilon, \sigma_{\epsilon}}^{-1} \circ g$ fixes not only $o$, but all vertices in $X$, whose $\ell^1$-distance to $o$ is at most $1$.

We continue by induction over the $\ell^1$-distance to $o$. Suppose $n \geq 1$ and suppose we have an isometry $g_n$ on $X$ that fixes all vertices in $X$ that satisfy $d_{\ell^1}(\cdot, o) \leq n$. For all $v \in X^{(0)}$ such that $d_{\ell^1}(v, o) = n$, we define $\sigma_v := \pre{v}{\sigma(g_n)}$. We claim that the isometry

\begin{equation} \label{eq:inductive.decomposition}
g_{n+1} := \left( \prod_{d_{\ell^1}(v,o) = n} A_{v, \sigma_v}^{-1} \right) \circ g_n
\end{equation}

is well-defined and fixes all vertices in $X$ that have distance $\leq n+1$ from $o$ with respect to $d_{\ell^1}$. (The remainder of the proof is largely dedicated to the proof of this claim.) We emphasize that the maps $A_{v, \sigma_v}$ do not commute in general and one has to choose some order for them. However, we will show that, for every vertex $w \in X^{(0)}$ with $d_{\ell^1}(w,o) \leq n+1$, the order of multiplication does not matter and thus any choice of order produces an element $g_{n+1}$ that satisfies our claim.

§.§.§ Step 2: For every $v$ with $d_{\ell^1}(o,v) = n$, $A_{v,\sigma_v}$ is well-defined.

In Constructions <ref> and <ref>, we chose isometries $A_{v, \sigma}$ for all $\sigma$ that appear at $v$ in $G_o$. We thus need to show that $\sigma_v$ appears at $v$ in $G_o$ so that $A_{v, \sigma_v}$ is well-defined. Since $n \geq 1$, we know that $v \neq \epsilon$ and we are entirely concerned with the choices made in Construction <ref>. By assumption, every $\sigma_v$ that is compatible with $v$ appears at $v$ in $G_o$. Thus, it suffices to show that $\sigma_v$ is compatible with $v$ for all $v$ that have $\ell^1$-distance $n$ from $o$. In other words, we need to show that, for every $\alpha \in \Sigma_v$ such that $v\alpha$ is reducible and for every $\beta \in \Sigma_v$ such that $[\alpha, \beta] = 1$, we have $\sigma_v(\alpha) = \alpha$ and $\sigma_v(\beta) = \beta$.

So let $v = \alpha_1 \dots \alpha_m \in \mathcal{L}_o \cong X^{(0)}$ such that $d_{\ell^1}(o,v) = n$. Let $\alpha, \beta \in \Sigma_v$ such that $v\alpha$ is reducible and $[\alpha, \beta]=1$. We can split $\alpha$ into two parts by writing

\[ \alpha_{inv} := \{ s \in \alpha \vert vs \text{ is reducible} \}, \]

\[ \alpha_{par} := \alpha \setminus \alpha_{inv}. \]

We claim that we can assume without loss of generality that $\alpha = \alpha_{inv}$. Indeed, $\alpha_{inv} \neq \emptyset$, as otherwise $v\alpha$ would be reduced. Since $v\alpha_{inv}$ is reducible and $[\alpha_{inv}, \alpha_{par}] = 1$, we cover the letter $\alpha_{par}$ when we discuss $\beta$ and we may assume without loss of generality that $\alpha = \alpha_{inv}$.

Since $\alpha = \alpha_{inv}$, the vertex $v' := v\alpha$ satisfies $d_{\ell^1}(o,v') < d_{\ell^1}(o,v) = n$. Therefore, the induction assumption on $g_n$ implies that

\[ \pre{v'}{\sigma(g_n)} = \id. \]

We can apply the second equation of Lemma <ref> to the pair $v, v' = v\alpha$ to compute

\[ \sigma_{v}(\alpha) = \pre{v}{\sigma(g_n)}(\alpha) = \pre{v'}{\sigma(g_n)}(\alpha^{-1})^{-1} = {\alpha^{-1}}^{-1} = \alpha. \]

Now consider $\beta$. Since $[\alpha, \beta] = 1$, there exists some $s \in \alpha$ such that $vs$ is reducible and $[s, \beta] = 1$. We denote $v'$ to be the endpoint of $vs$. Since $vs$ is reducible, $v$ is reduced and $d_{\ell^1}(o,v) = n$, we conclude that $d_{\ell^1}(o,v') = n-1$. By induction assumption, this implies that $\pre{v'}{\sigma(g_n)} = \id$. Using property $par$ on the pair $s, \beta$, we compute

\[ \sigma_v(\beta) = \pre{v}{\sigma(g_n)}(\beta) = \pre{v'}{\sigma(g_n)}(\beta) = \beta. \]

We conclude that $\sigma_v$ is compatible with $v$. By assumption, this means that $\sigma_v$ appears at $v$ in $G_o$ and there exists a choice of $A_{v,\sigma_v} \in G_o$. We conclude that the elements in the product $\prod_{ d_{\ell^1}(o,v) = n} A_{v, \sigma_v}$ exist.

§.§.§ Step 3: Every $w$ with $d_{\ell^1}(o,w) = n+1$ is moved by at most one $A_{v, \sigma_v}$.

Let $w \in \mathcal{L}_o$ such that $d_{\ell^1}(o,w) = n+1$. There exists $v \in \mathcal{L}_o$ with $d_{\ell^1}(o,v) = n$ and $s \in \Sigma_{0,v}$ such that $w \equiv vs$. (Note that $vs$ has to be reduced because $v$ is reduced, $d_{\ell^1}(o,v) = n$, and $d_{\ell^1}(o,vs) = d_{\ell^1}(o,w) = n+1$.) Suppose the pair $(v,s)$ is not unique and we find $v' \in \mathcal{L}_o \setminus \{ v \}$ with $d_{\ell^1}(o,v') = n$ and $s' \in \Sigma_{0,v'}$ such that $w \equiv v's'$. (Again, $v's'$ is reduced.) Note that $s \neq s'$, as there is only one outgoing edge at $w$ labeled $s^{-1}$ and it leads to $v$; so $s = s'$ implies $v = v'$, but we chose $v' \neq v$. Since both $vs$ and $v's'$ are reduced, the hyperplane that is crossed by $s'$ in the path $v's'$ has to be crossed by $vs$ at some point. As $s \neq s'$, it cannot be crossed by the edge $s$. Thus, this hyperplane has to be crossed by $v$. It follows that $vss'^{-1}$ is reducible and we find subwords $u_1, u_2$ of $v$ and some $\alpha' \in \Sigma_{u_1} \cup \{ \emptyset \}$ such that:

\[ \alpha' \cup \{ s' \} \in \Sigma_{u_1}, \quad vs = u_1 (\alpha' \cup \{ s' \} ) u_2 s, \quad [s', u_2 s] = 1. \]

Since $v = u_1 (\alpha' \cup \{ s' \}) u_2$ is reduced, we see from this equation that $d_{\ell^1}(o,u_1) \leq n-1$. As $g_n$ fixes all vertices with $d_{\ell^1}(o,\cdot) \leq n$ by induction assumption, we know that $\pre{u_1}{\sigma(g_n)} = \Id$. Applying property $par$ to $s'$ along the word $\alpha' u_2 s$ implies that

\[ s' = \pre{u_1}{\sigma(g_n)}(s') = \pre{u_1 \alpha' u_2 s}{\sigma(g_n)}(s') = \sigma_{v'}(s'), \]

where we used the fact that $v' \equiv vs s'^{-1} \equiv u_1 \alpha' u_2 s$ in the last equation. We conclude that $\sigma_{v'}(s') = s'$. Swapping $(v,s)$ and $(v',s')$ in the argument, we obtain $\sigma_{v}(s) = s$. We conclude that

\[ g_n(w) = g_n(v) \pre{v}{\sigma(g_n)}(s) = v \sigma_{v}(s) = vs = w \]

and for all $v, s$ such that $w = vs$

\[ A_{v, \sigma_v}(w) = A_{v, \sigma_v}(v) \sigma_v(s) = v s = w. \]

Now suppose $v' \in \mathcal{L}_o$ such that $d_{\ell^1}(o, v') = n$ and there exists no $s'$ such that $v's' \equiv w$. We claim that $w$ is of Type 1 relative to $(o,v')$ and thus $A_{v', \sigma_{v'}}(w) = w$. Indeed, suppose $w$ was of Type 2 or 3 relative to $(o,v')$. Then we would find a reduced word $w'$ over $\Sigma_0$ such that $w \equiv v'w'$ with $v'w'$ being reduced. However, since $d_{\ell^1}(o,w) = n+1$, any reduced word over $\Sigma_0$ equivalent to $w$ has to have edge-length $n+1$. Since $d_{\ell^1}(o,v') = n$, this implies that $w'$ has length $1$ and $(v',w')$ would be a pair $(v',s')$ such that $v's' \equiv w$, contradicting our assumption on $v'$. Thus, $w$ is of Type 1 relative to $(o,v')$. Since, by construction, $A_{v', \sigma_{v'}}$ fixes vertices that are of Type 1 relative to $(o,v')$, we obtain $A_{v', \sigma_{v'}}(w) = w$.

Combining the computations for the different cases above, we obtain in particular that, for every $w \in \mathcal{L}_o$ such that $d_{\ell^1}(o,w) = n+1$ and there is more than one pair $(v,s)$ such that $vs \equiv w$, we have

\begin{equation}\label{eq:Fixing.vertices.at.distance.n+1.part.one}
g_{n+1}(w) = \left( \prod_{d_{\ell^1}(v,o) = n} A_{v, \sigma_v}^{-1} \right) \circ g_n(w) = w.
\end{equation}

We conclude that, if $v, w \in \mathcal{L}_o$ such that $d_{\ell^1}(o,v) = n$ and $d_{\ell^1}(o,w) = n+1$, then $A_{v, \sigma_v}(w) = w$ unless there exists a unique pair $(v,s)$ such that $w \equiv vs$, in which case $v$ is the unique vertex with $d_{\ell^1}(o,v) = n$ such that $A_{v, \sigma_v}$ may not fix $w$. This proves Step 3.

§.§.§ Step 4: If $d_{\ell^1}(o,vs) = n+1$ and $A_{v,\sigma_v}(vs) \neq vs$, then for all $v' \neq v$ with $d_{\ell^1}(o,v') = n$, $A_{v', \sigma_{v'}}$ fixes the vertex $A_{v, \sigma_v}(vs)$.

Let $w \in X^{(0)}$ such that $d_{\ell^1}(o,w) = n+1$ and $(v,s)$ a pair as in Step 3 such that $w \equiv vs$. Suppose $A_{v, \sigma_v}(vs) \neq vs$. By Step 3, this implies that $(v,s)$ is the unique pair such that $w \equiv vs$. Since $A_{v, \sigma_v}$ is bijective on vertices, this implies that

\[ A_{v, \sigma_v}(vs) \neq A_{v, \sigma_v} \circ A_{v, \sigma_v}(vs) = A_{v, \sigma_v}( v \sigma_v(s) ). \]

Using Step 3 again, this implies that $(v, \sigma_v(s))$ is the unique pair such that $v\sigma_v(s) \equiv A_{v, \sigma_v}(w)$ and for all $v' \neq v$ with $d_{\ell^1}(o,v') = n$, we have $A_{v', \sigma_{v'}}( v\sigma_v(s) ) = v\sigma_v(s)$. This proves Step 4 and, combining Step 3 and Step 4, we conclude that the restriction of the product

\[ \prod_{ d_{\ell^1}(o,v) = n } A_{v, \sigma_v} \]

to the set

\[ \{ w \in X^{(0)} \vert d_{\ell^1}(o,w) = n+1 \} \]

does not depend on the ordering in the product. Note that this set is in fact preserved under the product, since we are multiplying automorphisms that fix $o$. Therefore, we conclude that the restriction of the inverse of this product to the same set does not depend on the ordering either.

§.§.§ Step 5: Finishing the induction step.

We are left to show that the map

\[ g_{n+1} = \left( \prod_{d_{\ell^1}(o,v) = n} A_{v, \sigma_v}^{-1} \right) \circ g_n \]

fixes all vertices in the $(n+1)$-ball around $o$ with respect to $d_{\ell^1}$. Suppose $d_{\ell^1}(o,w) \leq n$. By Remark <ref>, any such $w$ is of Type 1 or 2 relative to $(o,v)$ for any $v \in \mathcal{L}_o$ that satisfies $d_{\ell^1}(o,v) = n$. Thus, $w$ is fixed by all $A_{v, \sigma_v}$ appearing in the product (and thus by $A_{v, \sigma_v}^{-1}$). Furthermore, $w$ is fixed by $g_n$ due to the induction assumption, which yields $g_{n+1}(w) = w$.

Now suppose $d_{\ell^1}(w) = n+1$. If there is more than one pair $(v,s)$ such that $w \equiv vs$, then $g_{n+1}(w) = w$ by equation (<ref>). If there is a unique $(v,s)$ such that $w \equiv vs$, then

\[ g_n(w) = g_n(v) \pre{v}{\sigma(g_n)}(s) = v \sigma_v(s),\]

\[ A_{v, \sigma_v}(v s ) = A_{v, \sigma_v}(v) \pre{v}{A_{v, \sigma_v}}(s) = v \sigma_v(s),\]

\[ \forall v' \neq v : A_{v', \sigma_{v'}}( v s ) = v s \text{ by Step 3}, \]

\[ \forall v' \neq v : A_{v', \sigma_{v'}}( v \sigma(s) ) =  v \sigma_v(s) \text{ by Step 4}. \]

We conclude that

\[ g_{n+1}(w) = \left( \prod_{d_{\ell^1}(o,v) = n} A_{v, \sigma_v}^{-1} \right) \circ g_n(w) = w. \]

This construction implies that, for any $g \in G_o$, we can use induction over $n$ to create a product of elements of the form $A_{v, \sigma_v}$ such that

\[ \left( \prod_{ d_{\ell^1}(o,v) \leq n } A_{v, \sigma_v}^{-1} \right) \circ g \]

fixes all vertices within the $(n+1)$-ball around $o$ with respect to $d_{\ell^1}$ and thus fixes the cubical subcomplex in $X$ spanned by these vertices. As $n$ tends to infinity, these compact subcomplexes cover more and more of $X$. Therefore, these products provide a sequence of elements in the subgroup generated by $\{ A_{v, \sigma} \}_{v, \sigma}$ that converges to $g$ in compact-open topology. Thus, the $A_{v,\sigma}$ generate a dense subgroup of $G_o$.

§.§ Generating a dense subgroup of $\Aut(X)$

In this subsection, we prove Theorem <ref>. Our strategy is to conjugate elements $A_{v, \sigma}$ by elements of $\Gamma$ and show that these conjugates have useful properties. Recall from section <ref> that $\Gamma$ acts on $X$ by translations. We start by introducing some terminology and establishing some properties of the interaction between translations and elements of $\{ A_{v, \sigma} \}_{v, \sigma}$.

Let $v$ be a reduced word in $\mathcal{L}_o^{cube}$. We define the reductions of $v$ to be the elements of the set

\[ \reduc(v) := \{ \alpha \in \Sigma_v \vert v\alpha \text{ is reducible} \}. \]

Let $v, v'$ be two reduced words in $\mathcal{L}_o^{cube}$. We say $v$ and $v'$ have the same reductions if $\reduc(v) = \reduc(v')$ and we write $v \sim v'$. This defines an equivalence-relation on the set of reduced words in $\mathcal{L}_o^{cube}$.

Note that the empty word $\epsilon$ forms its own, somewhat special equivalence class, which is why we remove it from some of the upcoming statements and treat it separately.

Suppose two reduced words $v, v'$ in $\mathcal{L}_o^{cube}$ have the same reductions, and $\sigma : \Sigma_{0,v} \rightarrow \Sigma_{0}$ is a label-morphism.

* If $s, t \in \Sigma_{0,v}$ such that $vs$ and $vt$ are reducible, then $[s, t] = 1$.

* If $s \in \Sigma_{0,v}$ such that $vs$ is reducible and $s^{-1} \in \Sigma_{0,v}$, then $vs^{-1}$ is reduced.

* $\sigma$ is compatible with $v$ if and only if it is compatible with $v'$.

(1) Let $s \neq t \in \Sigma_0$ such that $vs$ and $vt$ are both reducible. The path $v$ thus crosses the hyperplanes $\hat{h}(s)$ and $\hat{h}(t)$. Let $w$ be a reduced edge path such that $v \equiv w$. Since $v$ and $w$ cross the same hyperplanes, $w$ crosses $\hat{h}(s)$ and $\hat{h}(t)$ as well. Since $w$ is an edge path, it crosses one of them before the other and there lies a vertex on $w$ between the two crossings. Without loss of generality, $w$ crosses $\hat{h}(s)$ first and let $u$ be a vertex on $w$ after crossing $\hat{h}(s)$ and before crossing $\hat{h}(t)$. The vertices $o, u, w, ws$ now lie in four different connected components of $X \setminus (\hat{h}(s) \cup \hat{h}(t))$. Thus $\hat{h}(s)$ and $\hat{h}(t)$ intersect transversally. Since $s, t \in \Sigma_{0,v}$, they appear at the same vertex and thus $[s,t] = 1$.