January 01, 1970
We prove that a finite, connected, singly connected, locally convex higher-rank tree whose \(1\)-skeleton is planar and which is non-degenerate, in the sense that every edge of each colour forms a commuting square with every other colour, has rank at most four. Under these hypotheses this establishes the planarity conjecture stated in [1]. The obstruction side of the argument uses only the non-planarity of \(K_5\); it makes no appeal to the four-colour theorem. The engine is a monotonicity property of the set of colours emitted at a vertex (“backward propagation”), which forces, in any finite singly connected non-degenerate \(k\)-graph, a single vertex emitting all \(k\) colours; once \(k\ge 5\), local convexity manufactures a subdivision of \(K_5\) at such a vertex.
Higher-rank graphs (\(k\)-graphs) were introduced by Kumjian and Pask [2] as a combinatorial model for a class of \(C^\ast\)-algebras. A rank-one graph is an ordinary directed graph, and its path category is free; for \(k\ge 2\) the category is genuinely two-dimensional, encoded by a \(k\)-coloured \(1\)-skeleton together with a complete collection of bi-coloured commuting squares [3], [4]. Recent work has studied \(k\)-graphs from a combinatorial and topological standpoint, in particular their fundamental group and groupoid [5]–[7], and the class of higher-rank trees: connected \(k\)-graphs with trivial fundamental group.
In [1] a family of higher-rank trees is constructed from polyhedral graphs. Each member is connected, singly connected, locally convex, acyclic, embeds in its fundamental groupoid, and is planar: its \(1\)-skeleton \(\operatorname{Sk}\) is a planar graph, satisfying the higher-rank analogue \(\lvert\operatorname{Sk}^1\rvert-2\lvert\operatorname{Sk}^0\rvert+4=0\) of the elementary identity \(\lvert T^1\rvert-\lvert T^0\rvert+1=0\) for a finite rank-one tree. The construction produces ranks \(2,3,4\) only; the bound \(4\) enters through the four-colour theorem applied to the faces of the polyhedron. In [1] we asked for intrinsic criteria guaranteeing planarity, noting that degenerate examples exist in arbitrarily high rank — examples in which “not all edges of each degree have relations between them” — and conjecture a theorem “akin to the four-colour theorem” bounding the rank of a planar higher-rank tree.
The purpose of this note is to prove such a bound. We isolate the non-degeneracy condition implicit in the wording of [1] — that every edge of each colour forms a commuting square with every other colour, which we call edge-level non-degeneracy (Definition 7) — and prove the following.
Theorem 1. Let \(\Lambda\) be a finite, connected, singly connected, locally convex higher-rank tree of rank \(k\ge 5\). If \(\Lambda\) is edge-level non-degenerate, then its \(1\)-skeleton \(\operatorname{Sk}_\Lambda\) is non-planar. Consequently, a finite, connected, singly connected, locally convex, edge-level non-degenerate higher-rank tree with planar \(1\)-skeleton has rank at most \(4\).
The proof has two ingredients, both elementary. First, the colours emitted at a vertex satisfy a monotonicity law: along any edge the edges emitted by the range vertex contains that of the source vertex (Lemma 3). Combined with edge-level non-degeneracy this forces, in a finite singly connected \(k\)-graph, a single vertex that emits all \(k\) colours (Proposition 2); the finite acyclicity supplied by single connectedness completes the argument. Second, at a vertex emitting five colours, local convexity closes all ten of their pairwise squares, and single connectedness makes the resulting ten “diagonal” vertices distinct, producing a subdivision of \(K_5\) in the \(1\)-skeleton (Lemma 4). Non-planarity then follows from Kuratowski’s theorem [8], [9].
We stress that the obstruction uses only that \(K_5\) is non-planar; the full four-colour theorem [10] is required for the achievability of rank \(4\) in the construction of [1], not for the bound proved here. The hypotheses and their role are discussed in Section 5.
We follow the conventions of [2]–[4]. Let \(\mathbb{N}^k\) be the free abelian monoid with generators \(\varepsilon_1,\dots,\varepsilon_k\), ordered coordinatewise.
Definition 1 (\(k\)-graph). A higher-rank graph of rank \(k\) (or \(k\)-graph) is a countable category \(\Lambda\) together with a functor \(d\colon\Lambda\to\mathbb{N}^k\) satisfying the factorisation property: whenever \(d(\lambda)=m+n\) there exist unique \(\mu,\nu\in\Lambda\) with \(d(\mu)=m\), \(d(\nu)=n\) and \(\lambda=\mu\nu\). We write \(\mathop{\mathrm{r}},\mathop{\mathrm{s}}\colon\Lambda\to\Lambda^0\) for the range and source maps and \(\Lambda^m:=d^{-1}(m)\). Composition \(\mu\nu\) is defined when \(\mathop{\mathrm{s}}(\mu)=\mathop{\mathrm{r}}(\nu)\), and then \(\mathop{\mathrm{r}}(\mu\nu)=\mathop{\mathrm{r}}(\mu)\), \(\mathop{\mathrm{s}}(\mu\nu)=\mathop{\mathrm{s}}(\nu)\).
Since a \(k\)-graph is a category, elements are composed from right to left. For this reason we often draw edges pointing from right to left (cf.Definition 2 below.
Elements of \(\Lambda^0\) are vertices; elements of \(\Lambda^{\varepsilon_i}\) are edges of colour \(i\). For \(v,w\in\Lambda^0\) and \(F\subseteq\Lambda\) we write, following [4], \(vF=\mathop{\mathrm{r}}^{-1}(v)\cap F\) and \(Fw=\mathop{\mathrm{s}}^{-1}(w)\cap F\). The \(1\)-skeleton \(\operatorname{Sk}_\Lambda\) is the \(k\)-coloured directed graph with vertices \(\Lambda^0\) and edges \(\bigcup_{i\le k}\Lambda^{\varepsilon_i}\), an edge \(f\) having colour \(i\) if and only if \(f\in\Lambda^{\varepsilon_i}\).
Definition 2 (Commuting squares). For \(i\ne j\), a commuting \((i,j)\)-square is an element \(\lambda\in \Lambda^{\varepsilon_i+\varepsilon_j}\), presented through its two factorisations \[\begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/tsulcbiz.png}\label{predisbk}\end{figure}\tag{1}\] Its four sides are the edges \(\mu,\nu,\nu',\mu'\). An edge \(e\) lies in (or borders) the square if \(e\in\{\mu,\nu,\nu',\mu'\}\).
By the factorisation property, \((i,j)\)-squares are in bijection with \(\Lambda^{\varepsilon_i+\varepsilon_j}\), and each square determines a unique cofactorisation; this is the completeness of the canonical set of squares of a \(k\)-graph [3].
Definition 3 (Local convexity [4]). \(\Lambda\) is locally convex if for all \(i\ne j\) and all \(e\in \Lambda^{\varepsilon_i}\), \(f\in\Lambda^{\varepsilon_j}\) with \(\mathop{\mathrm{r}}(e)=\mathop{\mathrm{r}}(f)\), there exist \(e'\in\mathop{\mathrm{s}}(f)\Lambda^{\varepsilon_i}\) and \(f'\in\mathop{\mathrm{s}}(e)\Lambda^{\varepsilon_j}\) with \(ef'=fe'\).
That is, two differently coloured edges with a common range close up to a commuting square.
Definition 4 (Connected, singly connected [1]). \(\Lambda\) is connected if the equivalence relation on \(\Lambda^0\) generated by \(\{(u,v):u\Lambda v\ne\emptyset\}\) is all of \(\Lambda^0\times\Lambda^0\), and singly connected if \(\lvert u\Lambda v\rvert\le 1\) for all \(u,v\).
Definition 5 (Higher-rank tree). A higher-rank tree is a connected \(k\)-graph with trivial fundamental group (in the sense of [5]). The trees of [1] are moreover singly connected and locally convex; these are the only structural properties we use below.
We call \(\Lambda\) finite if \(\Lambda^0\) is finite. The \(1\)-skeleton is planar if the underlying undirected graph of \(\operatorname{Sk}_\Lambda\) admits a planar embedding; this is the meaning of “planar” in [1]. The only graph with one vertex which is a tree consists of just that vertex (and is planar), so we shall assume \(\Lambda\) has more that one vertex.
The only fact we use about single connectedness is the absence of nontrivial directed loops.
Lemma 1 (No directed loops). If \(\Lambda\) is singly connected and \(\lambda\in v\Lambda v\) then \(d(\lambda)=0\), i.e. \(\lambda=\mathrm{id}_v\).
Proof. The identity \(\mathrm{id}_v\in v\Lambda v\) has \(d(\mathrm{id}_v)=0\). Any \(\lambda\in v\Lambda v\) with \(d(\lambda)\ne 0\) is a second element, contradicting \(\lvert v\Lambda v\rvert\le 1\). ◻
Throughout this section \(\Lambda\) is a \(k\)-graph.
Definition 6. For \(v\in\Lambda^0\) set \[\mathsf{E}(v)=\{\,i: v\Lambda^{\varepsilon_i}\ne\emptyset\,\}, \qquad \mathsf{R}(v)=\{\,i: \Lambda^{\varepsilon_i}v\ne\emptyset\,\}.\] We say \(v\) emits colour \(i\) if \(i\in\mathsf{E}(v)\) (there is a colour-\(i\) edge with range \(v\)) and receives colour \(i\) if \(i\in\mathsf{R}(v)\) (there is a colour-\(i\) edge with source \(v\)).
Definition 7 (Edge-level non-degeneracy). \(\Lambda\) is edge-level non-degenerate if for every colour \(i\), every \(e\in\Lambda^{\varepsilon_i}\), and every \(j\ne i\), the edge \(e\) lies in some commuting \((i,j)\)-square.
This is the condition that every edge of each colour “has relations with” every other colour; its negation is the degeneracy noted in [1].
Lemma 2 (Boxing criterion). Let \(e\in\Lambda^{\varepsilon_i}\) and \(j\ne i\). Then \(e\) lies in an \((i,j)\)-square if and only if \[j\in\mathsf{E}(\mathop{\mathrm{s}}(e))\cup\mathsf{R}(\mathop{\mathrm{r}}(e)).\] Consequently \(\Lambda\) is edge-level non-degenerate if and only if \[\mathsf{E}(\mathop{\mathrm{s}}(e))\cup\mathsf{R}(\mathop{\mathrm{r}}(e))\;\supseteq\;\{1,\dots,k\}\setminus\{i\} \qquad\text{for every }e\in\Lambda^{\varepsilon_i}.\]
Proof. By Definition 2, \(e\) is a colour-\(i\) side of an \((i,j)\)-square \(\lambda=\mu\nu=\nu'\mu'\) if and only if \(e=\mu\) or \(e=\mu'\).
If \(e=\mu\) then \(\lambda=e\nu\) with \(\nu\in\Lambda^{\varepsilon_j}\) and \(\mathop{\mathrm{s}}(e)=\mathop{\mathrm{r}}(\nu)\), so \(\nu\in\mathop{\mathrm{s}}(e)\Lambda^{\varepsilon_j}\); such a \(\lambda\) exists if and only if \(\mathop{\mathrm{s}}(e)\Lambda^{\varepsilon_j}\ne\emptyset\), i.e. \(j\in\mathsf{E}(\mathop{\mathrm{s}}(e))\).
If \(e=\mu'\) then \(\lambda=\nu'e\) with \(\nu'\in\Lambda^{\varepsilon_j}\) and \(\mathop{\mathrm{s}}(\nu')=\mathop{\mathrm{r}}(e)\), so \(\nu'\in\Lambda^{\varepsilon_j}\mathop{\mathrm{r}}(e)\); such a \(\lambda\) exists if and only if \(\Lambda^{\varepsilon_j}\mathop{\mathrm{r}}(e)\ne\emptyset\), i.e. \(j\in\mathsf{R}(\mathop{\mathrm{r}}(e))\).
Hence \(e\) lies in an \((i,j)\)-square if and only if \(j\in\mathsf{E}(\mathop{\mathrm{s}}(e))\cup\mathsf{R}(\mathop{\mathrm{r}}(e))\). The second statement is the conjunction over \(j\ne i\). ◻
Lemma 3 (Backward propagation). For every \(e\in\Lambda^{\varepsilon_i}\), \[\mathsf{E}(\mathop{\mathrm{r}}(e))\;\supseteq\;\{i\}\cup\mathsf{E}(\mathop{\mathrm{s}}(e)).\] For instance: \[\begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/ehuoyisl.png}\label{wxpojsnm}\end{figure}\tag{2}\]
Proof. Since \(e \in \mathop{\mathrm{r}}(e)\Lambda^{\varepsilon_i}\), we have \(i\in\mathsf{E}(\mathop{\mathrm{r}}(e))\). Let \(j\in\mathsf{E}(\mathop{\mathrm{s}}(e))\) with \(j\ne i\). Choose \(\nu\in\mathop{\mathrm{s}}(e)\Lambda^{\varepsilon_j}\); then \(\mathop{\mathrm{s}}(e)=\mathop{\mathrm{r}}(\nu)\), so \(e\nu\) is defined and \(d(e\nu)=\varepsilon_i+\varepsilon_j\). By the factorisation property \(e\nu=\nu'\mu'\) with \(d(\nu')=\varepsilon_j\), \(d(\mu')=\varepsilon_i\), and \(\mathop{\mathrm{r}}(\nu')=\mathop{\mathrm{r}}(e\nu)=\mathop{\mathrm{r}}(e)\). Thus \(\nu'\in\mathop{\mathrm{r}}(e)\Lambda^{\varepsilon_j}\), giving \(j\in\mathsf{E}(\mathop{\mathrm{r}}(e))\). The case \(j=i\) is already covered by the first sentence. ◻
Lemma 3 says that the emitted-colour set is monotone along edges, toward the range: emission can only grow as one moves along the arrows in \(\Lambda\).
Proposition 2 (Forced concentration). Let \(\Lambda\) be a finite, singly connected, edge-level non-degenerate \(k\)-graph with at least one edge. Then there is a vertex \(v_0\) with \(\mathsf{E}(v_0)=\{1,\dots,k\}\).
Proof. Suppose, for contradiction, that \(\lvert\mathsf{E}(v)\rvert\le k-1\) for every vertex \(v\).
Step 1: every emitting vertex receives an edge. Let \(e\in\Lambda^{\varepsilon_i}\). By Lemma 3, \(\mathsf{E}(\mathop{\mathrm{r}}(e))\supseteq\{i\}\cup\bigl(\mathsf{E}(\mathop{\mathrm{s}}(e))\setminus\{i\}\bigr)\), a disjoint union, so \[\bigl\lvert \mathsf{E}(\mathop{\mathrm{s}}(e))\setminus\{i\}\bigr\rvert \;\le\;\lvert\mathsf{E}(\mathop{\mathrm{r}}(e))\rvert-1\;\le\;k-2 .\] By Lemma 2 and edge-level non-degeneracy, \[\bigl(\mathsf{E}(\mathop{\mathrm{s}}(e))\setminus\{i\}\bigr)\cup\bigl(\mathsf{R}(\mathop{\mathrm{r}}(e))\setminus\{i\}\bigr) =\{1,\dots,k\}\setminus\{i\},\] so \(\lvert\mathsf{R}(\mathop{\mathrm{r}}(e))\setminus\{i\}\rvert\ge (k-1)-(k-2)=1\). In particular \(\mathsf{R}(\mathop{\mathrm{r}}(e))\ne\emptyset\): the range of every edge \(e \in \Lambda^{\varepsilon_i}\) receives some colour. Since every vertex which emits a colour is the range of an edge, every emitting vertex receives an edge.
Step 2: an infinite backward chain. As \(\Lambda\) has an edge, pick a vertex \(v_0\) that emits it. By Step 1 \(v_0\) receives an edge, so there is an edge \(f_1\) with \(\mathop{\mathrm{r}}(f_1)=v_0\). Put \(v_1:=\mathop{\mathrm{s}}(f_1)\); then \(v_1\) emits an edge (\(f_1\) witnesses it), so by Step 1 it receives an edge, giving \(f_2\) with \(\mathop{\mathrm{r}}(f_2)=v_1\), and so on. Inductively we obtain edges \(f_1,f_2,\dots\) and vertices \(v_0,v_1,\dots\) with \[\mathop{\mathrm{r}}(f_{n+1})=v_n=\mathop{\mathrm{s}}(f_n)\qquad(n\ge 1),\] so each composite \(f_{n}f_{n-1}\cdots f_{m+1}\) is defined.
Step 3: contradiction. Since \(\Lambda^0\) is finite, the sequence \((v_n)\) repeats: \(v_m=v_n=:y\) for some \(m>n\ge 0\). Then (reading from right to left) \[\sigma:=f_{n+1} \cdots f_{m-1} f_{m}\] satisfies \(\mathop{\mathrm{r}}(\sigma)=\mathop{\mathrm{r}}(f_{n+1})=v_{n+1}=y\) and \(\mathop{\mathrm{s}}(\sigma)=\mathop{\mathrm{s}}(f_{m})=v_m=y\), so \(\sigma\in y\Lambda y\), while \(d(\sigma)=\sum_{\ell=n+1}^{m}\varepsilon_{c(f_\ell)}\ne 0\). This contradicts Lemma 1. Hence some vertex emits all \(k\) colours. ◻
Lemma 4 (\(K_5\) at a \(5\)-emitting vertex). Let \(\Lambda\) be a singly connected, locally convex \(k\)-graph and \(v_0\) a vertex emitting (at least) five colours, say \(1,\dots,5\in\mathsf{E}(v_0)\). Then \(\operatorname{Sk}_\Lambda\) contains a subdivision of \(K_5\); in particular \(\operatorname{Sk}_\Lambda\) is non-planar.
Proof. For \(\ell\in \mathsf{E}(v_0) =\{1,\dots,5\}\) choose \(e_\ell\in v_0\Lambda^{\varepsilon_\ell}\) and put \(w_\ell:=\mathop{\mathrm{s}}(e_\ell)\). Fix \(\ell\ne m\). Since \(\mathop{\mathrm{r}}(e_\ell)=\mathop{\mathrm{r}}(e_m)=v_0\), local convexity (Definition 3) yields \(e'_m\in\mathop{\mathrm{s}}(e_\ell)\Lambda^{\varepsilon_m}\) and \(e'_\ell\in\mathop{\mathrm{s}}(e_m)\Lambda^{\varepsilon_\ell}\) with \(e_\ell e'_m=e_m e'_\ell\). Set \[d_{\ell m}:=\mathop{\mathrm{s}}(e_\ell e'_m)=\mathop{\mathrm{s}}(e'_m)=\mathop{\mathrm{s}}(e'_\ell).\] Then \(\mathop{\mathrm{r}}(e'_m)=\mathop{\mathrm{s}}(e_\ell)=w_\ell\) and \(\mathop{\mathrm{r}}(e'_\ell)=\mathop{\mathrm{s}}(e_m)=w_m\), so in \(\operatorname{Sk}_\Lambda\) the edges \(e'_m\) and \(e'_\ell\) form an undirected path \[w_\ell\;\text{---}\;d_{\ell m}\;\text{---}\;w_m\] of length two with interior vertex \(d_{\ell m}\).
Distinctness of the fifteen vertices \(w_\ell,d_{\ell m}\) \(1 \le \ell < m \le 5\), . The element \(e_\ell\) lies in \(v_0\Lambda w_\ell\) with \(d(e_\ell)=\varepsilon_\ell\), and \(e_\ell e'_m\in v_0\Lambda d_{\ell m}\) with degree \(\varepsilon_\ell+\varepsilon_m\). If two of the fifteen vertices coincided, then \(v_0\Lambda(\cdot)\) would contain two morphisms of distinct degrees (the relevant degrees among \(\varepsilon_\ell\) and \(\varepsilon_\ell+\varepsilon_m\) are pairwise distinct, and all are nonzero), contradicting single connectedness. Hence \(w_1,\dots,w_5\) and the ten \(d_{\ell m}\) are fifteen distinct vertices, and each \(d_{\ell m}\) lies on exactly one of the ten connecting paths.
Therefore \(w_1,\dots,w_5\) are the branch vertices of a subdivision of \(K_5\), each of its ten edges subdivided once by a distinct \(d_{\ell m}\). A graph containing a \(K_5\)-subdivision is non-planar by Kuratowski’s theorem [8], [9]. \[\begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/gtwafnip.png}\label{ribulakf}\end{figure}\tag{3}\] ◻
Proof of Theorem 1. Let \(\Lambda\) be finite, connected, singly connected, locally convex, edge-level non-degenerate, of rank \(k\ge 5\). By Proposition 2 some vertex \(v_0\) emits all \(k\) colours, in particular five of them; by Lemma 4 the \(1\)-skeleton \(\operatorname{Sk}_\Lambda\) contains a subdivision of \(K_5\) and is non-planar. The final assertion is the contrapositive: a tree of rank \(\ge 5\) with these properties cannot have planar \(1\)-skeleton, so a planar one has rank \(\le 4\). ◻
Remark 3 (Sharpness). The bound is sharp, and the argument in Proposition 2 explains why. Running Proposition 2 at rank \(k=4\) still forces a vertex emitting all four colours, but the corresponding construction in Lemma 4 produces only a subdivision of \(K_4\), which is planar. Thus the obstruction first appears at rank \(5\), where \(K_5\) becomes the smallest non-planar complete graph. This is exactly the threshold realised by the construction in [1], whose planar trees occur in ranks \(2,3,4\).
Remark 4 (Role of the four-colour theorem). The theorem above is purely an obstruction: it uses only that \(K_5\) is not planar [8]. The four-colour theorem [10] is used in [1] for the complementary, constructive statement that rank \(4\) is always attainable (four colours suffice to colour the faces of a planar polyhedron). The two ingredients of a complete “\(4\)-colour-type” theory for planar higher-rank trees are therefore independent: achievability of \(\le 4\) from the four-colour theorem, and impossibility of \(\ge 5\) from \(K_5\), the latter proved here.
Remark 5 (The two readings of planarity, and \(Q_4\)). One may instead read “planar” as planarity of the associated \(2\)-complex \(X_\Lambda\): the CW-complex whose \(0\)- and \(1\)-cells are the vertices and edges of \(\operatorname{Sk}_\Lambda\) and which carries one \(2\)-cell glued along each commuting square (the \(2\)-skeleton of the cubical complex of [7]). Since a complex embeds in the sphere only if its \(1\)-skeleton does, planarity of \(X_\Lambda\) is stronger than the hypothesis of Theorem 1, and the corresponding bound \[X_\Lambda\;\text{planar and}\;\Lambda\;\text{edge-level non-degenerate} \;\Longrightarrow\;\operatorname{rank}\Lambda\le 4\] is an immediate corollary: planarity of \(X_\Lambda\) forces planarity of \(\operatorname{Sk}_\Lambda\), whereupon Theorem 1 applies.
The two readings do not coincide, and in particular one cannot reduce the \(1\)-skeleton statement to the \(2\)-complex one by promoting \(1\)-skeleton planarity to planarity of \(X_\Lambda\). The obstruction is the hypercube \(Q_4=\Omega_{4,\mathbf{1}}\), regarded as a rank-\(4\) \(k\)-graph [1]. It is a connected, singly connected, locally convex higher-rank tree, and it is edge-level non-degenerate: an edge of \(Q_4\) varies a single coordinate and lies on the three two-faces obtained by additionally varying one of the other three coordinates, so it boxes with all three remaining colours. Each edge of \(Q_4\) therefore borders three commuting squares. But any subcomplex of \(S^2\) has at most two \(2\)-cells along a given edge, so \(X_{Q_4}\) is not a surface, hence not planar; correspondingly \(\operatorname{Sk}_{Q_4}\) already contains a subdivision of \(K_{3,3}\) and is non-planar [1]. Thus the surface condition “every edge borders at most two squares”, which planarity of \(X_\Lambda\) entails, is not a consequence of being a non-degenerate locally convex tree: \(Q_4\) disproves it. (As \(Q_4\) has rank \(4\), it is of course no counterexample to Theorem 1, which constrains only rank \(\ge 5\).) Consequently \(2\)-complex planarity is a strictly stronger and non-automatic hypothesis, and the \(1\)-skeleton reading adopted in Theorem 1 — that of [1] — is the substantive one.
Remark 6 (Necessity of edge-level non-degeneracy). The interaction-graph weakening of Definition 7 — requiring only that each pair of colours box somewhere — is too weak. Glue, for each pair \(\{i,j\}\subseteq\{1,\dots,5\}\), one \((i,j)\)-square, identifying all of their sink corners to a single vertex. The result is connected, singly connected, locally convex, has planar \(1\)-skeleton (ten quadrilaterals meeting at a point), is a tree, and realises every pair of colours, yet has rank \(5\). It is excluded by Definition 7, since each of its edges lies in exactly one square and so boxes with only one other colour. This matches the description of the degenerate high-rank examples in [1].
Remark 7 (Finiteness). Finiteness is used only in Step 2–3 of Proposition 2: in an infinite \(k\)-graph the backward chain \((v_n)\) may fail to repeat, and the conclusion can fail. The infinite case — relevant to the \(\tilde{A}_2\) examples of [1] — is not addressed here.
Remark 8 (On the hypotheses). Single connectedness is used twice, each time only through Lemma 1: to close the chain in Proposition 2 and to separate the fifteen vertices in Lemma 4. Local convexity is used only in Lemma 4, to close the ten pairwise squares at \(v_0\). Both hold for the trees of [1]. We did not use the triviality of the fundamental group beyond what single connectedness already supplies.
Declaration 9. I acknowledge the use of Claude to brainstorm ideas and proofread grammar. A full record of prompts and outputs is available upon request