Proof Convergence is obvious. For any interval $[a..b]\subset I$ , there is some $\gamma _{n}$ such that $\gamma |_{[a..b]}$ agrees with $\gamma _{n}|_{[a..b]}$ . Consequently, $b-a=\operatorname {\mathrm {d}}(\gamma (a),\gamma (b))$ . It follows that $\gamma $ is a geodesic.

Proof of Theorem 4.2 Let $\Gamma $ be as in the statement of the theorem. Because a translation-like $\mathbb {Z}$ action restricts to a translation-like ${\mathbb Z}$ action on each component, we may assume that $\Gamma $ is connected. Clearly, every component of a graph admitting a translation-like $\mathbb {Z}$ action is infinite, and so we may further assume that $\Gamma $ is infinite.

Let $(X_{i})_{i}$ be a sequence of finite subsets of $V(\Gamma )$ satisfying:

1. (i) $X_{i} \subset X_{i+1}$ ;

2. (ii) $V(\Gamma ) = \bigcup _{i=1}^{\infty } X_{i}$ ;

3. (iii) the graph $\Gamma (X_{i})$ spanned by $X_{i}$ is connected.

We prove the following two simple lemmas.

Proof Because $\Gamma $ is countably infinite and connected we may fix $v_{1},v_{2},\ldots ,$ an enumeration of $V(\Gamma )$ , so that for every $i>1$ there is $j<i$ so that $v_{j}$ and $v_{i}$ co-bound an edge. Set $X_{i}$ to be

$$ \begin{align*} X_{i} := \{v_{1},\ldots,v_{i}\} .\end{align*} $$

By construction and induction $(X_{i})_{i}$ satisfies conditions (i)–(iii) above.

For the next lemma, we need the following.

Proof Clearly $V(H_{1}^{3}) \subset V(H_{2}^{3})$ . For any $u,v \in V(H_{1}^{3})$ we have

$$ \begin{align*} \operatorname{\mathrm{d}}_{H_{2}}(u,v) \leq \operatorname{\mathrm{d}}_{H_{1}}(u,v) .\end{align*} $$

If follows that any edge of $H_{1}^{3}$ is an edge of $H_{2}^{3}$ . The lemma follows.

In [Reference KaraganisKar68], Karaganis proved that the cube of any finite connected graph with at least three vertices admits a Hamiltonian circuit. (In other words, the only finite connected graphs whose cube is not Hamiltonian is a single edge and, depending on one’s definition, a single vertex.) For each $i \geq 3$ , $\Gamma (X_{i})$ is such a graph; we apply Karaganis’ theorem and fix $H_{i}$ , a Hamiltonian circuit of $\Gamma (X_{i})^{3}$ .

We next construct a sequence denoted $(E_{i})_{i}$ satisfying the following five properties for each $i \geq 3$ :

1. (1) $E_{i}$ is a set of edges of $\Gamma ^{3}$ ;

2. (2) $E_{i}$ is finite;

3. (3) $E_{i} \subset E_{i+1}$ ;

4. (4) at every vertex $v_{j} \in X_{i}$ , we have that $E_{i}$ has exactly two edges adjacent to $v_{j}$ ;

5. (5) the edges of $E_{i}$ are contained in infinitely many of the $H_{k}$ .

Consider first $v_{1}$ . Because $v_{1}$ is covered by every $H_{k}$ and $\Gamma (X_{i})^{3}$ is locally finite, infinitely many of the $H_{k}$ traverse the same pair of edges, say $e_{1}$ and $e_{1}'$ , at $v_{1}$ . We subsequence $(H_{k})_{k}$ so that each $H_{k}$ in the subsequence traverses both $e_{1}$ and $e_{1}'$ (we do not rename the sequence). We set

$$ \begin{align*} E_{1} := \{e_{1},e_{1}'\} .\end{align*} $$

For each $i\ge 1$ , having constructed $E_{i}$ , we consider $v_{i+1}$ . Local finiteness implies that infinitely many of the (subsequenced) $H_{k}$ traverse the same edges at $v_{i+1}$ , say $e_{i+1}$ and $e_{i+1}'$ . We further subsequence $(H_{k})_{k}$ and so that each $H_{k}$ in the subsequence traverses both $e_{i+1}$ and $e_{i+1}'$ . We set

$$ \begin{align*} E_{i+1} := E_{i} \cup \{e_{i+1},e_{i+1}' \} \end{align*} $$

This completes the construction of the sequence $(E_i)$ satisfying the conditions above.

We set H to be the subgraph of $\Gamma ^{3}$ whose vertices are $V(\Gamma )$ , and whose edges are $\bigcup _{i} E_{i}$ .

We endow the paths H with an arbitrary orientation. Given a vertex $v \in V(\Gamma ^{3})$ , let $f(v)$ be the vertex that follows v in the appropriate path of H. By Claim 4.6 this defines a fixed-point free action of $\mathbb {Z}$ on $V(\Gamma ^{3}) = V(\Gamma )$ . Clearly, we have

$$ \begin{align*} \operatorname{\mathrm{d}}_{\Gamma}(v,f(v)) \leq 3\operatorname{\mathrm{d}}_{\Gamma^{3}}(v,f(v)) = 3 \end{align*} $$

Thus, f defines a translation-like $\mathbb {Z}$ action on $\Gamma $ , proving Theorem 4.2.

Proof In the proof of this proposition (which does not involve the SFT structure of $\Omega $ ), to be consistent with the left action of G on $\partial G$ , we define a left action of G on $\Omega _S$ given by

$$ \begin{align*} (g \cdot \omega)(g') := (\omega\cdot g^{-1})(g') = \omega (g^{-1} g'). \end{align*} $$

We proceed as follows. First, we describe a factor map $\pi :\Omega _S{\rightarrow }\partial G$ . We use this map, together with the fact the $\partial G$ is minimal as a G system, to show that every shortlex shelling includes states from ${{\mathcal A}_{\text {max}}}\cup {{\mathcal A}_{\text {big}}}$ . We then use a compactness argument to show that there exists a k such that ${{\mathcal A}_{\text {max}}}\cup {{\mathcal A}_{\text {big}}}$ states are k-dense in every shelling. Finally, we use the fact that the future of any $2\delta $ -ball contains a k-ball to conclude that such states are $2\delta $ -dense.

Coding the boundary. Given a shortlex shelling $X=(h,\operatorname{\textsf{state}},P)$ , consider

. The function $\gamma _X:n\mapsto P^n(1_G)$ satisfies $h \circ \gamma _{X}(n) = h(1_G)-n$ and defines a geodesic ray. This defines a map $\pi :\Omega _S \to \partial G$

We claim that $\pi $ is a factor map, that is, $\pi $ is continuous, equivariant, and surjective.

Continuity follows directly from the definitions. To see that $\pi $ is equivariant, fix $g \in G$ and let $g \cdot X := (h',\operatorname{\textsf{state}}',P')$ , so that

. We have that $\gamma _{g\cdot X}(n)=gP^n(g^{-1})$ because a simple induction shows that

$$ \begin{align*} P^{\prime n}(1_G)=gP^{n}(g^{-1}),\end{align*} $$

as $P^{\prime 0}(1_G)=gP^0(g^{-1})$ and the inductive hypothesis $P^{\prime n}(1_G)=gP^{n}(g^{-1})$ implies

By Corollary 5.6, we know that $\gamma :n\mapsto P^n(g^{-1})$ is asymptotic to $\gamma _X$ and, thus,

$$ \begin{align*}\pi(g\cdot X)=[\gamma_{g\cdot X}]=[g\cdot \gamma]=g\cdot\pi(X),\end{align*} $$

showing that $\pi $ is G-equivariant. Finally, by [Reference GromovGro87], the action of G on its boundary is minimal, so the image of $\pi $ must be all of $\partial G$ , because it is a closed, non-empty subset of $\partial G$ preserved by G.

Every shortlex shelling includes a state of maximal growth. Let $\Omega _S'$ consist of all such that $X'=(h',\operatorname{\textsf{state}}',P')$ is a shortlex shelling with $\operatorname{\textsf{state}}'(G)\subseteq {{\mathcal A}_{\text {min}}}.$ We wish to show that $\Omega _S'$ is empty, so suppose otherwise. By minimality of $\partial G$ and the fact that $\pi $ is a factor map, we see that every point of $\partial G$ may be represented by an element $\pi (\Omega _S')$ .

Because $\operatorname{\textsf{state}}_0$ realizes values in ${{\mathcal A}_{\text {big}}}$ at infinitely many points, by compactness, there exists a shortlex shelling $X=(h,\operatorname{\textsf{state}},P)$ such that $\operatorname{\textsf{state}}(1_G)\in {{\mathcal A}_{\text {max}}}\cup {{\mathcal A}_{\text {big}}}$ . Let $X'=(h',\operatorname{\textsf{state}}',P')$ be a shortlex shelling such that and . For $g\in P^{-n}(1_G)$ , we may form asymptotic geodesics $\gamma ,\gamma '$ based at g via $\gamma (n)=P^n(g)$ and $\gamma '(n)=P^{\prime n}(g)$ and apply Lemma 3.11 to see that $P^{\prime n}(g)$ is within $2\delta $ of $1_G$ . Hence, $\#P^{\prime -n}B(2\delta ,1_G)\geq \#P^{-n}\{1_G\}.$ Because $\operatorname{\textsf{state}}(1_G)\in {{\mathcal A}_{\text {max}}}\cup {{\mathcal A}_{\text {big}}}$ , we know

$$ \begin{align*}\log(\#P^{-n}\{1_G\})/n{\rightarrow} \log(\lambda),\end{align*} $$

but because $\operatorname{\textsf{state}}'(B(2\delta ,1_G))\subset {{\mathcal A}_{\text {min}}}$ , we have

$$ \begin{align*}\limsup\log(\#P^{\prime -n}B(2\delta,1_G))/n<\log(\lambda),\end{align*} $$

giving us a contradiction. We conclude that $\Omega _S'$ must be empty.

Maximal growth states are k-dense for some k. Finally, suppose there is no k such that states of ${{\mathcal A}_{\text {max}}}\cup {{\mathcal A}_{\text {big}}}$ occur k-densely in every . Then there exist shortlex shellings $X_k=(h_k,\operatorname{\textsf{state}}_k,P_k)$ and $g_k\in G$ such that $\operatorname{\textsf{state}}_k(B(k,g_k))\subset {{\mathcal A}_{\text {min}}}$ . Then subconverges to some , but we must have , which we have seen is impossible.

Maximal growth states are $2\delta $ -dense. Suppose that $\operatorname{\textsf{state}}(B(2\delta ,g))\subseteq {{\mathcal A}_{\text {min}}}$ . We have seen in the proof of Proposition 5.5 that there exists some $g'\in P^{-*}(g)$ such that $B(k,g')\subseteq P^{-*}B(2\delta ,g)$ . Because ${{\mathcal A}_{\text {min}}}$ states, by definition, can only lead to ${{\mathcal A}_{\text {min}}}$ states, we have $\operatorname{\textsf{state}}(B(k,g'))\subseteq {{\mathcal A}_{\text {min}}}$ . Because $G^+$ is k-dense, we know that this cannot be the case, so we conclude that $G^+$ is in fact $2\delta $ -dense.

Proof Let v be a element of H and let B be the $2\delta $ ball in G around v. The future of B contains arbitrarily large balls and, in particular, must contain elements of $G^+$ . Thus, B contains an element of $G^+$ , say $v'$ . Now $v'$ must have either a predecessor or successor $v"$ in $H^+$ . We have that

$$ \begin{align*}\operatorname{\mathrm{d}}(v',v^{\prime\prime})=|h(v')-h(v^{\prime\prime})|\leq 2\delta\end{align*} $$

and, thus, $\operatorname {\mathrm {d}}(v",v)\leq 4\delta .$

Proof Suppose for $g_1$ and $g_2$ in $H^+$ , there is some C with $\operatorname {\mathrm {d}}(P^{-n}\{g_1\},P^{-n}\{g_2\})<C$ for all n. Take some $n>C+2\delta $ . Let B be a ball containing $g_1,g_2,P^{-n}{g_1}$ , and $P^{-n}{g_2}$ . Because every ball lies in the 1-interior of some cone there exists some $g_0\in G$ and a constant $C'\in {\mathbb Z}$ such that for all $g'\in B$ , we have that $h(g')=\operatorname {\mathrm {d}}(g',g_0)-C'$ . Let $\gamma _i$ , $i=1,2$ , be a geodesic from $g_0$ to $g_i$ . Let $t=\operatorname {\mathrm {d}}(g_0,g_1)=\operatorname {\mathrm {d}}(g_0,g_2)$ . Then, by Lemma 3.8, $\operatorname {\mathrm {d}}(g_1,g_2)=\operatorname {\mathrm {d}}(\gamma _1(t),\gamma _2(t))\leq 2\delta $ .

Proof Without loss of generality, set $H=h^{-1}(0)$ , and let $\xi $ denote the point of $\partial G$ represented by the geodesic ray $n\mapsto P^n(1_G)$ . A deep result of Swarup (building on work of Bowditch) asserts that $\partial G\setminus \xi $ is connected because G is one-ended [Reference SwarupSwa96]. We use this to show that the divergence graph on $H^+$ is connected.

1. (i) By an X-geodesic, we mean a function $\gamma : {\mathbb Z} \to G^+$ satisfying ${\gamma (n) = P(\gamma (n+1))}$ . It follows that $\gamma $ is a geodesic and satisfies $h(\gamma (n+1)) = h(\gamma (n))+1$ . Therefore, $h \circ \gamma :{\mathbb Z} \to {\mathbb Z}$ is an order preserving bijection, and we reparameterize $\gamma $ so that $h(\gamma (n)) = n$ . By an X-ray we mean a function $\gamma :{\mathbb Z}_{\geq 0} \to G^+$ satisfying $\gamma (n) = P(\gamma (n+1))$ and $h(\gamma (0))=0$ .

2. (ii) It follows that there is a bijection between X-geodesics and X-rays: given an X-geodesic we obtain an X-ray by restricting its domain, and any X-ray $\gamma $ can be extended uniquely to an X-geodesic $\gamma '$ by setting $\gamma '(n) = \gamma (n)$ for $n \geq 0$ and $\gamma '(n) = P^{-n}(\gamma (0))$ for $n<0$ . By Corollary 5.6, the ends of $\gamma '$ are $\xi $ and $[\gamma ]$ .

3. (iii) If S is a subset of $H^+$ , let ${\Pi }(S)$ denote the subset of $\partial G$ consisting of all $[\gamma ]$ where $\gamma $ is an X-ray with $\gamma (0)\in S$ . We write ${\Pi }(g)$ for ${\Pi }(\{g\})$ .

Our strategy is as follows: starting with a component S of the divergence graph on $H^+$ , we show that ${\Pi }(S)$ and ${\Pi }(H^+ \setminus S)$ decompose $\partial G \setminus \{\xi \}$ as the disjoint union of closed sets, with ${\Pi }(S)$ non-empty. Therefore, ${\Pi }(H^+ \setminus S) = \emptyset $ , which will imply that S is the entire divergence graph.

To that end, we claim the following conditions are satisfied.

(1) For any $g\in H^+$ , ${\Pi }(g)\neq \emptyset $ . We define an X-ray from g recursively as follows: starting with $\gamma (0) = g$ , having defined $\gamma (n) \in H_n^+$ , choose $\gamma (n+1) \in P^{-1}(\gamma (n))\cap H^+_{(n+1)}$ , which is not empty because $\gamma (n) \in H_n^+$ . Then $[\gamma ] \in {\Pi }(g)$ , establishing the claim.

(2) ${\Pi }(H^+)=\partial {G}\setminus \xi $ . Let $\gamma $ be an X-ray defining $[\gamma ] \in {\Pi }(H^+)$ . Then $\gamma $ can be extended to an X-geodesic connecting $\xi $ to $[\gamma ]$ . Because no geodesic connects $\xi $ to itself, we have that $[\gamma ] \neq \xi $ . This shows that $\xi \notin {\Pi }(H^+)$ .

Let $\eta \in \partial {G}\setminus \xi $ . We must show that $\eta $ is represented by an X-geodesic. Let $\gamma $ be a bi-infinite geodesic connecting $\xi $ and $\eta $ , which exists by [Reference Bridson and HaefligerBH99, Lemma III.H.3.2]. Now, for $n \in {\mathbb Z}$ we define a geodesic $\gamma _n$ as follows: pick $g_n \in B(2\delta ,\gamma (n))\cap G^+ $ (which is not empty by Proposition 6.5). Set $\gamma _n(t) := P^{n-t}(\gamma (n))$ . Similar to condition (1), set $\gamma _n(t+1)$ to be in $G^+ \cap P^{-1}(\gamma _n(t))$ for $t> n$ . By Lemma 3.11, we have that $\gamma _n(t) \in B(10\delta ,\gamma (t))$ for any $t \leq n$ . Therefore, for any $t \in {\mathbb Z}$ , $(\gamma _n(t))_{n\in {\mathbb N}}$ takes finitely many values. By Lemma 3.1 we have that $\gamma _n$ converges to a geodesic $\gamma '$ , and it is clear that (after reparametrizing) $\gamma '$ is an X-geodesic connecting $\xi $ at $-\infty $ to $\eta $ at $\infty $ . Thus $\eta \in {\Pi }(H^+)$ .

(3) ${\Pi }(S)\cap {\Pi }(H^+\setminus S) = \emptyset $ . By our definitions, any points $p, q\in H^+$ with ${\Pi }(p)\cap {\Pi }(q)\neq \emptyset $ share an edge in the divergence graph, and S is a component.

(4) For any $A \subset H$ we have that ${\Pi }(A)$ is closed in $\partial G \setminus \{\xi \}$ . Let $\eta ,\eta _{n} \in \partial G \setminus \{\xi \}$ satisfy:

1. (a) $\eta _{n} \to \eta $ ;

2. (b) $\eta _{n} \in {\Pi }(A)$ .

We will prove condition (4) by showing that $\eta \in {\Pi }(A)$ . Convergence on $\partial G$ is defined using the Gromov product, see Definitions [Reference Bridson and HaefligerBH99, III.H.1.20, 3.15]. By Remark [Reference Bridson and HaefligerBH99, III.H.3.17(6)] we have that $\eta _{n} \to \eta $ is equivalent to

$$ \begin{align*} \lim_{n \to \infty }(\eta_{n},\eta)_{p} = \infty \end{align*} $$

(where here $p \in G$ is an arbitrary basepoint) and, in particular, for any geodesic $\delta _{n}$ connecting $\eta _{n}$ to $\eta $ , we have that

$$ \begin{align*} \lim_{n \to \infty} \operatorname{\mathrm{d}}(p,\delta_{n}) = \infty .\end{align*} $$

We choose X-rays $\gamma ,\gamma _{n}$ so that:

1. (i) $[\gamma _{n}] = \eta _{n}$ and $\gamma _{n}(0) \in A$ ( $\gamma _{n}$ exists by assumption);

2. (ii) $[\gamma ] = \eta $ ( $\gamma $ exists by condition (2) above).

We will use $p := \gamma (0)$ as our basepoint. For sufficiently large n, we have that

$$ \begin{align*} \operatorname{\mathrm{d}}(\gamma(0),\delta_{n})> {3} \delta .\end{align*} $$

We now extend $\gamma $ and $\gamma _{n}$ to X-geodesics $\hat \gamma , \hat \gamma _{n}$ and consider the ideal triangle

$$ \begin{align*} \hat\gamma,\delta_{n}, \hat\gamma_{n} .\end{align*} $$

Because by Lemma 3.14 ideal geodesic triangles are ${3} \delta $ -slim, $\gamma (0) = \hat \gamma (0)$ is within ${3}\delta $ of $\hat \gamma _{n} \cup \delta _{n}$ , and we conclude that for some $t \in {\mathbb Z}$

$$ \begin{align*} \operatorname{\mathrm{d}}(\hat \gamma(0),\hat\gamma_{n}(t)) \leq {3}\delta .\end{align*} $$

Because $\hat \gamma , \hat \gamma _n$ are X-geodesics, we can conclude $t\le {3} \delta $ . Furthermore $\operatorname {\mathrm {d}}(\hat \gamma _n(t),\hat \gamma _n(0))=t$ and so we have that $ \operatorname {\mathrm {d}}(\gamma (0),\gamma _{n}(0)) = \operatorname {\mathrm {d}}(\hat \gamma (0),\hat \gamma _{n}(0)) \leq {6}\delta $ . Because this holds for every sufficiently large n, we conclude that $\{\gamma _{n}(0) \}$ is finite.

Therefore, $\gamma _n$ subconverges to some X-ray $\tilde \gamma $ with $\tilde \gamma (0)\in \{\gamma _n(0)\}\subset A$ . By Lemma 3.13, $[\gamma ]=[\tilde \gamma ]\in {\Pi }(A)$ . In particular, ${\Pi }(S)$ and ${\Pi }(H^+\setminus S)$ are closed in $\partial G\setminus \{\xi \}$ .

As noted above, [Reference SwarupSwa96] shows that $\partial {G}\setminus \xi $ is connected. Consequently, by conditions (2), (3), and (4), one of ${\Pi }(H^+\setminus S)$ or ${\Pi }(S)$ is empty. By condition (1) we have that ${\Pi }(S)$ is not empty and, hence, ${\Pi }(H^+\setminus S)=\emptyset $ . Applying condition (1) again, we conclude that $H^+\setminus S$ is empty, completing the proof.