Proof By a standard result in linear algebra, $\mathbf {n}$ is orthogonal to the expanding real eigenplane corresponding to the eigenvalue $\lambda $ in $M_\sigma $ with complex eigenvector $\mathbf {v}$ . Let $\mathbf {n} = \mathrm{Re} (\mathbf {v}) \times \mathrm{Im} (\mathbf {v})$ . By the Perron–Frobenius theorem there is an eigenvector $\mathbf {u}$ for the Perron–Frobenius eigenvalue $\omega $ of $C_2(M_\sigma )$ . Since $M_\sigma $ is complex non-Pisot, Lemma 3.1 implies $\omega = \lambda \overline {\lambda }$ . We know $\mathbf {u} = \mathrm{Re} (\mathbf {v})\wedge \mathrm{Im} (\mathbf {v}) = [P_1 \; P_2 \; P_3 ]^T,$ where $P_i = \det ([\mathrm{Re} (\mathbf {v}) \; \mathrm{Im} (\mathbf {v})][i])$ . However, we know that $\mathbf {n} = \mathrm{Re} (\mathbf {v}) \times \mathrm{Im} (\mathbf {v}) = [P_1 \; { -P_2 } \; P_3 ]^T$ . Since $\mathbf {u}> 0$ , it follows that $\mathbf {n}$ must have the asserted sign pattern.

Proof Since $\mathbf {u} = [u_1 \; u_2 \; u_3 ]^T$ has linearly independent entries over $\mathbb {Q}$ , it follows that $u_3 \neq 0$ , and thus the matrix $[ \mathbf { v_1} \; \mathbf {v}_2 ]$ is invertible. Define $A = [\mathbf {v}_1 \; \mathbf {v}_2]^{-1}$ and $\mathbf {w}_3 = A \mathbf {v}_3$ . It is easy to show that the entries of $\mathbf {w}_3$ have ratio ${u_1}/{u_2}$ , which must be irrational as $u_1$ and $u_2$ are rationally independent. Since A is a change of basis matrix on $\mathbb {R}^2$ , it is enough to show that $\operatorname {\mathrm {Span}}_{\mathbb {Z}}\{\mathbf {e}_1,\mathbf {e}_2, \mathbf {w}_3\}$ is dense in $\mathbb {R}^2$ . Define $T:\mathbb {T}^2 \to \mathbb {T}^2$ by $T(\mathbf {x}) = (\mathbf {x} + \mathbf {w}_3 )\mod 1$ , and note that T is strictly ergodic and thus minimal. Now choose any $(x,y) \in \mathbb {R}^2$ and fix any $\epsilon> 0$ . Since $\operatorname {\mathrm {Orb}}_{T}(\mathbf {0})$ is dense in $\mathbb {T}^2$ , there is an $n \in \mathbb {N}$ such that $T^n(\mathbf {0}) =( n\mathbf {w}_3)\mod 1 \in B_\epsilon ( (x,y)\mod 1)$ . Take $ [k \; l ]^T = [ \lfloor x \rfloor \; \lfloor y \rfloor ]^T - \lfloor n \mathbf {w}_3\rfloor $ . Then $ [k \; l ]^T + n\mathbf {w}_3 = k \mathbf {e}_1 + l \mathbf {e}_2 + n\mathbf {w}_3 \in B_{\epsilon }(x,y)$ , which gives us that $\operatorname {\mathrm {Span}}_{\mathbb {Z}}\{\mathbf {e}_1, \mathbf {e}_2, \mathbf {w}_3 \}$ is dense in $\mathbb {R}^2$ .

Proof Let K be compact with $L+K = \mathbb {R}^3$ . Choose $a> 0$ such that $[-a,a]^3 \supseteq K$ . Fix $\epsilon> 0$ . By compactness, $\text { there exist } \mathbf {v}_1, \mathbf {v}_2, \ldots , \mathbf {v}_n \in L$ such that $[-a,a]^2 \subseteq \bigcup _{i=1}^n ([-\epsilon ,\epsilon ]^2 + (\mathbf {v}_i)_{x,y})$ . Choose N so that $K' = \bigcup _{i=1}^n ((\mathbf {v}_i)_z +[-a,a]) \subseteq [-N,N]$ . To see that $([-\epsilon , \epsilon ]^2 \times [-N,N])+L = \mathbb {R}^3$ , choose any $\mathbf {w} \in \mathbb {R}^3$ . We know there exists $\mathbf {u} \in L$ such that $\mathbf {w} \in K+\mathbf {u}$ . There exists $\mathbf {v}_i$ such that $(\mathbf {w})_{x,y} - (\mathbf {u})_{x,y} \in [-\epsilon ,\epsilon ]^2 + (\mathbf {v}_i)_{x,y}$ . Now we note that $(\mathbf {v}_i)_{z} - (\mathbf {u})_{z} - \mathbf {w}_z \in [-a,a] \subseteq [-N,N].$ Thus, $\mathbf {w} \in K + \mathbf {u}$ .

Proof Let $K = [-a,a]\times [-b,b] \times [-c,c]$ and choose any $\mathbf {v} \in \mathbb {R}^3$ . Since $K+L = \mathbb {R}^3$ , we know there is an $\mathbf {l} \in L$ such that $\mathbf {v} \in \mathbf {l}+K$ . Since $\mathbf {0} \in K$ , we know that $\mathbf {l} \in \mathbf {l}+K$ . Now we consider $\mathbf {v} + K$ . As before, we see that $\mathbf {v} \in \mathbf {v} + K$ . Thus $(\mathbf {l}+K) \cap (\mathbf {v} + K) \neq \emptyset $ and $\mathbf {l} - \mathbf {v} = \mathbf {k} \in K$ . So, $\mathbf {l} = \mathbf {v} + \mathbf {k} \in \mathbf {v} + K$ .

Proof Fix $\epsilon> 0$ . By Lemma 3.6, if $0 < \delta < \epsilon /\sqrt {2}$ then there is an $N> 0$ such that $K + L = \mathbb {R}^3$ , satisfying $K = ([-\delta , \delta ]^2 \times [-N, N])$ . Choose any $x,y \in \mathbb {R}$ . By Lemma 3.7, we know that $ ( [ x \;\; y \;\; 0 ]^T + K ) \cap L \neq \emptyset $ . Let $\mathbf {v}$ be in that intersection. Thus $(s,t) = \pi _{\mathbb {R}^2}(\mathbf {v}) \in \pi _{\mathbb {R}^2}((\mathbb {R} \times [-N,N]) \cap L)$ and $(s,t) \in ([-\delta ,\delta ]^2+[ x \; y ]^T) $ . Since $0 < \delta < \epsilon /\sqrt {2}$ , it follows that $[-\delta ,\delta ]^2 \subseteq B_\epsilon (\mathbf {0})$ . This gives us that $(s,t) \in B_\epsilon ( [ x \; y ]^T )$ . Therefore, we see that $\Vert (x,y)-(s,t)\Vert < \epsilon $ .

Proof Assume $\{\mathbf {v}_1,\mathbf {v}_2,\mathbf {n}\}$ are orthonormal. Then $A = [\mathbf {v}_1\; \mathbf {v}_2 \; \mathbf {n}]^T$ . So, if $\mathbf {v}_1 = [ a_1 \; a_2 \; a_3]^T, \mathbf {v}_2 = [b_1 \; b_2 \; b_3 ]^T$ and $\mathbf {n} = [ n_1 \; n_2 \; n_3 ]^T$ , then $\mathbf {w}_1 = \pi _{\mathbb {R}^2}(A\mathbf {e}_1) = [a_1 \; b_1 ]^T, \mathbf {w}_2 = \pi _{\mathbb {R}^2}(A \mathbf {e}_2)= [ a_2 \; b_2]^T$ and $\mathbf {w}_3 = \pi _{\mathbb {R}^2}(A \mathbf {e}_2) = [ a_3 \; b_3 ]^T$ . Thus, $C_2([ \mathbf {v}_1 \; \mathbf {v}_2 ]) = \mathbf {v}_1 \times \mathbf {v}_2 = \mathbf {n}.$ Since by Lemma 3.3 $\mathbf {n}$ has linearly independent entries over $\mathbb {Q}$ , it follows from Lemma 3.4 that $\operatorname {\mathrm {Span}}_{\mathbb {Z}}\{\mathbf {w}_1,\mathbf {w}_2,\mathbf {w}_3\}$ is dense in $\mathbb {R}^2$ .

Proof It is easily shown that $\Theta (P_\sigma )$ fixes $P_\sigma $ if $q - 2 \geq r$ . For reference, Figure 5 shows $\Theta (P_\sigma )$ for the case $q = 3, r = 1$ . The rest of the proof is just induction.

Figure 5.

This is $\Theta ^3(P_\sigma )$ for $q = 3, r = 1$ .

Proof All that is required is to show that $P_\sigma $ is in some power of $t_a$ . This is worked out in Figure 7.

Proof Let $v \in \Sigma $ and $w \in \mathcal {F}(\{a,b,c\}),$ where $\ell (w) = \mathbf {v}$ , and assume w is not a good word. Then we can find a smallest subword of w of the form $iw'i^{-1}$ , where $w'$ is a good word. We will prove that we can always replace $iw'i^{-1}$ with a word $w_1i w_2 i^{-1}$ (or $iw_2i^{-1}w_1$ ) in w to get a new word u where $|w_2| < w'$ and $\ell (u) = \ell (w)$ . This tells us that we can eventually cancel any i and $i^{-1}$ pairs from w and still have $\ell (w) = \mathbf {v}$ . In this proof we will consider one of several cases, as each case works out similarly. Assume that $i = a$ . If $w'$ contained a or $a^{-1}$ , then there would be a smaller subword of the given form, which is a contradiction. If w only contains one type of letter, say b, then $abb \ldots ba^{-1}$ would define a sequence of $a \wedge b$ tiles in which the path goes from the lower left of the first tile to the upper left of the last tile. Then we could replace $abb \ldots ba^{-1}$ in w with $bb \ldots b$ which eliminates the bad subword, and we still have $\ell (w) = \mathbf {v}$ . Now assume $w'$ contains the letters b and c and that $w'$ starts with the letter b. Then $w'$ will have some sequence of bs eventually followed by a c. Notice that the edges corresponding to a and b can be tiled in two ways, either with one $a \wedge b$ tile or with an $a \wedge c$ and $b \wedge c$ tile. If it is tiled in the second way, then the tiling of $abb \ldots bc$ is fixed, since we now have a consecutive $c^{-1}$ and b edge. This forces all remaining tiles to be $b \wedge c$ tiles. If we start with $a \wedge b$ , then we continue to see $a \wedge b$ tiles until the last tile, which must be $a \wedge c$ . Lastly, it is possible to see a sequence of $a \wedge b$ followed by $a \wedge c$ . Again, as in the first case, all remaining tiles are fixed and must be $b \wedge c$ . See Figure 8. If we have the first case, we can substitute $cabb \ldots b$ for $abb \ldots bc$ . If we have the second case, we can substitute $bb \ldots bca$ . In the third case, we can substitute $bb \ldots bcabb \ldots b$ . In all three cases, a and $a^{-1}$ are closer together than before and the result follows. If $\mathbf {v} \in \Lambda $ , then $\mathbf {v} \in \Sigma - \mathbf {u}$ for some $\mathbf {u} \in \Sigma $ . We would follow the same proof in this case as before, except replacing paths along $\Sigma - \mathbf {u}$ for paths along $\Sigma $ .

Proof First, notice that $\pi _{\mathcal {U}}(\Sigma ) \subseteq \mathbb {Z}^2$ . Since the angle between the planes $P_{n_\theta }$ and $P_{n_{\mathcal {U}}}$ is less than $\pi /2$ , and $\pi _{E_\lambda }$ maps $\Sigma $ onto $P_{n_\theta }$ , it follows that $\pi _{\mathcal {U}}(\Sigma ) = \mathbb {Z}^2$ . All that remains is to show that $\pi _{\mathcal {U}}$ is one-to-one on $\Sigma $ . Suppose not. Then there exist $\mathbf {v}, \mathbf {w} \in \Sigma $ such that $\mathbf {v} - \mathbf {w} = k \mathbf {n_{\mathcal {U}}}$ . If $k = 1$ , then there is a good word w corresponding to $k \mathbf {n_{\mathcal {U}}}$ which is a permutation of the word $ab^{-1}c$ . However, it is easy to verify using the angle relations (see Figure 10) that w cannot lead to a tiling $T_\sigma $ , so w is not allowed. In the case $k> 1$ , again we have a good word w where $\ell (w) = k \mathbf {n_{\mathcal {U}}}$ . But such a word must have a subword $\alpha \beta ^{-l} \gamma $ , where $\alpha , \beta ,\gamma $ are distinct characters. Assume the subword has the form $ab^{-l}c$ . In this case, since $b^{-1}$ follows a, the angle between the edges must be acute, which implies that $ab^{-1}$ must be on the top side of the tile $a \wedge b$ . Therefore, we have a tiling similar to Figure 11, which implies again we have a path $ab^{-1}c$ , giving a contradiction.

Proof We will only prove this for $\mathbf {e}_1$ and $-\mathbf {e}_2$ . The proof of all other cases follows similarly with Figure 13 changed. When a path from $(0,0)$ to $(x,y)$ is shifted (in this case by either $-\mathbf {e}_1$ or $\mathbf {e}_2$ , then two SWONT graphs are added to the end of any new path (with $(0,0)$ being appropriately translated). See Figure 14. Therefore, Figure 13 tells us that the difference in height is not more than $\pm 1$ .

Proof Fix any $\epsilon> 0$ . By Lemma 3.8, we know that there is an $R> 0$ such that $\pi _{\mathbb {R}^2}((\mathbb {R}^2 \times [-R,R]) \cap L)$ is $\epsilon $ -dense. From the properties of f and g, we know there exists an $N> 0$ such that for all $\parallel\!\mathbf {x}\!\parallel\, > N$ , $f(\mathbf {x})> R$ and $g(\mathbf {x}) < -R$ . Therefore, $\pi _{\mathbb {R}^2}((\mathbb {R}^2 \setminus \overline {B_N(\mathbf {0})} \times [-R,R]) \cap L)$ is $\epsilon $ -dense in $\mathbb {R}^2 \setminus \overline {B_N(\mathbf {0})}$ , and thus $\pi _{\mathbb {R}^2}(\Gamma \cap L)$ is eventually dense.

Proof For any $\mathbf {u} \in \Lambda $ there are two vectors $\mathbf {v}, \mathbf {w} \in \Sigma $ such that $\mathbf {v} - \mathbf {w} = \mathbf {u}$ . Since $\mathbf {v}, \mathbf {w} \in \Sigma $ , there are facet tiles $t^{\prime }_{\mathbf {v}}$ and $t^{\prime }_{\mathbf {w}}$ satisfying $t^{\prime }_{\mathbf {v}} - \mathbf {v}, t^{\prime }_{\mathbf {w}}- \mathbf {w} \in \{t_{a}', t_{b}', t_{c}'\}$ . Thus, $\mathbf {v} \in \mathcal {L}_{t^{\prime }_{\mathbf {v}}}(\tilde {\Sigma })$ and $\mathbf {w} \in \mathcal {L}_{t^{\prime }_{\mathbf {w}}}(\tilde {\Sigma })$ , which implies that $\mathbf {u} \in \mathcal {L}_{t_{\mathbf {v}}}(\tilde {\Sigma }) - \mathcal {L}_{t_{\mathbf {w}}}(\tilde {\Sigma }) \subseteq \bigcup _{i,j \in \{t_{a}', t_{b}', t_{c}'\}}( \mathcal {L}_i(\tilde {\Sigma }) - \mathcal {L}_j(\tilde {\Sigma }) ) .$

Proof For any $M^k \mathbf {v} \in M^k \Lambda $ , there is a $\mathbf {u} \in \Sigma $ such that $\mathbf {v} - \mathbf {u} \in \Sigma $ . Choose $w \in \mathcal {F}(\mathcal {A})$ such that $\ell (w) = \mathbf {v} - \mathbf {u}$ . Since $T_\sigma $ is fixed by $\Theta $ , it follows that $f_{\sigma (w)}$ is also a path in $T_\sigma $ and thus $f_{\sigma (w)}'$ is a path in $\Sigma .$ Since $M^k(\ell (w)) = M^k\mathbf {v} - M^k\mathbf {u},$ it follows that $M^k\mathbf {v} - M^k\mathbf {u}$ is also in $\Sigma $ . The same argument proves that $M^k\mathbf u \in \Sigma $ , so $M^k\mathbf {v} \in \Lambda $ .

Proof Choose patches $P_1, P_2 \in \mathcal {P}_\sigma ^*$ . By primitivity, there exists $k \in \mathbb {N}$ such that $P_1, P_2 \subset \Theta ^k(t_i)$ , for all $t_i \in \{t_a, t_b, t_c\}$ . It suffices to show that the displacement vectors between $\Theta ^k(t_i)$ and $\Theta ^k(t_j)$ , for all $t_i, t_j \in \{t_a, t_b, t_c\}$ , are eventually dense. Equivalently, it is enough to prove $\pi _{E_\lambda }(\Upsilon )$ is eventually dense in $\mathbb {R}^2$ , where $\Upsilon = \bigcup _{t^{\prime }_i,t^{\prime }_j \in \{t_a', t_b', t_c'\}}\Psi ((\Theta ')^k(t^{\prime }_i),(\Theta ')^k(t^{\prime }_j)).$

To show $M^k\Lambda \subseteq \Upsilon $ , choose any $\mathbf {z}\in \Lambda $ and let $\mathbf {z} = \mathbf {u} - \mathbf {v}$ , where $\mathbf {u}, \mathbf {v} \in \Sigma .$ We can find $\mathbf {u}, \mathbf {v} \in \Sigma $ such that $\mathbf {u} - \mathbf {v} = \mathbf {z}$ and there exist facet tiles $t_1'$ and $t_2'$ such that $(\mathbf {u},t_1')$ and $(\mathbf {v},t_2')$ are contained in $\tilde {\Sigma }$ . Thus, there is a path which starts at $\mathbf {u}$ and ends at $\mathbf {v}$ . In other words, there is a word $w \in \mathcal {F}(\mathcal {A})$ such that $f_w'([0,|w|])$ is a path on $\tilde {\Sigma } - \mathbf {u}$ . Notice that $\ell (w) = \mathbf {z}$ , so $\ell (\sigma ^k(w)) = M^k\mathbf {z}$ . Since $f_w$ is contained in $T_\sigma - \pi _{E_\lambda }(\mathbf {u})$ , and by Lemma 4.6 ${\Theta ^k(T_\sigma - \pi _{E_\lambda }(\mathbf {u})) = T_\sigma - \lambda ^k \pi _{E_\lambda }(\mathbf {u})}$ , it follows that $f_{\sigma ^k(w)}'([0,|\sigma ^k(w)|])$ is a path from $(\Theta ')^k(t_1')$ to $(\Theta ')^k(t_2')$ . Since $f_{\sigma ^k(w)}'(|\sigma ^k(w)|) = \lambda ^k \mathbf {z}$ and $\pi _{E_\lambda }$ is one-to-one, it follows that $M^k\mathbf {z} \in \Upsilon $ .

Finally, we prove that $\pi _{E_\lambda }(M^k\Lambda )$ is eventually dense. Assume $\{\mathbf {v}_1, \mathbf {v}_2\}$ is a basis for $P_{\mathbf {n}}$ . Let A be the matrix which takes $\mathbf {v}_1$ to $\mathbf {e}_1$ , $\mathbf {v}_2$ to $\mathbf {e}_2$ , and $\mathbf {n}$ to $\mathbf {e}_3$ . It follows from Lemma 3.4 that $A M^k\mathbb {Z}^3$ is an irrational lattice, with $AM^k\Lambda $ as a subset. Since $\tilde \Lambda $ is unbounded, it follows that $AM^k\Lambda $ is unbounded. By Theorem 5.5, $\pi _{\mathbb {R}^2}(AM^k\Lambda )$ is eventually dense. Since $\pi _{P_{\mathbf {n}}}(M^k\Lambda ) = \pi _{\mathbb {R}^2}(AM^k\Lambda )$ , it follows that $(X_{T_{\sigma ,\mathbf {n}}},\mathbb {R}^2)$ is topologically mixing.

Proof First we prove equation (6.1). Fix any $j \in \{a,b,c,a^{-1},b^{-1},c^{-1}\}$ and write $\ell (j) = a_1^{(j)} \lambda \mathbf {v_\lambda } + a_2^{(j)} \overline {\lambda }\mathbf {\overline {v_\lambda }} + a_3^{(j)} \theta \mathbf {n_\theta }$ , where $a_1^{(j)}, a_2^{(j)} \in {\mathbb {C}}$ and $a_3^{(j)} \in \mathbb {R}$ . For any $k \geq 1$ , $\ell (\sigma ^k(j)) = (M_\sigma )^k(\ell (j)) = a_1^{(j)} \lambda ^k \mathbf {v} + \overline {a_1^{(j)}} \overline {\lambda }^k\mathbf {\overline {v}} + a_3^{(j)} \theta ^k\mathbf {n_\theta }.$ It follows that $(\ell (\sigma ^k(j)))_{x,y} =a_1^{(j)} \lambda ^k (\mathbf {v_\lambda })_{x,y} + \overline {a_1^{(j)}} \overline {\lambda }^k(\mathbf {\overline {v_\lambda }})_{x,y} + a_3^{(j)} \theta ^k(\mathbf {n_\theta })_{x,y}. $ Therefore,

$$ \begin{align*}\frac{\parallel\!(\ell(\sigma^k(j)))_{x,y}\!\parallel}{|\lambda|^k} = \bigg |\hspace{-1pt} \bigg | a_1^{(j)} \frac{\lambda^k}{|\lambda|^k} (\mathbf{v_\lambda})_{x,y} + \overline{a_1^{(j)}} \frac{\overline{\lambda}^k}{|\lambda|^k}(\mathbf{\overline{v_\lambda}})_{x,y} + a_3^{(j)} \bigg(\frac{\theta}{|\lambda|} \bigg)^k(\mathbf{n_\theta})_{x,y}\bigg |\hspace{-1pt} \bigg |.\end{align*} $$

Let us consider $ | | a_1^{(j)} ({\lambda ^k}/{|\lambda |^k}) (\mathbf {v_\lambda })_{x,y} + \overline {a_1^{(j)}} ({\overline {\lambda }^k}/{|\lambda |^k})(\mathbf {\overline {v_\lambda }})_{x,y} | |.$ Assume that $(\mathbf {v_\lambda })_{x,y} = \mathbf {f}_1 + \mathbf {f}_2 i$ , $\arg (\lambda )= e^{i\varphi }$ , and $a_1^{(j)} = r e^{i\gamma }$ . Applying Euler’s formula and simplifying, we obtain

$$ \begin{align*}\bigg |\hspace{-1pt} \bigg | a_1^{(j)} \frac{\lambda^k}{|\lambda|^k} (\mathbf{v_\lambda})_{x,y} + \overline{a_1^{(j)}} \frac{\overline{\lambda}^k}{|\lambda|^k}(\mathbf{\overline{v_\lambda}})_{x,y}\bigg |\hspace{-1pt} \bigg | &= 2r \parallel\! \mathbf{f}_1 \cos(k\varphi+\gamma) - \mathbf{f}_2\sin(k\varphi+\gamma)\!\parallel\\ &\leq 2 r (\parallel\! \mathbf{f}_1\!\parallel + \parallel\! \mathbf{f}_2 \!\parallel). \end{align*} $$

A simple calculation show that if $\parallel\! \mathbf {f}_1 \cos (k\varphi +\gamma ) - \mathbf {f}_2\sin (k\varphi +\gamma )\!\parallel $ is not bounded away from zero, then $f_1$ and $f_2$ are linearly dependent. But this is impossible since $E_\lambda $ is not perpendicular to the $xy$ -plane. Thus, there must exist $\epsilon> 0$ such that $2r \parallel\! \mathbf {f}_1 \cos (k\varphi +\gamma ) - \mathbf {f}_2\sin (k\varphi +\gamma )\!\parallel\, > \epsilon $ for all k.

For each $k \in \mathbb {N}$ , define

$$ \begin{align*}\gamma_k = \bigg |\hspace{-1pt} \bigg | a_1^{(j)} \frac{\lambda^k}{|\lambda|^k} (\mathbf{v_\lambda})_{x,y} + \overline{a_1^{(j)}} \frac{\overline{\lambda}^k}{|\lambda|^k}(\mathbf{\overline{v_\lambda}})_{x,y}\bigg |\hspace{-1pt} \bigg |\end{align*} $$

and

$$ \begin{align*}\beta_k = \bigg |\hspace{-1pt} \bigg | a_1^{(j)} \frac{\lambda^k}{|\lambda|^k} (\mathbf{v_\lambda})_{x,y} + \overline{a_1^{(j)}} \frac{\overline{\lambda}^k}{|\lambda|^k}(\mathbf{\overline{v_\lambda}})_{x,y} + a_3^{(j)} \bigg(\frac{\theta}{|\lambda|} \bigg)^k(\mathbf{n_\theta})_{x,y}\bigg|\hspace{-1pt} \bigg|.\end{align*} $$

Since $({\theta }/{|\lambda |} )^k$ goes to zero as $k \to \infty $ , it follows that $\lim _{k \to \infty } (\gamma _k - \beta _k) = 0$ . Thus, there exists $N \in \mathbb {N}$ such that for all $k \geq N$ we have $|\gamma _k - \beta _k| < \epsilon /2,$ which is to say $0 < \epsilon /2 \leq \gamma _k - \epsilon /2 < \beta _k < \gamma _k + \epsilon /2.$ By assumption $\beta _k> 0$ , for all $k \in \mathbb {N}$ , and since there are just finitely many $k \leq N$ , we can choose an $L> 0$ such that $L \leq \beta _k$ for all k.

Also, since $\gamma _k \leq 2 r (\parallel\!\mathbf {f}_1\!\parallel + \parallel\!\mathbf {f}_2\!\parallel )$ for all $k \in \mathbb {N}$ , and there are at most finitely many $k \leq N$ , we can choose $L'> 0$ such that $\beta _k \leq L'$ , for all $k \in \mathbb {N}$ . Since $\beta _k = {\parallel\!(\ell (\sigma ^k(j)))_{x,y}\!\parallel }/{|\lambda |^k},$ we obtain equation (6.1).

To prove equation (6.2), we note that $\pi _{E^{\prime }_\lambda }^\perp (a_1^{(j)}\lambda ^k \mathbf {v_\lambda } + \overline {a_1^{(j)}}\overline {\lambda }^k\mathbf {\overline {v_\lambda }} ) = 0$ . So, $|\pi _{E^{\prime }_\lambda }^\perp (\ell (\sigma ^k(i)))| = |a_3^{(j)}||\theta |^k|\pi _{E^{\prime }_\lambda }^\perp (n_\theta )|$ , which gives us the desired result.

To prove the final statement, note that $\mathbf {e}_2 = a_1^{(b)}\mathbf {v_\lambda } + \overline {a_1^{(b)}}\mathbf {\overline {v_\lambda }} + a_3^{(b)}\mathbf {n_\theta },$ and $\pi _{E^{\prime }_\lambda }^\perp (\mathbf {v_\lambda }) = \pi _{E^{\prime }_\lambda }^\perp (\mathbf {\overline {v_\lambda }}) = 0.$ Thus, $\pi _{E^{\prime }_\lambda }^\perp (\mathbf {e}_2) = (\mathbf {n^{\prime }_\theta })_y < 0,$ by Lemma 3.2. Therefore, $\pi _{E^{\prime }_\lambda }^\perp (\ell (\sigma ^k(b))) = \pi _{E^{\prime }_\lambda }^\perp (M^k(\mathbf {e}_2)) = a_3^{(3)}\theta ^k\langle \mathbf {n^{\prime }_\theta }, \mathbf {n_\theta }\rangle < 0,$ since we are assuming $\theta> 0$ .

Proof Without loss of generality, assume $\mathbf {v} = \mathbf {0}$ . Suppose $\ell _a(w_{\mathbf {v'}}) < 0$ . Since $w_{\mathbf {v'}}$ is a good word, this implies $w_{\mathbf {v'}}$ contains at least one $a^{-1}$ but no occurrences of a. By its relative position in the tiling, $w_{\mathbf {v'}}$ contains at least one b and at least one $c^{-1}$ but no occurrences of $b^{-1}$ or c. Therefore, by a similar argument to the proof of Lemma 5.2 there is a path which cannot be tiled. This is a contradiction. A similar argument shows $(\mathbf {v})_y < (\mathbf {v})_{y'}$ .

Proof Since $u_c$ is contained entirely in K, then choose two vertices $\mathbf {v}_l$ (left vertex) and $\mathbf {v}_r$ (right vertex) on the bottom side of $u_c$ with corresponding good words $w_l$ and $w_r$ . By path independence, it follows that $0 \leq \ell _a(w_l) < \ell _a(w_r)$ . Since $(\mathbf {v}_r)_x < \delta $ , it follows from Lemma 6.3 that $1 \leq \ell _a(w_r) \leq \ell _a(w).$

Proof Choose $(d,e) \in \mathbb {Z}^2$ , and fix $k \in \mathbb {N}$ . Want to prove there exists a $(d,e,z) \in \Lambda $ such that $z \geq k$ . Choose $\delta> 0$ , and define K and $K'$ as in the paragraph above. It follows from Lemma 6.4 that there must be a vertex $\mathbf {v} \in \mathbb {Z}^2 \cap K'$ with good word w such that $\ell _a(w) = d$ and $\ell _b(w) = e$ . Since $K'$ is bounded, it is clear that there exists $\gamma> 0$ where $h(x,y) < \gamma $ for all $(x,y) \in K' \cap \mathbb {Z}^2$ . Let $\mathbf {v}_1 \in \mathcal {U}_\sigma $ be a vertex that has at least $z + \gamma $ consecutive c edges. Notice that our choice of $\delta $ is independent of the shift of $\mathcal {U}_\sigma $ . Thus, in $\mathcal {U}_\sigma - \mathbf {v}_1$ , we can find a vertex $\mathbf {v}_2 \in K' \cap \mathbb {Z}^2$ such that there is a good word w in which $\ell _a(w) = -d$ and $\ell _b(w) = -e$ . Thus, the vertex $-\mathbf {v}_2$ in $\mathcal {U}_\sigma - \mathbf {v}_1 - \mathbf {v}_2$ will have at least $\gamma +z$ consecutive c edges. Notice that $\mathbf {v}_1 - \mathbf {v}_2 \in \mathbb {Z}^2$ , and thus the lift of $-\mathbf {v}_2$ must be contained in $\Lambda $ . Therefore, the vertex $-\mathbf {v}_2 + (\gamma +z)[ -1 \; 1 ]^T$ will have a corresponding good word $w'$ in which $\ell (w') = [d \; e \; z' ]^T$ , where $z'> \gamma + z - \gamma = z$ . This proves no upper bound for any $(d,e) \in \mathbb {Z}^2$ , and a similar argument shows no lower bound.

Proof Let w be a good word for $\mathbf {v}$ . Assume that the maximum number of consecutive c edges is k. It follows that we can place at most k occurrences of c (or $c^{-1}$ ) between the letters a or $a^{-1}$ , and b or $b^{-1}$ . Since there are $|i|+|j|+1$ locations to place the k occurrences of c, it follows that $|(\mathbf {v})_z| \leq |i| k + |j| k + k$ . Therefore, $\parallel\! \pi _{E_\lambda }(\mathbf {v})\!\parallel \leq |i| + |\lambda | |j| + |i| k|\lambda |^2 + |j| k|\lambda |^2 + k |\lambda |^2 \leq |\lambda ^2|((|i|+|j|)(1+k) + k) \leq 2|\lambda |^2(|i|+|j|)(1+k).$ Since $\parallel\!( \mathbf {v})_{x,y}\!\parallel\, > 0$ , the relative growth rate of $|i|+|j|$ and $\sqrt {i^2+j^2}$ is equal, and k only depends on $\sigma $ , the result follows.

Proof Assume that $\parallel\!(\mathbf {v})_{x,y}\!\parallel\, > 0$ . Let $\pi _{E_\lambda }(\mathbf {v}) = w \in {\mathbb {C}}$ . There exists a $u \in {\mathbb {C}}$ such that $w \in \operatorname {\mathrm {Vert}}(T_\sigma ) - u$ . Define $S_k = \lambda ^k \operatorname {\mathrm {Vert}}(T_\sigma ) - u$ , for all $k \geq 0$ . Lemma 4.7 implies that $S_0 \supseteq S_1 \supseteq S_2 \supseteq \cdots $ . Choose the largest $k \in \mathbb {N}$ such that $\parallel\!\ell (\sigma ^k(i))_{x,y}\!\parallel\ \leq\ \parallel\!(\mathbf {v})_{x,y}\!\parallel $ , for all $i \in \mathcal {A}$ . Choose the smallest l such that $|\lambda |^l\parallel\!(\mathbf {v})_{x,y}\!\parallel\ \geq |w|$ . Note that Lemma 6.6 implies that $l \leq n$ , where n depends only on the substitution and not on $\mathbf {v}$ . Lemma 6.2 implies that $L_1'|\lambda |^{k+l+1}| \, \geq \,\, \parallel\!\ell (\sigma ^{k+1+1}(i))_{x,y}\!\parallel \,\, \geq \, |\lambda |^l \parallel\!(\mathbf {v})_{x,y}\!\parallel \,\, \geq \, | w |,$ for some $i \in \mathcal {A}$ . Recall that $L_1'$ only depends on $\sigma $ . Notice that $\lambda T_\sigma - u$ is a blown-up version of the tiling of $T_\sigma - u$ . Consider the tiling $\lambda ^{l+k+1}T_\sigma - u$ . Suppose $(x_{k}, \lambda ^{k} t_{k})$ contains $0$ and $(y_{k}, \lambda ^{k}t_{k}')$ contains w. Select the vertex $a_{k} \in (x_{k}, \lambda ^{k}t_{k})$ closest to w and the vertex $b_{k}$ in $(y_{k},\lambda ^{k}t_{k}')$ closest to $0$ . Now choose the shortest edge path $\gamma _{k}$ on $\lambda ^{k}T_\sigma - u$ between $a_{k}$ and $b_{k}$ . Since there is a one-to-one correspondence between finite paths on a tiling and finite words, there is a word $w_k \in \mathcal {F}(\mathcal {A})$ such that $\lambda ^k f_{w_k}([0,|w_{k}|])+a_{k} = \gamma _{k}$ . We will set $c_{k} = w_k$ . Our careful choice of k ensures that there exists an $i \in \mathcal {A}$ such that $\parallel\!\ell (\sigma ^k(i))_{x,y}\!\parallel\, \leq\, \parallel\!(\mathbf {v})_{x,y}\!\parallel{\!.}$ Note that the number of choices for $c_k$ is finite by the local finite condition on $T_\sigma $ .

Now consider $\lambda ^{k-1}T_\sigma - u$ . In a similar fashion as earlier, we can find tiles $(x_{k-1},\lambda ^{k-1}t_{k-1})$ and $(y_{k-1}, \lambda ^{k-1}t_{k-1})$ that contain $0$ and w. Choose $a_{k-1}$ and $b_{k-1}$ such that $a_{k-1}$ is the closest vertex to w and $b_{k-1}$ is the vertex closes to $0$ . Since $a_{k},b_{k} \in S_{k}$ , and $S_{k} \subseteq S_{k-1}$ , there must be an edge path in $\lambda ^{k-1}T_\sigma -u$ connecting $a_{k}$ to $a_{k-1}$ , and $b_{k}$ to $b_{k-1}$ . Again, choose the shortest path. We refer to the word corresponding to the path from $a_{k-1}$ to $a_{k-2}$ as $s_{k-2}$ , and the path from $b_{k-1}$ to $b_{k-2}$ as $p_{k-2}$ . We continue in this manner until we have words $s_0, \ldots , s_{k-1}, c_{k}, p_{k-1}, \ldots , p_0 \in \mathcal {F}(\mathcal {A})$ . Let $t = s_0\sigma (s_1)\cdots \sigma ^{k-1}(s_{k-1})\sigma ^{k}(c_{k})\sigma ^{k-1}(p_{k-1})\cdots p_0 \in \mathcal {F}(\mathcal {A});$ it follows that $f_{t}(|t|) = w$ , and the lemma follows.

Proof Choose any $\mathbf {w} \in \Lambda $ . By Lemma 6.7, there is a word w of the form $w = s_0 \sigma (s_1) \cdots \sigma ^{k-1}(s_{k-1})\sigma ^{k}(c_{k})\sigma ^{k-1}(p_{k-1})\cdots \sigma (p_1) p_0,$ for which $f^{\prime }_{w}(|w|) = \mathbf {w}$ . For any $\sigma ^i(s_i)$ , where $0 \leq i \leq k-1$ (this holds similarly for all $\sigma ^{i}(p_i)$ s and $\sigma ^{k}(c_{k}))$ , it follows that $\ell (\sigma ^i(s_i)) = b_1 \lambda ^i \mathbf {v_\lambda } + b_2 \overline {\lambda }^i \mathbf {\overline {v_\lambda }} + b_3 \theta ^i \mathbf {n_\theta }.$ Since $\pi ^\perp _{E^{\prime }_\lambda }(\mathbf {v_\lambda }) = 0$ , it follows that $|\pi _{E^{\prime }_\lambda }^\perp (\ell (\sigma ^i(s_i)))| = |b_3| |\theta |^n |\pi _{E^{\prime }_\lambda }^\perp (\mathbf {n_\theta })|.$ Notice that $b_3$ only depends on $s_i$ and not on the power of $\sigma $ . Since the possible choices of the words $c_{k}, p_i, s_i$ are finite, we can choose a number C such that $|\pi _{E^{\prime }_\lambda }^\perp (\ell (\sigma ^i(s_i)))|\leq C|\theta ^i| \,|\pi _{E^{\prime }_\lambda }^\perp (\mathbf {n_\theta })|,$ for all words $c_{k-1}, p_i$ and $s_i$ . Define $C' = C|\pi _{E^{\prime }_\lambda }^\perp (\mathbf {n_\theta })|$ . Applying the triangle inequality, we get $|\pi _{E^{\prime }_\lambda }^\perp (\ell (w))| \leq 2C' \sum _{i = 0}^{k} |\theta |^i \leq 2C' {|\theta |^{k}}/({|\theta |-1}).$

Notice that $C'$ depends on the generalized substitution but not on the path. Lemma 6.7 also tells us that $\Vert (\mathbf {w})_{x,y}\Vert = \Vert \ell (w)_{x,y}\Vert \geq \Vert \ell (\sigma ^{k}(a))_{x,y}\Vert .$ By Lemma 6.2, there exits a constant $C"$ such that $C"\Vert (\mathbf {w})_{x,y} \Vert \geq ({2C'}/{|\theta |-1} )^{1/\alpha } |\lambda |^{k}.$ Hence, $ |\pi _{E^{\prime }_\lambda }^\perp (\mathbf {w}) | \leq {2C'}/({|\theta |-1}) |\lambda |^{\alpha k}\leq C"\Vert (\mathbf {w})_{x,y}\Vert ^\alpha .$

Proof Let $\gamma $ be a path on $\Sigma $ that takes $\mathbf {v}$ to $\mathbf {w}$ . Then along the path from $\mathbf {u}+\mathbf {v}$ to $\mathbf {u} + \mathbf {w}$ the path $\gamma $ must intersect $\Sigma $ at a point $\mathbf {z}$ . Let $\gamma '$ be the subpath of $\gamma $ from $\mathbf {u}+\mathbf {v}$ to $\mathbf {u}+\mathbf {w}$ that intersects $\Sigma $ . Then following the path $\gamma '$ from $\mathbf {v}$ gives us $\mathbf {v'}$ which is necessarily still in $\Sigma $ and also $\mathbf {v'}+ \mathbf {u}$ is also in $\Sigma $ . Thus, $\mathbf {u} = \mathbf {u}+\mathbf {v'} - \mathbf {v'} \in \Sigma - \Sigma $ .

Proof Suppose that $\pi ^\perp _{E^{\prime }_\lambda }(\mathbf {v}) < 0$ and there is no $\mathbf {w} \in \Sigma $ such that $\mathbf {v}+\mathbf {w}$ is on or below $\Sigma $ (it works similarly if $\pi ^\perp _{E^{\prime }_\lambda }(\mathbf {v})> 0$ ). Since $\mathbf {0} \in \Sigma $ , we know that $\mathbf {v}$ is above $\Sigma $ . There exists $\mathbf {t}_1 \in \Sigma $ such that $(\mathbf {t}_1)_{x,y} = (\mathbf {v})_{x,y}, \mbox { and } (\mathbf {t}_1)_z < (\mathbf {v})_z.$ Since $\mathbf {t}_1 + \mathbf {v}$ is above $\Sigma $ , there must a vector $\mathbf {t}_2 \in \Sigma $ such that $(\mathbf {t}_2)_{x,y} = (\mathbf {t}_1)_{x,y} + (\mathbf {v})_{x,y} = 2 (\mathbf {v})_{x,y} \mbox { and } (\mathbf {t}_2)_z < (\mathbf {t}_1)_z + (\mathbf {v})_z < 2 (\mathbf {v})_z.$ Iterating this provides us with a sequence of vectors $\mathbf {t}_m \in \Sigma $ such that $(\mathbf {t}_m)_{x,y} = m (\mathbf {v})_{x,y} \mbox { and } (\mathbf {t}_m)_z < m (\mathbf {v})_z.$ Since we know that $\pi _{E_\lambda }^\perp (\mathbf {u}) = \langle \mathbf {u}, \mathbf {n}\rangle $ for any vector $\mathbf {u}$ , and $(\mathbf {n})_z> 0$ , it follows that $\pi _{E^{\prime }_\lambda }^\perp (\mathbf {t}_m) < m \pi _{E^{\prime }_\lambda }^\perp (\mathbf {v}) = m \pi _{E^{\prime }_\lambda }^\perp (\mathbf {v}) {\Vert ( \mathbf {v} )_{x,y}\Vert }/{\Vert ( \mathbf {v} )_{x,y}\Vert } = \Vert ( \mathbf {t}_m )_{x,y}\Vert {-|\pi _{E^{\prime }_\lambda }^\perp (\mathbf {v})|}/{\Vert ( \mathbf {v} )_{x,y}\Vert }.$ Lemma 6.8 implies that there is a constant C such that $\pi _{E^{\prime }_\lambda }^\perp (t_m) \geq -C\Vert (\mathbf {t}_m)_{x,y} \Vert ^\alpha $ . This gives us $-C\Vert (\mathbf {t}_m)_{x,y} \Vert ^\alpha \leq \pi _{E^{\prime }_\lambda }^\perp (t_m) \leq -C" \Vert ( \mathbf {t}_m )_{x,y}\Vert .$ Or simply we see that

(6.4) $$ \begin{align}C \Vert (\mathbf{t}_m)_{x,y} \Vert^\alpha \geq C" \Vert (\mathbf{t}_m)_{x,y} \Vert, \end{align} $$

where $C" = {|\pi _{E^{\prime }_\lambda }^\perp (\mathbf {v})|}/{\Vert ( \mathbf {v} )_{x,y}\Vert }$ is a constant that depends only on $\mathbf {v}$ . However, for any $m> N$ , where $N = ({C}/{C"})^{1/(1 - \alpha )}\Vert \mathbf {v}_{x,y} \Vert ^{- 1}$ , we see that

$$ \begin{align*}C\Vert (\mathbf{t}_m)_{x,y}\Vert^\alpha &= C m^{\alpha} \Vert (\mathbf{v})_{x,y} \Vert^\alpha = C \frac{m}{m^{1-\alpha}} \Vert (\mathbf{v})_{x,y} \Vert ^\alpha\\ &< C m\bigg( \bigg(\dfrac{C}{C"} \bigg)^{1/(1 - \alpha)}\Vert (\mathbf{v})_{x,y} \Vert^{- 1}\bigg )^{\alpha-1} \Vert (\mathbf{v})_{x,y} \Vert^\alpha,\end{align*} $$

since $0< \alpha < 1$ . Continuing, we see that

$$ \begin{align*}C m \bigg( \bigg(\dfrac{C}{C"} \bigg)^{1/(1 - \alpha)}\Vert (\mathbf{v})_{x,y} \Vert^{- 1}\bigg )^{\alpha-1} \Vert (\mathbf{v})_{x,y} \Vert^\alpha = mC" \Vert (\mathbf{v})_{x,y} \Vert = C" \Vert (\mathbf{t}_m)_{x,y}\Vert.\end{align*} $$

Hence, $C\Vert (\mathbf {t}_m)_{x,y}\Vert ^\alpha < C" \Vert (\mathbf {t}_m)_{x,y} \Vert $ whenever $m> N$ , which contradicts equation (6.4). Therefore, we can find a $\mathbf {w}$ such that $\mathbf {v}+\mathbf {w}$ is below $\Sigma $ .

We now need to show that there is a $\mathbf {w'} \in \Sigma $ such that $\mathbf {v}+\mathbf {w'}$ is above $\Sigma $ . Suppose there is no such $\mathbf {w'} \in \Sigma $ . Again, since $\mathbf {0} \in \Sigma $ , we know that $\mathbf {v}$ is below $\Sigma $ . Thus, there is a vector $\mathbf {t}_1 \in \Sigma $ such that $(\mathbf {t}_1)_{x,y} = (\mathbf {v})_{x,y} \mbox { and } (\mathbf {t}_1)_z> (\mathbf {v})_z.$ Iterating as previously, we get a sequence of vectors $\mathbf {t}_m \in \Sigma $ such that $(\mathbf {t}_m)_{x,y} = m ( \mathbf {v})_{x,y} \mbox { and } (\mathbf {t}_m)_z> m (\mathbf {v})_z.$ Let us define a sequence $\mathbf {\zeta }a_n = \ell (\sigma ^n(b))$ , for all $n \geq 1$ . It is clear that each $\mathbf {\zeta }a_n$ belongs to $\Sigma $ . By Lemma 6.2, there are constants $L_1, L_1', L_2$ such that

(6.5) $$ \begin{align}L_1 |\lambda|^k \leq \Vert (\mathbf{\zeta}a_k)_{x,y} \Vert \leq L_1'|\lambda|^k\end{align} $$

and

(6.6) $$ \begin{align} \pi_{E^{\prime}_\lambda}^\perp(\zeta_k) \leq -L_2 |\theta|^k = -L_2|\lambda|^{k\alpha}. \end{align} $$

Since the projections are linear, we can see that $\pi _{E^{\prime }_\lambda }^\perp (\mathbf {\zeta }a_k) = \pi _{E^{\prime }_\lambda }^\perp (\mathbf {\zeta }a_k - \mathbf {t}_m) + \pi _{E^{\prime }_\lambda }^\perp (\mathbf {t}_m).$ Lemma 6.8 implies there is a $C>0$ such that

(6.7) $$ \begin{align}|\pi_{E^{\prime}_\lambda}^\perp(\mathbf{\zeta}a_k-\mathbf{t}_m)| \leq C\Vert (\mathbf{\zeta}a_k-\mathbf{t}_m)_{x,y}\Vert^\alpha = C\Vert (\mathbf{\zeta}a_k - m \mathbf{v})_{x,y}\Vert^\alpha. \end{align} $$

Let us define $L_3 = ({L_2}/{2C |\lambda |^\alpha })^{1/\alpha }$ and $C' = CL_3/L_1'.$ We define k to be the integer satisfying

(6.8) $$ \begin{align} L_3 |\lambda|^k \leq \Vert (\mathbf{v})_{x,y} \Vert < L_3 |\lambda|^{k+1}. \end{align} $$

Choose the largest m satisfying $ m \Vert ( \mathbf {v} )_{x,y} \Vert \leq \Vert ( \zeta _k )_{x,y} \Vert $ , and so that

(6.9) $$ \begin{align} \Vert (\zeta_k - m \mathbf{v})_{x,y} \Vert < \Vert (\mathbf{v})_{x,y}\Vert .\end{align} $$

Applying (6.8) and (6.9) to (6.7) gives us

(6.10) $$ \begin{align} C\Vert (\mathbf{\zeta}a_k - m \mathbf{v})_{x,y}\Vert^\alpha < C \Vert (\mathbf{v})_{x,y} \Vert^\alpha < C(L_3 |\lambda|^{k+1})^\alpha = C L_3^\alpha |\lambda|^{(k+1)\alpha}. \end{align} $$

Since $0 < |\pi ^\perp _{E_\lambda }| < C' \| (\mathbf {v})_{x,y}\|^\alpha $ , it follows that ${\pi _{E^{\prime }_\lambda }^\perp (\mathbf {v})> -C'\Vert (\mathbf {v})_{x,y}\Vert ^\alpha .}$ Thus

$$ \begin{align*}\pi_{E^{\prime}_\lambda}^\perp(\mathbf{t}_m)>-C'\Vert (\mathbf{v})_{x,y}\Vert^\alpha > -C' m (L_3|\lambda|^{k+1})^\alpha \geq -C'\frac{\Vert (\mathbf{\zeta}a_k)_{x,y}\Vert}{\Vert (\mathbf{v})_{x,y}\Vert} L_3^\alpha |\lambda|^{(k+1)\alpha},\end{align*} $$

where the last inequality comes from our assumption on m. Continuing, equation (6.5) tells us that $-\Vert (\zeta _k)_{x,y}\Vert \geq -L_1' |\lambda |^k$ and equation (6.8) tells us that $\Vert (\mathbf {v})_{x,y} \Vert \geq L_3|\lambda |^k$ , which gives us that

(6.11) $$ \begin{align} -C'\frac{\Vert (\mathbf{t}_m)_{x,y}\Vert }{\Vert (\mathbf{v})_{x,y}\Vert } L_3^\alpha |\lambda|^{(k+1)\alpha} \geq -C'\frac{L_1'|\lambda|^k}{L_3|\lambda|^k } L_3^\alpha |\lambda|^{(k+1)\alpha} = -CL_3^\alpha|\lambda|^{(k+1)\alpha}. \end{align} $$

Now we can use the inequalities (6.10) and (6.11) to give us

(6.12) $$ \begin{align} \pi_{E^{\prime}_\lambda}^\perp(\mathbf{\zeta}a_k) = \pi_{E^{\prime}_\lambda}^\perp(\mathbf{\zeta}a_k - \mathbf{t}_m) + \pi_{E^{\prime}_\lambda}^\perp(\mathbf{t}_m)> -2CL_3^\alpha |\lambda|^{(k+1)\alpha} = - L_2 |\lambda|^{k\alpha}. \end{align} $$

Hence, equation (6.12) contradicts equation (6.6). So, $\mathbf {v} \in \Sigma - \Sigma $ by Lemma 6.9.

Proof of Theorem 1.1 Proof of Theorem 1.1

If the number of edges corresponding to c is unbounded, then $\Lambda = \mathbb {Z}^3$ and Theorem 5.8 immediately implies the desired result. Otherwise, we need to use Lemma 6.10. Notice that $\mathbf {n_{\mathcal {U}}}$ is not parallel to the real eigenplane $E^{\prime }_\lambda $ . Therefore, any line in the direction of $\mathbf {n}_{\mathcal {U}}$ must intersect both $P_r$ and $P_{-r}$ , for any $r> 0$ . Fix r and construct $D'$ and F as in the previous paragraph. Since $D'$ is compact, we can construct a cylinder $D"$ lying between $P_r$ and $P_{-r}$ whose sides are parallel to $\mathbf {n_{\mathcal {U}}}$ , and which contains $D'$ . Choose any point $\mathbf {x} \in (F \setminus D")\cap P_r$ ; by the construction of $D"$ there must be a point $\mathbf {x'} \in (F\setminus D")\cap P_{-r}$ such that $\mathbf {x} - \mathbf {x'} = k \mathbf {n}_{\mathcal {U}}$ . Thus, $\Vert \mathbf {x} - \mathbf {x'} \Vert \geq 2r$ , which means that, by Lemma 6.10, if $\sigma $ satisfies (1.1), then $\Lambda $ is unbounded. Recall that $\mathbf {n}$ is a vector with entries that are linearly independent over $\mathbb {Q}$ . Therefore, by Theorem 5.8, the substitution tiling dynamical system $(X_{T_{\sigma ,\mathbf {n}}},\mathbb {R}^2)$ , where $T_{\sigma ,\mathbf {n}} = \pi _{P_{\mathbf {n}}}(\iota (T_\sigma ))$ is topologically mixing.