Geometrical Models for a Class of Reducible Pisot Substitutions

Abstract

We set up a geometrical theory for the study of the dynamics of reducible Pisot substitutions. It is based on certain Rauzy fractals generated by duals of higher dimensional extensions of substitutions. We obtain under certain hypotheses geometric representations of stepped surfaces and related polygonal tilings, as well as self-replicating and periodic tilings made of Rauzy fractals. We apply our theory to one-parameter family of substitutions. For this family, we analyze and interpret in a new combinatorial way the codings of a domain exchange defined on the associated fractal domains. We deduce that the symbolic dynamical systems associated with this family of substitutions behave dynamically as first returns of toral translations.

Appendices

Appendix A: Proof of Lemma 5.9

Lemma A.1

Let $n=p+m$, $M\in {\mathbb {R}}^{n\times n}$ and $\{\mathbf{e}_i\}_{i=1,\ldots ,n}$ be the canonical basis of ${\mathbb {R}}^n$. Moreover, let $\underline{a}=a_1\wedge \cdots \wedge a_m\in {\mathcal {O}}_m$ and $\mathbf{x}_1,\ldots ,\mathbf{x}_p\in {\mathbb {R}}^n$. Then

$$\begin{aligned}&\det (M\mathbf{x}_1,\ldots ,M\mathbf{x}_p,\mathbf{e}_{a_1},\ldots \mathbf{e}_{a_m}) \\&\quad =\sum _{\underline{b}=b_1\wedge \cdots \wedge b_{m}\in {\mathcal {O}}_m} \Big (\bigwedge _{i=1}^{p}M_\sigma \Big )_{\underline{a}^*,\underline{b}^*}(-1)^{\underline{a}+\underline{b}} \det (\mathbf{x}_1,\ldots ,\mathbf{x}_{p},\mathbf{e}_{b_1},\ldots ,\mathbf{e}_{b_{m}}). \end{aligned}$$

Proof

We write $\mathbf{x}_k=\sum _{j=1}^n\mathbf{x}_k^{(j)}{} \mathbf{e}_j$ for $k\in \{1,\ldots ,p\}$, $M=(m_{ij})_{1\le i,j\le n}$ and compute the determinant:

$$\begin{aligned}&\det (M\mathbf{x}_1,\ldots ,M\mathbf{x}_p,\mathbf{e}_{a_1},\ldots \mathbf{e}_{a_m})\\&= \det \Big (\sum _{i_1=1}^n\sum _{j_1=1}^nm_{i_1j_1}\mathbf{x}_1^{(j_1)}{} \mathbf{e}_{i_1},\ldots , \sum _{i_p=1}^n\sum _{j_p=1}^nm_{i_pj_p}{} \mathbf{x}_p^{(j_p)}{} \mathbf{e}_{i_p}, \mathbf{e}_{a_1},\ldots \mathbf{e}_{a_m}\Big ) \\&=\sum _{1\le j_1,\ldots ,j_p\le n}{} \mathbf{x}_1^{(j_1)}\cdots \mathbf{x}_p^{(j_p)} \det \Big (\sum _{i_1=1}^nm_{i_1j_1}\mathbf{e}_{i_1},\ldots ,\sum _{i_p=1}^nm_{i_pj_p}{} \mathbf{e}_{i_p},\mathbf{e}_{a_1},\ldots \mathbf{e}_{a_m}\Big )\\&=\sum _{\begin{array}{c}1\le j_1,\ldots ,j_p\le n,\\ \#\,\{j_1,\ldots , j_p\}=p\end{array}}{} \mathbf{x}_1^{(j_1)}\cdots \mathbf{x}_p^{(j_p)} \det \Big (\sum _{i_1=1}^nm_{i_1j_1}\mathbf{e}_{i_1},\ldots ,\sum _{i_p=1}^nm_{i_pj_p}{} \mathbf{e}_{i_p},\mathbf{e}_{a_1},\ldots \mathbf{e}_{a_m}\Big ). \end{aligned}$$

Remember that $\big (\bigwedge _{i=1}^{p}M\big )_{\underline{a}^*,\underline{b}^*}$ is the $p\times p$ minor of M obtained by deleting the rows $a_1,\ldots ,a_m$ and the columns $b_1,\ldots ,b_m$ of M, where $\underline{a}=a_1\wedge \cdots \wedge a_m$ and $\underline{b}=b_1\wedge \cdots \wedge b_m$. Therefore, by definition,

$$\begin{aligned} \det \Big (\sum \limits _{i_1=1}^nm_{i_1j_1}\mathbf{e}_{i_1},\ldots ,\sum _{i_p=1}^nm_{i_pj_p}{} \mathbf{e}_{i_p},\mathbf{e}_{a_1},\ldots \mathbf{e}_{a_m}\Big )\\ =(-1)^{\underline{a}+((p+1)+\cdots +n)} \Big (\bigwedge _{i=1}^{p}M\Big )_{\underline{a}^*,\underline{j}'^*} {\mathrm{sgn}}(\tau ), \end{aligned}$$

where $\underline{j}'=j_1'\wedge \cdots \wedge j_m'\in {\mathcal {O}}_m$ such that $\{j_1,\ldots ,j_p\}\cup \{j_1',\ldots ,j_m'\}=\{1,\ldots ,n\}$, $\tau$ is the permutation of $\{1,\ldots ,n\}$ satisfying $\tau (j_1')=j_1',\ldots ,\tau (j_m')=j_m'$ and $\tau (j_1)\le \cdots \le \tau (j_p)$ and ${\mathrm{sgn}}(\tau )$ its signature. We denote the set of permutations on $\{1,\ldots ,n\}$ by ${\mathrm{Per}}_n$. In this way, we obtain

$$\begin{aligned}&\det (M\mathbf{x}_1,\ldots ,M\mathbf{x}_p,\mathbf{e}_{a_1},\ldots \mathbf{e}_{a_m}) \\&\quad = \sum _{\begin{array}{c}1\le j_1,\ldots ,j_p\le n,\\ \#\,\{j_1,\ldots ,j_p\}=p\end{array}}\Big (\bigwedge _{i=1}^{p}M\Big )_{\underline{a}^*,\underline{j}'^*}(-1)^{\underline{a}+((p+1)+\cdots +n)} {\mathrm{sgn}}(\tau )\mathbf{x}_1^{(j_1)}\cdots \mathbf{x}_p^{(j_p)}\\&\quad = \sum \limits _{\underline{j}=j_1\wedge \ldots \wedge j_p\in {\mathcal {O}}_p}\Big (\bigwedge _{i=1}^{p}M\Big )_{\underline{a}^*,\underline{j}'^*}(-1)^{\underline{a}+((p+1)+\cdots +n)}\\&\qquad \times \sum _{\begin{array}{c}\tau \in {\mathrm{Per}}_n,\\ \tau (j_k')=j_k'\\ (1\le k\le m)\end{array}}{\mathrm{sgn}}(\tau )\mathbf{x}_1^{(\tau (j_1))}\cdots \mathbf{x}_p^{(\tau (j_p))}. \end{aligned}$$

Note that

$$\begin{aligned}&\sum _{\begin{array}{c}\tau \in {\mathrm{Per}}_n,\\ \tau (j_k')=j_k'\\ (1\le k\le m)\end{array}}{\mathrm{sgn}}(\tau )\mathbf{x}_1^{(\tau (j_1))}\cdots \mathbf{x}_p^{(\tau (j_p))}\\&\quad =(-1)^{\underline{j}'+((p+1)+\cdots +n)}\det (\mathbf{x}_1,\ldots ,\mathbf{x}_p,\mathbf{e}_{j_1'},\ldots ,\mathbf{e}_{j_m'}), \end{aligned}$$

which leads to the desired equality, after renaming $\underline{j}'\leftrightarrow \underline{b}$:

$$\begin{aligned}&\det (M\mathbf{x}_1,\ldots ,M\mathbf{x}_p,\mathbf{e}_{a_1},\ldots \mathbf{e}_{a_m}) \\&\quad =\sum \limits _{\underline{b}=b_1\wedge \ldots \wedge b_m\in {\mathcal {O}}_m}\Big (\bigwedge _{i=1}^{p}M\Big )_{\underline{a}^*,\underline{b}^*}(-1)^{\underline{a}+\underline{b}} \det (\mathbf{x}_1,\ldots ,\mathbf{x}_p,\mathbf{e}_{b_1},\ldots ,\mathbf{e}_{b_m}).\qquad \qquad \end{aligned}$$

$\square$

Proof of Lemma 5.9

Let $\mathbf{u}_{d+1},\ldots ,{\mathbf {u}}_{n}$ be a basis of ${\mathbb {K}}_n$ made of eigenvectors of $M_\sigma$ for the associated eigenvalues $\zeta _{d+1},\ldots ,\zeta _n$. Also, let $\mathbf{v}_{d+1},\ldots ,{\mathbf {v}}_{n}$ be eigenvectors of ${}^tM_\sigma$ for the associated eigenvalues $\zeta _{d+1},\ldots ,\zeta _n$, normalized to have ${\mathbf {u}}_k\cdot {\mathbf {v}}_k=1$ for $k\in \{d+1,\ldots ,n\}$. Since ${\mathbf {v}}_1,{\mathbf {v}}_{d+1},\ldots ,{\mathbf {v}}_n$ are all orthogonal to the contracting space ${\mathbb {K}}_c$, the following equality holds for every $\underline{a}=a_1\wedge \cdots \wedge a_{d-1}\in {\mathcal {O}}_{d-1}$:

$$\begin{aligned} \lambda _{d-1}(\pi _c\overline{({\mathbf {0}},\underline{a})})= \frac{\left| \det \big ({\mathbf {v}}_1,{\mathbf {v}}_{d+1},\ldots ,{\mathbf {v}}_{n},\pi _c(\mathbf{e}_{a_1}),\ldots ,\pi _c(\mathbf{e}_{a_{d-1}})\big )\right| }{\lambda _{\bar{n}}\big ({\mathbf {v}}_1,{\mathbf {v}}_{d+1},\ldots ,{\mathbf {v}}_{n}\big )}, \end{aligned}$$

(A.1)

where $\lambda _p$ denotes the Lebesgue measure on the p-dimensional space and

$$\begin{aligned} \lambda _{\bar{n}}\left( {\mathbf {v}}_1,{\mathbf {v}}_{d+1},\ldots ,{\mathbf {v}}_{n}\right) =:1/K_0 \end{aligned}$$

is the measure of the parallelotope generated by the vectors ${\mathbf {v}}_1,{\mathbf {v}}_{d+1},\ldots ,{\mathbf {v}}_{n}$.

Note that for all $\mathbf{x}\in {\mathbb {R}}^n$, we have the decomposition

$$\begin{aligned} \mathbf{x}=\langle \mathbf{x},\mathbf{v}_1\rangle \mathbf{u}_1+\sum _{i=d+1}^n\langle \mathbf{x},\mathbf{v}_i\rangle \mathbf{u}_i+\pi _c(\mathbf{x}). \end{aligned}$$

(A.2)

We use the above decomposition for the vectors ${\mathbf {v}}_1,{\mathbf {v}}_{d+1},\ldots ,{\mathbf {v}}_{\bar{n}}$ and expand the determinant by multilinearity to obtain:

$$\begin{aligned} \begin{array}{rcl} \lambda _{d-1}(\pi _c\overline{({\mathbf {0}},\underline{a})})&{}=&{} \underbrace{K_0\det \big (\langle \mathbf{v}_k,\mathbf{v}_l\rangle _{k,l=1,d+1,\ldots ,n}\big )}_{=:K}\\ &{}&{}\cdot \;\big |\det (\mathbf{u}_1,\mathbf{u}_{d+1},\ldots ,\mathbf{u}_{n},\pi _c(\mathbf{e}_{a_1}),\ldots ,\pi _c(\mathbf{e}_{a_{d-1}}))\big |. \end{array} \end{aligned}$$

The terms containing $\pi _c(\mathbf{v}_k)$ vanished for each value of $k\in \{1,d+1,\ldots ,n\}$), since the d vectors $\pi _c(\mathbf{v}_k),\pi _c(\mathbf{e}_{a_1}),\ldots ,\pi _c(\mathbf{e}_{a_{d-1}})$ are linearly dependent in the $(d-1)$-dimensional space ${\mathbb {K}}_c$. Using now (A.2) for the vectors $\mathbf{e}_{a_1},\ldots ,\mathbf{e}_{a_{d-1}}$, we can simplify the last expression to

$$\begin{aligned} \lambda _{d-1}(\pi _c\overline{({\mathbf {0}},\underline{a})})= K\big |\det (\mathbf{u}_1,\mathbf{u}_{d+1},\ldots ,\mathbf{u}_{n},\mathbf{e}_{a_1},\ldots ,\mathbf{e}_{a_{d-1}})\big |. \end{aligned}$$

(A.3)

Now, by (N), $g(0)=1=\Pi _{k=d+1}^n\zeta _k$ and therefore

$$\begin{aligned}&\det (\mathbf{u}_1,\mathbf{u}_{d+1},\ldots ,\mathbf{u}_{n},\mathbf{e}_{a_1},\ldots ,\mathbf{e}_{a_{d-1}}) \\&\quad = \frac{1}{\beta }\det (\beta \mathbf{u}_1,\zeta _{d+1}\mathbf{u}_{d+1},\ldots ,\zeta _n\mathbf{u}_{n},\mathbf{e}_{a_1},\ldots ,\mathbf{e}_{a_{d-1}})\\ \\&\quad =\frac{1}{\beta }\det (M_\sigma \mathbf{u}_1,M_\sigma \mathbf{u}_{d+1},\ldots ,M_\sigma \mathbf{u}_{n},\mathbf{e}_{a_1},\ldots ,\mathbf{e}_{a_{d-1}}). \end{aligned}$$

By Lemma A.1, we can relate the last determinant to $\bigwedge _{i=1}^{\bar{n}}M_\sigma$: for all $\underline{a}=a_1\wedge \cdots \wedge a_{d-1}\in {\mathcal {O}}_{d-1}$, we have

$$\begin{aligned}&\det (M_\sigma \mathbf{u}_1,M_\sigma \mathbf{u}_{d+1},\ldots ,M_\sigma \mathbf{u}_{n},\mathbf{e}_{a_1},\ldots ,\mathbf{e}_{a_{d-1}}) \\&\quad =\sum \limits _{\underline{b}=b_1\wedge \cdots \wedge b_{d-1}\in {\mathcal {O}}_{d-1}} \Big (\bigwedge _{i=1}^{\bar{n}}M_\sigma \Big )_{\underline{a}^*,\underline{b}^*}(-1)^{\underline{a}+\underline{b}} \det (\mathbf{u}_1,\mathbf{u}_{d+1},\ldots ,\mathbf{u}_{n},\mathbf{e}_{b_1},\ldots ,\mathbf{e}_{b_{d-1}})\\&\quad =\sum \limits _{\underline{b}\in {\mathcal {O}}_{d-1}} \Big (\bigwedge _{i=1}^{\bar{n}}{}^tM_\sigma \Big )_{\underline{b}^*,\underline{a}^*}(-1)^{\underline{a}^*+\underline{b}^*} \det (\mathbf{u}_1,\mathbf{u}_{d+1},\ldots ,\mathbf{u}_{n},\mathbf{e}_{b_1},\ldots ,\mathbf{e}_{b_{d-1}})\\&\quad =\sum \limits _{\underline{b}\in {\mathcal {O}}_{d-1}}(M_{d-1})_{\underline{b},\underline{a}}\det (\mathbf{u}_1,\mathbf{u}_{d+1},\ldots ,\mathbf{u}_{n},\mathbf{e}_{b_1},\ldots ,\mathbf{e}_{b_{d-1}}). \end{aligned}$$

We used here that $(-1)^{\underline{a}+\underline{b}}=(-1)^{\underline{a}^*+\underline{b}^*}$. By the above computation, this means that the vector

$$\begin{aligned} \mathbf{V}=\big (\det (\mathbf{u}_1,\mathbf{u}_{d+1},\ldots ,\mathbf{u}_{n},\mathbf{e}_{a_1},\ldots ,\mathbf{e}_{a_{d-1}})\big )_{\underline{a}\in {\mathcal {O}}_{d-1}} \end{aligned}$$

is an eigenvector of ${}^tM_{d-1}$ for the eigenvalue $\beta$. It follows that

$$\begin{aligned} \beta |{\mathbf {V}}|=|{}^tM_{d-1}{\mathbf {V}}|\le |{}^tM_{d-1}||{\mathbf {V}}|. \end{aligned}$$

Now, it follows from (4.5) and (P) that $|{}^tM_{d-1}|=\bigwedge _{i=1}^{\bar{n}}M_\sigma$ is a primitive matrix. Therefore, by Lemma 5.8, $\beta$ is its Perron–Frobenius eigenvalue. Consequently, the inequality $\beta |{\mathbf {V}}|\le |{}^tM_{d-1}||{\mathbf {V}}|$ is an equality and

$$\begin{aligned} \big (\lambda _{d-1}(\pi _c\overline{({\mathbf {0}},\underline{a})})\big )_{\underline{a}\in {\mathcal {O}}_{d-1}}= K|\mathbf{V}| \end{aligned}$$

is an eigenvector of $|{}^tM_{d-1}|$ for the eigenvalue $\beta$. $\square$

Appendix B: Proof of Lemma 7.1

The classical Rauzy fractal subtiles $\mathcal {R}(a)$ can be described via Dumont–Thomas numeration (see [17]) as

$$\begin{aligned} \mathcal {R}(a) = \Bigg \{\sum _{i\ge 0} \pi _c (M_\sigma ^i\,{\mathbf {l}}(p_i)): (p_i)_{i\ge 0} \in {\mathcal {G}}_p(a)\Bigg \}, \end{aligned}$$

(B.1)

where ${\mathcal {G}}_p(a)$ denotes the set of labels of infinite paths in the prefix graph of the substitution ending at $a\in {\mathcal {A}}$ (for more details see e.g. [13, 15]). We can follow as well infinite paths in the suffix graph. If we do this we get instead

$$\begin{aligned} -\mathcal {R}(a)-\pi _c({\mathbf {e}}_a) = \Bigg \{\sum _{i\ge 0} \pi _c (M_\sigma ^i\,{\mathbf {l}}(s_i)): (s_i)_{i\ge 0} \in {\mathcal {G}}_s(a)\Bigg \}, \end{aligned}$$

(B.2)

where ${\mathcal {G}}_s(a)$ denotes the set of labels of infinite paths in the suffix graph of the substitution ending at $a\in {\mathcal {A}}$ (see [15, Sect. 5]).

We can use Dumont–Thomas numeration to describe also the subtiles ${\mathcal {R}}(a\wedge b)$ using the ${\mathbf {E}}^2(\sigma )$ -suffix graph that is defined as follows.

Definition B.1

The ${\mathbf {E}}^2(\sigma )$-suffix graph has set of vertices $\{\underline{a}^*:\underline{a}\in \wedge ^3{\mathcal {A}}\}$, and there is an edge $\underline{a}^* \xrightarrow {\mathbf {s}}\underline{b^*}$ if and only if $\sigma (\underline{a}) = {\mathbf {p}}\underline{b}\mathbf {s}$, or equivalently if and only if $(M_\sigma ^{-1}{\mathbf {l}}(\mathbf {s}),\underline{a}^*) \in {\mathbf {E}}^2(\sigma )({\mathbf {0}},\underline{b}^*)$.

In Fig. 15 the ${\mathbf {E}}^2(\sigma _t)$-suffix graph is depicted. Observe that this graph with reversed edges describes the images of every face by ${\mathbf {E}}^2(\sigma _t)$.

Fig. 15

The ${\mathbf {E}}^2(\sigma _t)$-suffix graph. Note that an edge labeled by $5,\ldots ,1^{t-1}5$ denotes that there exist t edges, one labeled by 5, one by 15, and so on, until $1^{t-1}5$ (if $t=0$ then there is no edge of this type). The same is valid for the edges labeled by $2,\ldots ,1^{t}2$ (for $t=0$ there is only one edge labeled by 2)

Proposition B.2

We have

$$\begin{aligned} \mathcal {R}(a\wedge b) = \Bigg \{ \sum _{i\ge 0}\pi _c(M_\sigma ^i {\mathbf {l}}(\mathbf {s}_i)) : (\mathbf {s}_i)_{i\ge 0} \in {\mathcal {G}}_{s}(a\wedge b)\Bigg \} \end{aligned}$$

where ${\mathcal {G}}_s(a\wedge b)$ denotes the set of labels of infinite walks in the ${\mathbf {E}}^2(\sigma )$-suffix graph ending at state $a\wedge b$.

Proof

This is a direct consequence of Proposition 6.2 and of the definition of ${\mathbf {E}}^2(\sigma )$. $\square$

By abuse of notation we will write 0 instead of $\epsilon$ by reading labels of walks in the suffix or ${\mathbf {E}}^2(\sigma _0)$-suffix graphs.

We will relate the elements $\sum _{i\ge 0} \pi _c(M_\sigma ^i {\mathbf {l}}(\mathbf {s}_i))$ with $(\mathbf {s}_i)_{i\ge 0}\in {\mathcal {G}}_{s}(a\wedge b)$ with those $\sum _{i\ge 0} \pi _c(M_\sigma ^i {\mathbf {l}}(\mathbf {s}_i))$ for $(\mathbf {s}_i)_{i\ge 0} \in {\mathcal {G}}_s(a)$.

For the Hokkaido substitution $\sigma _0$ we have

$$\begin{aligned} {\mathcal {G}}_s(a) = \big \{ (\mathbf {s}_i)_{i\ge 0} = \cdots 0^5 2^{k_2} 0^5 2^{k_1} 0^{a-1} : 0\le k_i \le \infty \big \}. \end{aligned}$$

Proof of Lemma  7.1

Observe that

$$\begin{aligned} \pi _e(M_\sigma ^3{\mathbf {e}}_2) = \pi _e(M_\sigma {\mathbf {e}}_2) + \pi _e({\mathbf {e}}_2) \quad \text {i.e.}\quad \delta (2000.) = \delta (0022.), \end{aligned}$$

(B.3)

where $\delta (w) = \sum _{i=0}^{|w|} \pi _e(M_\sigma ^i {\mathbf {l}}(w_i))$. Notice that we can extend $\delta$ to infinite strings $(s_i)_{i\ge 0}$. We will prove using (B.3) that $\delta (\mathcal {G}_s(2\wedge 3))=\delta ({\mathcal {G}}_s(1)\cup {\mathcal {G}}_s(4))$. For this reason we will write $w =_\delta w'$ if $\delta (w)=\delta (w')$. The cycle

in the graph of Fig. 15 produces strings of type $0020^3(20)^k002=_\delta 0^5 2^{2k+2}$. Starting from state $2\wedge 5$ we get strings $0^5 2^{2k+1}$. Walking from the first node $2\wedge 5$ to the second $2\wedge 5$ returns $0^5 2^{2k}$ and extending this walk to the left starting from $2\wedge 3$ we obtain $0^5 2^{2k+3}$. Walking in Fig. 15 from $2\wedge 3$ to $2\wedge 5$ we get the word $0022002=_\delta 2 0^5 2$. Thus, $\delta (\mathcal {G}_s(2\wedge 3)) = \delta (\cdots 0^52^{k_2}0^52^{k_1})$, with $0\le k_i\le \infty$, i.e., $\delta ({\mathcal {G}}_s(1))\subseteq \delta (\mathcal {G}_s(2\wedge 3))$. Strings ending with $0^3$ are obtained following the loop $2\wedge 3 {\mathop {\rightarrow }\limits ^{00}}4\wedge 5{\mathop {\rightarrow }\limits ^{2}}2\wedge 5{\mathop {\rightarrow }\limits ^{2}}2\wedge 3$. Since these are all possible non-trivial paths ending at $2\wedge 3$ we have proven that $\delta (\mathcal {G}_s(2\wedge 3))=\delta ({\mathcal {G}}_s(1)\cup {\mathcal {G}}_s(4))$ which implies ${\mathcal {R}}(2\wedge 3) = (-{\mathcal {R}}(1)-\pi _c({\mathbf {e}}_1)) \cup (-{\mathcal {R}}(4)-\pi _c({\mathbf {e}}_4))$ by (B.2).

Since $2\wedge 3$ goes to $3\wedge 4$ by reading a 0 we deduce immediately that all the strings ending at $3\wedge 4$ are equivalent under $\delta$ to those in ${\mathcal {G}}_s(2)\cup {\mathcal {G}}_s(5)$. Hence by (B.2) we get ${\mathcal {R}}(3\wedge 4) = (-{\mathcal {R}}(2)-\pi _c({\mathbf {e}}_2)) \cup (-{\mathcal {R}}(5)-\pi _c({\mathbf {e}}_5))$.

Starting from $2\wedge 5$ and going to $2\wedge 4$ passing by $1\wedge 3$ we read 00 and by the above reasonings we get then all possible strings $\cdots 2^{k_2}0^52^{k_1}00$ belonging to ${\mathcal {G}}_s(3)$. From $2\wedge 5 {\mathop {\rightarrow }\limits ^{00}} 2\wedge 4{\mathop {\rightarrow }\limits ^{0}}3\wedge 5{\mathop {\rightarrow }\limits ^{2}}2\wedge 4$ we get all expansions in ${\mathcal {G}}_s(5)+2$, where with the latter we mean the set of $(s_i)_{i\ge 0}\in {\mathcal {G}}_s(5)$ such that $s_0=2$. Walking k times through the loop $2\wedge 4\rightarrow 3\wedge 5 \rightarrow 2\wedge 4$ and extending to the left with $2\wedge 5$ we get strings $0^2 (02)^k$. Subtracting $v_4 = \delta (200.)$ we get $0^2(02)^{k-2}0002 =_\delta 0^5 2^{2k-3}$. Walking through the loop $2\wedge 4 \rightarrow 3\wedge 5 \rightarrow 2\wedge 5 \rightarrow 2\wedge 4$ and then once into $2\wedge 4\rightarrow 3\wedge 5 \rightarrow 2\wedge 4$ we read the string $020^5 =_\delta 0^42^3$ and, after subtracting $v_4$, we get $0^5 2^2$. Repeating this loop we get arbitrary large strings ending with an even number of 2s. Thus we have shown we get strings in ${\mathcal {G}}_s(1)+4$.

So by (B.2) we just proved that ${\mathcal {R}}(2\wedge 4)$ is made of the domains $\delta ({\mathcal {G}}_s(3)) = -{\mathcal {R}}(3)-\pi _c({\mathbf {e}}_3)$, $\delta ({\mathcal {G}}_s(5)+2) = -{\mathcal {R}}(5)-\pi _c({\mathbf {e}}_5)+\pi _c({\mathbf {e}}_2) = -{\mathcal {R}}(5)-\pi _c({\mathbf {e}}_3)$ and $\delta ({\mathcal {G}}_s(1)+4) = -{\mathcal {R}}(1)-\pi _c({\mathbf {e}}_1) + \pi _c({\mathbf {e}}_4) = -{\mathcal {R}}(1)-\pi _c({\mathbf {e}}_3)$, since $\pi _c({\mathbf {e}}_1)=\pi _c({\mathbf {e}}_3)+\pi _c({\mathbf {e}}_4)$ and $\pi _c({\mathbf {e}}_5)=\pi _c({\mathbf {e}}_2)+\pi _c({\mathbf {e}}_3)$. $\square$