Equivalence between Digital Well-Composedness and Well-Composedness in the Sense of Alexandrov on n-D Cubical Grids

Abstract

Among the different flavors of well-composednesses on cubical grids, two of them, called, respectively, digital well-composedness (DWCness) and well-composedness in the sense of Alexandrov (AWCness), are known to be equivalent in 2D and in 3D. The former means that a cubical set does not contain critical configurations, while the latter means that the boundary of a cubical set is made of a disjoint union of discrete surfaces. In this paper, we prove that this equivalence holds in n-D, which is of interest because today images are not only 2D or 3D but also 4D and beyond. The main benefit of this proof is that the topological properties available for AWC sets, mainly their separation properties, are also true for DWC sets, and the properties of DWC sets are also true for AWC sets: an Euler number locally computable, equivalent connectivities from a local or global point of view. This result is also true for gray-level images thanks to cross section topology, which means that the sets of shapes of DWC gray-level images make a tree like the ones of AWC gray-level images.

Appendices

Proofs of Section 3

Lemma 1Let c be a value in $({\mathbb {Z}}/2) \setminus {\mathbb {Z}}$, and let y be a value in ${\mathbb {Z}}$. Then,

$$\begin{aligned} y \in \left\{ c - \frac{1}{2}, c + \frac{1}{2} \right\} \Leftrightarrow \beta ({\mathcal {H}} (y)) \subseteq \beta ({\mathcal {H}} (c)). \end{aligned}$$

In other words, when ${\mathcal {H}} (c)$is a 0-face of ${\mathbb {H}}^{1}$, y is a neighbor of c in $({\mathbb {Z}}/2)$iff the opening of ${\mathcal {H}} (y)$is included in the opening of ${\mathcal {H}} (c)$.

Proof

When c belongs to $({\mathbb {Z}}/2) \setminus {\mathbb {Z}}$,

$$\begin{aligned} {\mathcal {H}} (c) = \left\{ c + \frac{1}{2} \right\} \in {\mathbb {H}}^{1} _0, \end{aligned}$$

and then we have:

$$\begin{aligned} \beta ({\mathcal {H}} (c)) = \{\{c - 1/2,c+1/2\},\{c + 1/2\},\{c + 1/2, c + 3/2\}\}. \end{aligned}$$

Also, when $y \in {\mathbb {Z}}$, ${\mathcal {H}} (y) = \{y,y+1\} \in {\mathbb {H}}^{1} _1$, and then:

$$\begin{aligned} \beta ({\mathcal {H}} (y)) = \{\{y,y+1\}\}. \end{aligned}$$

If y belongs to $\{c - \frac{1}{2}, c + \frac{1}{2} \}$, we obtain that

$$\begin{aligned} \beta ({\mathcal {H}} (y)) \subseteq \beta ({\mathcal {H}} (c)). \end{aligned}$$

Conversely, if $\{\{y,y+1\}\}$ is included into

$$\begin{aligned} \{\{c - 1/2,c+1/2\},\{c + 1/2\},\{c + 1/2, c + 3/2\}\}, \end{aligned}$$

it means that $y \in \{c - 1/2,c+1/2\}$. $\square$

Proposition 4Let S be a block in ${\mathbb {Z}} ^n$, and let c be its center in $\left( \frac{{\mathbb {Z}}}{2}\right) ^n$. Then, S and c are related this way:

$$\begin{aligned} S = {\mathcal {Z}} _n (\beta ({\mathcal {H}} _n (c)) \cap {\mathbb {H}}^{n} _n). \end{aligned}$$

Proof

This proof is depicted in Fig. 13. Now, we can remark that:

$$\begin{aligned} S = \left\{ c + \sum _{i \in \frac{1}{2} (c)} \lambda _i e^i \; ; \; \forall i \in \frac{1}{2} (c), \lambda _i \in \left\{ -\frac{1}{2},\frac{1}{2} \right\} \right\} . \end{aligned}$$

Then, for any $y \in S$,

• if $i \in \llbracket 1,n \rrbracket \setminus \frac{1}{2} (c)$, then $y_i = c_i$,

• if $i \in \frac{1}{2} (c)$ such that $\lambda _i = 1/2$, then $y_i = c_i + 1/2$ with $c_i \in ({\mathbb {Z}}/2) \setminus {\mathbb {Z}}$,

• and if $i \in \frac{1}{2} (c)$ such that $\lambda _i = -1/2$, hence $y_i = c_i - 1/2$ with $c_i \in ({\mathbb {Z}}/2) \setminus {\mathbb {Z}}$.

Then, for any $i \in \llbracket 1,n \rrbracket$, by Lemma 1, ${\mathcal {H}} (y_i) \in \beta ({\mathcal {H}} (c_i))$, and then ${\mathcal {H}} _n (y) \in \beta ({\mathcal {H}} _n (c))$. Because $y \in {\mathbb {Z}} ^n$, ${\mathcal {H}} _n (y) \in {\mathbb {H}}^{n} _n$, and then ${\mathcal {H}} _n (y) \in \beta ({\mathcal {H}} _n (c)) \cap {\mathbb {H}}^{n} _n$, which leads to $y \in {\mathcal {Z}} _n (\beta ({\mathcal {H}} _n (c)) \cap {\mathbb {H}}^{n} _n)$.

Conversely, let us assume that $y \in {\mathcal {Z}} _n (\beta ({\mathcal {H}} _n (c)) \cap {\mathbb {H}}^{n} _n)$. Then, ${\mathcal {H}} _n (y) \in \beta ({\mathcal {H}} _n (c)) \cap {\mathbb {H}}^{n} _n$, which means that $y \in {\mathbb {Z}} ^n$, and ${\mathcal {H}} _n (y) \in \beta ({\mathcal {H}} _n (c))$. In other words, for any $i \in \llbracket 1,n \rrbracket$, ${\mathcal {H}} (y_i) \in \beta ({\mathcal {H}} (c_i))$. Two cases are then possible: $c_i \in {\mathbb {Z}}$, hence $y_i = c_i$, or $c_i \in ({\mathbb {Z}}/2) \setminus {\mathbb {Z}}$ and thus by Lemma 1, $y_i \in \{c_i-\frac{1}{2},c_i+\frac{1}{2} \}$. This way, $y \in S$. $\square$

Lemma 2Let a, b be two elements of ${\mathbb {H}}^{n}$. Then, $\alpha (a) \cap \alpha (b) \ne \emptyset$iff the $a \wedge b$is WD. Furthermore, when $a \wedge b$is WD, we can switch the operators $\times$and $\wedge$in this way:

$$\begin{aligned} a \wedge b = (\times _{i \in \llbracket 1,n \rrbracket } a_i) \wedge (\times _{i \in \llbracket 1,n \rrbracket } b_i) = \times _{i \in \llbracket 1,n \rrbracket } (a_i \wedge b_i), \end{aligned}$$

and we obtain $\alpha (a \wedge b) = \alpha (a) \cap \alpha (b)$.

Fig. 52

Study case-by-case showing that, when $a,b \in {\mathbb {H}}^{1}$, $a \wedge b$ exists iff $\alpha (a) \cap \alpha (b) \ne \emptyset$. The closures $\alpha (a)$ and $\alpha (b)$ are depicted in red, and the face encircled in blue corresponds to $a \wedge b$ when it exists

Proof

Let $a_1,b_1$ be two elements of ${\mathbb {H}}^{1}$, then it is easy to show by a case-by-case study (see Fig. 52 for the different possible cases) that:

$$\begin{aligned} \left\{ \begin{array}{l} \left\{ \alpha (a_1) \cap \alpha (b_1) \ne \emptyset \right\} \Leftrightarrow \left\{ a_1 \wedge b_1 \text { is WD}\right\} ,\\ ~\\ \left\{ a_1 \wedge b_1 \text { WD} \right\} \Rightarrow \left\{ \alpha (a_1 \wedge b_1) = \alpha (a_1) \cap \alpha (b_1)\right\} . \end{array} \right. \end{aligned}$$

Now let us treat the case where $a,b \in {\mathbb {H}}^{n}$. For this aim, let us remark that:

$$\begin{aligned} \alpha (a) \cap \alpha (b)&= \alpha (\times _{i \in \llbracket 1,n \rrbracket } a_i) \cap \alpha (\times _{i \in \llbracket 1,n \rrbracket } b_i),\\&= \otimes _{i \in \llbracket 1,n \rrbracket } \alpha (a_i) \cap \otimes _{i \in \llbracket 1,n \rrbracket } \alpha (b_i),\\&= \otimes _{i \in \llbracket 1,n \rrbracket } \left( \alpha (a_i) \cap \alpha (b_i)\right) .\\ \end{aligned}$$

Then, when $\alpha (a) \cap \alpha (b) \ne \emptyset$, we obtain that for any $i \in \llbracket 1,n \rrbracket$, $\alpha (a_i) \cap \alpha (b_i)$ is not empty, which implies that $a_i \wedge b_i$ is WD and that we have:

$$\begin{aligned} \alpha (a_i) \cap \alpha (b_i) = \alpha (a_i \wedge b_i). \end{aligned}$$

This way, $\alpha (a) \cap \alpha (b)$ is equal to $\otimes _{i \in \llbracket 1,n \rrbracket } \alpha (a_i \wedge b_i)$, and then is equal to:

$$\begin{aligned} \alpha (\times _{i \in \llbracket 1,n \rrbracket } (a_i \wedge b_i)), \end{aligned}$$

and then the supremum of $\alpha (a) \cap \alpha (b)$ is $\times _{i \in \llbracket 1,n \rrbracket } (a_i \wedge b_i)$. We can then denote by $a \wedge b$ this last term. Furthermore, it satisfies $\alpha (a \wedge b) = \alpha (a) \cap \alpha (b)$.

Conversely, when $a \wedge b$ is WD, the supremum of $\alpha (a) \cap \alpha (b)$ exists and thus $\alpha (a) \cap \alpha (b)\ne \emptyset$. $\square$

Lemma 3Let x, y be two elements of ${\mathbb {Z}} ^n$. Then, x and y are antagonists in a block of ${\mathbb {Z}} ^n$of dimension $k \in \llbracket 0,n \rrbracket$iff:

$$\begin{aligned} { \left\{ \begin{array}{lr} \text {Card} \left\{ m \in \llbracket 1,n \rrbracket \; ; \; x_m = y_m \right\} = n - k, &{} (1)\\ and\\ \text {Card} \left\{ m \in \llbracket 1,n \rrbracket \; ; \; |x_m - y_m| = 1 \right\} = k. &{} (2)\\ \end{array} \right. } \end{aligned}$$

In other words, x and y are k-antagonists iff they have $(n-k)$equal coordinates and that the remaining coordinates differ from 1.

Proof

Let x, y be two elements of ${\mathbb {Z}} ^n$ satisfying (1) and (2) with $k \in \llbracket 0,n \rrbracket$. Now, let us take $c \in {\mathbb {Z}} ^n$ such that $\forall i \in \llbracket 1,n \rrbracket$, $c_i := \min (x_i,y_i)$, and let us define:

$$\begin{aligned} {\mathcal {I}}_x:= \left\{ i \in \llbracket 1,n \rrbracket \; ; \; c_i \ne x_i\right\} , \end{aligned}$$

and

$$\begin{aligned} {\mathcal {I}}_y:= \left\{ i \in \llbracket 1,n \rrbracket \; ; \; c_i \ne y_i\right\} . \end{aligned}$$

By (1), there are $(n-k)$ coordinates i such that $x_i = y_i = c_i$, and then k coordinates i such that $x_i \ne c_i$ or $y_i \ne c_i$. Then,

$$\begin{aligned} \text {Card} ({\mathcal {I}}_x \cup {\mathcal {I}}_y) = k. \end{aligned}$$

Since by (2) we have:

$$\begin{aligned} \left\{ \begin{array}{lr} x &{} = c + \sum _{i \in {\mathcal {I}}_x} e^i,\\ y &{} = c + \sum _{i \in {\mathcal {I}}_y} e^i,\\ \end{array} \right. \end{aligned}$$

then x and y belong to $S(c,{\mathcal {F}})$ where

$$\begin{aligned} {\mathcal {F}}:= \left\{ e^i \in {\mathbb {B}} \; ; \; i \in {\mathcal {I}}_x \cup {\mathcal {I}}_y \right\} \end{aligned}$$

is of cardinality k. Furthermore, the $L^1$ norm of $x-y$ is equal to k, and thus x and y maximize the $L^1$-distance between two points into $S(c,{\mathcal {F}})$. So, x and y are antagonists in $S(c,{\mathcal {F}})$.

Conversely, let us assume that $x, y \in {\mathbb {Z}} ^n$ are antagonists in a block $S(c,{\mathcal {F}} )$ of dimension $k \in \llbracket 0,n \rrbracket$. For any $i \in \llbracket 1,n \rrbracket$, $e^i$ belongs to ${\mathcal {F}}$ and hence $|x_i - y_i|=1$, or it does not belong to ${\mathcal {F}}$ and hence $x_i = y_i$. Since $\text {Card} ({\mathcal {F}} ) = k$ by hypothesis, this concludes the proof. $\square$

Lemma 4$\forall p,p' \in {\mathbb {Z}} ^n$, p and $p'$are k-antagonists, $k \in \llbracket 0,n \rrbracket$, iff ${\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p')$is WD and belongs to ${\mathbb {H}}^{n} _{n-k}$.

Proof

The intuition of the proof is depicted in Fig. 14 for the 3D case. Let $p,p'$ be defined in ${\mathbb {Z}} ^n$ and $k \in \llbracket 0,n \rrbracket$ such that p and $p'$ are antagonists in a block of dimension $k \in \llbracket 0,n \rrbracket$. By Lemma 3, there exists a family ${\mathfrak {I}} \subseteq \llbracket 1,n \rrbracket$ of k coordinates s.t. $\forall i \in {\mathfrak {I}}$, $|p_i - p'_i| = 1$ and $\forall i \in \llbracket 1,n \rrbracket \setminus {\mathfrak {I}}$, $p_i = p'_i$. Since for each $i \in \llbracket 1,n \rrbracket$, we have $p_i, p'_i \in {\mathbb {Z}}$, then ${\mathcal {H}} (p_i) = \{p_i,p_i+1\}$, and ${\mathcal {H}} (p'_i) = \{p'_i,p'_i+1\}$. Let us denote $z_i = {\mathcal {H}} (p_i)$, and $z'_i = {\mathcal {H}} (p'_i)$, then $z_i, z'_i \in {\mathbb {H}}^{1} _1$.

When i is in ${\mathfrak {I}}$, $p'_i = p_i - 1$, and $\alpha (z_i) \cap \alpha (z'_i) = \{\{p_i\}\}$, and then $z_i \wedge z'_i = \{p_i\} \in {\mathbb {H}}^{1} _0$, or $p'_i = p_i + 1$, and $\alpha (z_i) \cap \alpha (z'_i) = \{\{p'_i\}\}$ and then $z_i \wedge z'_i = \{p'_i\} \in {\mathbb {H}}^{1} _0$. When i belongs to $\llbracket 1,n \rrbracket \setminus {\mathfrak {I}}$, $z_i = z'_i$ and $\alpha (z_i) \cap \alpha (z'_i) = \alpha (z_i)$ and then $z_i \wedge z'_i = z_i \in {\mathbb {H}}^{1} _1$. It follows then that $\times _{i \in \llbracket 1,n \rrbracket } (z_i \wedge z'_i)$ belongs to ${\mathbb {H}}^{n} _{n-k}$.

Also, since $\alpha (z_i) \cap \alpha (z'_i) \ne \emptyset$ for any $i \in \llbracket 1,n \rrbracket$, $\alpha ({\mathcal {H}} _n (p)) \cap \alpha ({\mathcal {H}} _n (p'))$ is equal to $\otimes _{i \in \llbracket 1,n \rrbracket } (\alpha (z_i) \cap \alpha (z'_i))$ which is non-empty, and then, by Lemma 2, ${\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p')$ exists and is equal to $\times _{i \in \llbracket 1,n \rrbracket } (z_i \wedge z'_i)$, which belongs to ${\mathbb {H}}^{n} _{n-k}$.

Let us now prove the converse implication. Let $p, p'$ be two points of ${\mathbb {Z}} ^n$, and $z = {\mathcal {H}} _n (p), z' = {\mathcal {H}} _n (p')$ such that $z \wedge z'$ is WD and belongs to ${\mathbb {H}}^{n} _{n-k}$. Then, we define ${\mathfrak {I}} = \{ i \in \llbracket 1,n \rrbracket \; ; \; z_i \wedge z'_i \in {\mathbb {H}}^{1} _0\}$, whose cardinality is equal to k thanks to Lemma 2. Now, let us observe that, for any $i \in \llbracket 1,n \rrbracket$, $p_i \in \{p'_i-1,p'_i+1\}$ iff $z_i \wedge z'_i \in {\mathbb {H}}^{1} _0$, then p and $p'$ have exactly k different coordinates, and they differ from one. Then, p and $p'$ are antagonists in a block of dimension k by Lemma 3. $\square$

Lemma 5Let $p,p'$be two elements of ${\mathbb {Z}} ^n$such that p and $p'$are $(3^n-1)$-neighbors in ${\mathbb {Z}} ^n$or equal. Then,

$$\begin{aligned} {\mathcal {H}} _n \left( \frac{p+p'}{2} \right) = {\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p'). \end{aligned}$$

Proof

Since p and $p'$ are $(3^n-1)$-neighbors in ${\mathbb {Z}} ^n$ or equal, they are antagonists in a block of dimension $k \in \llbracket 0,n \rrbracket$, and then by Lemma 4, ${\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p')$ is WD.

Now, let us prove that:

$$\begin{aligned} \frac{p+p'}{2} = {\mathcal {Z}} _n ({\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p')). \end{aligned}$$

This is equivalent to say that for any $i \in \llbracket 1,n \rrbracket$, we have $\frac{p_i+p'_i}{2} = {\mathcal {Z}} ({\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i))$ by Lemma 2. Let us rename the following equality:

$$\begin{aligned} {\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i) = \{p_i,p_i+1\} \wedge \{p'_i,p'_i+1\} \ \ \ (P). \end{aligned}$$

Since p and $p'$ are $(3^n-1)$-neighbors in ${\mathbb {Z}} ^n$ or equal, they satisfy for any $i \in \llbracket 1,n \rrbracket$ that $p_i \in \{p'_i-1,p'_i,p'_i+1\}$. Then, we have 3 possible cases:

• $p_i = p'_i - 1$, and then by (P):

  $$\begin{aligned} {\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i) = \{p'_i-1,p'_i\} \wedge \{p'_i,p'_i+1\} = \{p'_i\}, \end{aligned}$$

  and then ${\mathcal {Z}} ({\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i)) = \left( p'_i - \frac{1}{2} \right) = \frac{p_i+p'_i}{2}$,

• or we have $p'_i = p_i - 1$, and then a symmetrical reasoning leads to the same result,

• or $p'_i = p_i$, and then the result is immediate.

The proof is done. $\square$

Proposition 5Let S be a block and let $p,p' \in S$be any two antagonists in S. Then, the center of the block S is equal to $\frac{p+p'}{2}$. Furthermore, its image by ${\mathcal {H}} _n$in ${\mathbb {H}}^{n}$is equal to ${\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p')$.

Proof

Starting from the two antagonists $p,p'$ in S, we can compute $z \in {\mathbb {Z}} ^n$ and ${\mathcal {F}} \subseteq {\mathbb {B}}$ such that $S = S(z,{\mathcal {F}})$. In fact, for all $i \in \llbracket 1,n \rrbracket$, $z_i = \min (p_i,p'_i)$, and ${\mathcal {F}} = \{e^i \; ; \; i \in \llbracket 1,n \rrbracket , \; p_i \ne p'_i\}$. Then, it is clear that:

$$\begin{aligned} p = (p-z) + z = z + \sum _{p_i \ne z_i} e^i, \end{aligned}$$

and in the same manner:

$$\begin{aligned} p' = (p'-z) + z = z + \sum _{p'_i \ne z_i} e^i. \end{aligned}$$

Then,

$$\begin{aligned} p + p' = 2 z + \sum _{f \in {\mathcal {F}}} f, \end{aligned}$$

which shows that $\frac{p+p'}{2}$ is the center of S in $\left( {\mathbb {Z}}/2\right) ^n$. The second part of the proposition follows from Lemma 5. $\square$

Lemma 6Let p be an element of ${\mathbb {Z}} ^n$, then we can reformulate the squared closure of ${\mathcal {H}} _n (p)$in the following manner:

$$\begin{aligned} \alpha ^{\square } ({\mathcal {H}} _n (p)) = \bigcup _{v \in {\mathcal {N}}^*_{3^n - 1} (p)} \alpha ({\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (v)). \end{aligned}$$

Proof

This proof is depicted in Fig. 15. Let p be an element of ${\mathbb {Z}} ^n$. Then, let us compute the value of $\alpha ({\mathcal {H}} _n (p))$:

$$\begin{aligned} { \begin{array}{rl} \alpha ({\mathcal {H}} _n (p)) &{} =\alpha \left( {\mathcal {H}} _n ( \times _{i \in \llbracket 1,n \rrbracket } p_i) \right) ,\\ &{} =\alpha \left( \times _{i \in \llbracket 1,n \rrbracket } {\mathcal {H}} (p_i) \right) ,\\ &{} =\otimes _{i \in \llbracket 1,n \rrbracket } \alpha \left( {\mathcal {H}} (p_i) \right) ,\\ &{} =\otimes _{i \in \llbracket 1,n \rrbracket } \alpha \left( \{p_i,p_i+1\} \right) ,\\ &{} =\otimes _{i \in \llbracket 1,n \rrbracket } \{\{p_i\},\{p_i,p_i+1\},\{p_i+1\}\},\\ &{} =\otimes _{i \in \llbracket 1,n \rrbracket } \left\{ {\mathcal {H}} \left( p_i - \frac{1}{2} \right) , {\mathcal {H}} (p_i), {\mathcal {H}} \left( p_i+\frac{1}{2} \right) \right\} ,\\ &{} =\otimes _{i \in \llbracket 1,n \rrbracket } {\mathcal {H}} (\{p_i - \frac{1}{2}, p_i, p_i+\frac{1}{2} \}),\\ &{} = {\mathcal {H}} _n \left( \otimes _{i \in \llbracket 1,n \rrbracket } \{p_i - \frac{1}{2}, p_i, p_i+\frac{1}{2} \} \right) ,\\ &{} = {\mathcal {H}} _n \left( \left\{ q \in \left( \frac{{\mathbb {Z}}}{2}\right) ^n \; ; \; \Vert q - p\Vert _{\infty } \le \frac{1}{2} \right\} \right) . \end{array}} \end{aligned}$$

We can deduce that:

$$\begin{aligned} \begin{array}{rl} \alpha ^{\square } ({\mathcal {H}} _n (p)) &{} = \left\{ {\mathcal {H}} _n (q) \; ; \; q \in \left( \frac{{\mathbb {Z}}}{2}\right) ^n, \Vert q - p\Vert _{\infty } = \frac{1}{2} \right\} ,\\ ~\\ &{} = \left\{ f \in {\mathbb {H}}^{n} \; ; \; \Vert {\mathcal {Z}} _n (f) - p\Vert _{\infty } = \frac{1}{2} \right\} . \end{array} \end{aligned}$$

However, $\Vert {\mathcal {Z}} _n (f) - p\Vert _{\infty } = \frac{1}{2}$ is equivalent to:

$$\begin{aligned} \Vert v- p\Vert _{\infty } = 1 \text { with }v:= 2 {\mathcal {Z}} _n (f) - p. \end{aligned}$$

Then, $\alpha ^{\square } ({\mathcal {H}} _n (p))$ is equal to the set of elements $f \in {\mathbb {H}}^{n}$ satisfying $v \in {\mathcal {N}}^*_{3^n - 1} (p)$ and $f = {\mathcal {H}} _n \left( \frac{v + p}{2}\right)$. By Lemma 5, we obtain that:

$$\begin{aligned} \alpha ^{\square } ({\mathcal {H}} _n (p)) = \left\{ {\mathcal {H}} _n (v) \wedge {\mathcal {H}} _n (p) \in {\mathbb {H}}^{n} \; ; \; v \in {\mathcal {N}}^*_{3^n - 1} (p)\right\} , \end{aligned}$$

which leads to the required formula by applying the $\alpha$ operator on both sides. $\square$

Lemma 7Let S be a block in ${\mathbb {Z}} ^n$of dimension $k \ge 2$. Now, let $p,p'$be two antagonists in S, and v be a 2n-neighbor of p in S. Then, we have the following relation:

$$\begin{aligned} {\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p') \in \alpha ({\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (v)). \end{aligned}$$

Proof

Since p and v are 2n-neighbors, they are antagonists in a block of dimension 1. Then, by Lemma 4, ${\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (v)$ is WD. For the same reason, ${\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p')$ is WD. By Lemma 2, the first term of the relation is equal to:

$$\begin{aligned} \times _{i \in \llbracket 1,n \rrbracket } ({\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i)). \end{aligned}$$

Likewise, the second term is equal to:

$$\begin{aligned} \otimes _{i \in \llbracket 1,n \rrbracket } \left( \alpha ({\mathcal {H}} (p_i)) \cap \alpha ({\mathcal {H}} (v_i)) \right) . \end{aligned}$$

Then, it is sufficient to show that for all $i \in \llbracket 1,n \rrbracket$:

$$\begin{aligned} {\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i) \in \alpha ({\mathcal {H}} (p_i)) \cap \alpha ({\mathcal {H}} (v_i)). \ \ \ (P) \end{aligned}$$

Let us define ${\mathfrak {I}} = \left\{ i \in \llbracket 1,n \rrbracket \; ; \; p_i \ne p'_i \right\} .$ Since v is a 2n-neighbor of p into S, there exists an index $i^*$ in ${\mathfrak {I}}$ such that $v_{i^*} \ne p_{i^*}$, i.e., $v_{i^*} = p'_{i^*}$, and $\forall i \in \llbracket 1,n \rrbracket \setminus \{i^*\}, v_i = p_i$. When $i \in \llbracket 1,n \rrbracket \setminus {\mathfrak {I}}$ or when $i = i^*$, the property (P) is obviously true. When $i \in {\mathfrak {I}} \setminus \{i^*\}$, then $v_i = p_i$, which implies:

$$\begin{aligned} \alpha ({\mathcal {H}} (p_i)) \cap \alpha ({\mathcal {H}} (v_i))&\,=\, \alpha ({\mathcal {H}} (p_i)) \\&\,=\, \left\{ \{p_i\},\{p_i+1\},\{p_i,p_i+1\} \right\} . \end{aligned}$$

When $p'_i = p_i - 1$, we obtain ${\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i) = \{p_i\}$, and when $p'_i = p_i + 1$, we obtain ${\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i) = \{p_i + 1\}$. Then, in both cases,

$$\begin{aligned} {\mathcal {H}} (p_i) \wedge {\mathcal {H}} (p'_i) \in \alpha ({\mathcal {H}} (p_i)) \cap \alpha ({\mathcal {H}} (v_i)). \end{aligned}$$

The proof is done. $\square$

Lemma 8Let S be a block of ${\mathbb {Z}} ^n$, and let $z^* \in {\mathbb {H}}^{n}$be the image by ${\mathcal {H}} _n$of the center of S. For all $y \in {\mathbb {Z}} ^n$,

$$\begin{aligned} \{ y \not \in S\} \Rightarrow \{ \alpha ({\mathcal {H}} _n (y)) \cap \beta (z^*) = \emptyset \}. \end{aligned}$$

Proof

This proof can be observed in Fig. 16. Let us proceed by counterposition. Let y be an element of ${\mathbb {Z}} ^n$ such that:

$$\begin{aligned} \alpha ({\mathcal {H}} _n (y)) \cap \beta (z^*) \ne \emptyset . \end{aligned}$$

Then, for all $i \in \llbracket 1,n \rrbracket$, $\alpha ({\mathcal {H}} (y_i)) \cap \beta (z^* _i)$ is not empty. Now, let us show that y belongs to S. Since there exists $p_i \in \alpha ({\mathcal {H}} (y_i)) \cap \beta (z^* _i)$, then ${\mathcal {H}} (y_i) \in \beta (p_i)$ and $p_i \in \beta (z^* _i)$, which leads to ${\mathcal {H}} (y_i) \in \beta (z^* _i)$ by transitivity of the operator $\beta$, and then ${\mathcal {H}} _n (y) \in \beta (z^*)$. Since $y \in {\mathbb {Z}} ^n$, ${\mathcal {H}} _n (y) \in {\mathbb {H}}^{n} _n$, and then ${\mathcal {H}} _n (y) \in \beta (z^*) \cap {\mathbb {H}}^{n} _n$, which is equivalent to $y \in {\mathcal {Z}} _n \left( \beta (z^*) \cap {\mathbb {H}}^{n} _n\right)$, which is the reformulation of a block centered at $z^*$ by Proposition 4. $\square$

Proposition 6Let $n \ge 1$and $k \in \llbracket 0,n\rrbracket$be two integers. Let $|X| = (X,\alpha _X)$and $|Y| = (Y,\alpha _Y)$be two k-surfaces in ${\mathbb {H}}^{n}$. Then, if |X| is a suborder of |Y|, then $|X|=|Y|$.

Proof

Let us proceed by induction on k.

$\underline{\mathrm{Initialization}\,(k = 0):}$ when |X| and |Y| are two 0-surfaces, the inclusion $X \subseteq Y$ implies directly that $X = Y$ since they have the same finite cardinality, and then

$$\begin{aligned} |X| = |Y|. \end{aligned}$$

$\underline{\mathrm{Heredity}\,(k \ge 1):}$ we assume that when two $(k-1)$-surfaces satisfy an inclusion relationship, that is, when they are nested, then they are equal. Now, let |X| and |Y| be two k-surfaces, $k \ge 1$, such that |X| is a suborder of |Y|. Then, for all $x \in X$, $x \in Y$ and so we can write:

$$\begin{aligned} \theta ^{\square } _X(x) = \theta ^{\square } (x) \cap X \subseteq \theta ^{\square } (x) \cap Y = \theta ^{\square } _Y(x) \end{aligned}$$

because $X \subseteq Y$. However, $|\theta ^{\square } _X(x)|$ and $|\theta ^{\square } _Y(x)|$ are $(k-1)$-surfaces and $|\theta ^{\square } _X(x)|$ is a suborder of $|\theta ^{\square } _Y(x)|$, then we have thanks to the induction hypothesis:

$$\begin{aligned} |\theta ^{\square } _X(x)| = |\theta ^{\square } _Y(x)|. \end{aligned}$$

Now, let us assume that we have:

$$\begin{aligned} X \subsetneq Y \ \ \ (P). \end{aligned}$$

Then, let x be a point of X and y a point of $Y \setminus X$. Since |Y| is connected (because it is an k-surface with $k \ge 1$), it is path-connected, and so $x,y \in Y$ implies that there exists a path $\pi$ joining them into Y. This way, there exist $x' \in X$ and $y' \in Y \setminus X$ such that:

$$\begin{aligned} y' \in \theta ^{\square } (x'). \end{aligned}$$

In other words, $y' \in \theta ^{\square } _Y(x')$, where $\theta ^{\square } _Y(x')$ is equal to $\theta ^{\square } _X(x')$ since $x' \in X$. This leads to $y' \in X$. We obtain a contradiction on (P). Thus, we have $X = Y$. Supplying these two posets with $\subseteq$, we obtain $|X| = |Y|$.

By applying the induction on k until n, this property is true for any finite $k \in \llbracket 0,n\rrbracket$. $\square$

Corollary 1Let $|X_1|$and $|X_2|$be two k-surfaces, $k \ge 0$, with $X_1 \cap X_2 = \emptyset$. Then, $|X_1 \sqcup X_2|$is not a k-surface. In other words, the disjoint union of two k-surfaces, $k \ge 0$, is not a k-surface.

Proof

Let $|X_1|$ and $|X_2|$ be two disjoint k-surfaces in ${\mathbb {H}}^{n}$ with $k \in \llbracket 0,n\rrbracket$. If we assume that $|X_1 \sqcup X_2|$ is a k-surface, then $X_1 \subseteq X_1 \sqcup X_2$ implies by Proposition 6 that $X_1 = X_1 \sqcup X_2$, which is a contradiction since $X_2$ is non-empty. Then, $|X_1 \sqcup X_2|$ is not a k-surface. $\square$

Proposition 7Let a, b be two elements of ${\mathbb {H}}^{n}$with $a \in \beta ^{\square } (b)$. Then, $|\alpha ^{\square } (a) \cap \beta ^{\square } (b)|$is a $(\textit{dim} (a) - \textit{dim} (b) - 2)$-surface.

Proof

Since $|{\mathbb {H}}^{n} |$ is an n-surface, then $|\alpha ^{\square } (a)|$ is a $(\rho (a,|{\mathbb {H}}^{n} |) - 1)$-surface by Proposition 3, and then is a $(\textit{dim} (a)-1)$-surface. Now, we can remark that because b belongs to $\alpha ^{\square } (a)$, we can write:

$$\begin{aligned} \alpha ^{\square } (a) \cap \beta ^{\square } (b) = \beta ^{\square } _{\alpha ^{\square } (a)}(b), \end{aligned}$$

and then, again by Proposition 3, $|\alpha ^{\square } (a) \cap \beta ^{\square } (b)|$ is a $((\textit{dim} (a) - 1) - \rho (b,|\alpha ^{\square } (a)|) - 1)$-surface. Since we have:

$$\begin{aligned} \rho (b,|\alpha ^{\square } (a)|) = \rho (b, |{\mathbb {H}}^{n} |) = \textit{dim} (b), \end{aligned}$$

the proof is done. $\square$

Proofs of Section 4

Proposition 8The sets $\alpha ({\mathcal {X}})$and $\alpha ({\mathcal {Y}})$are regular closed sets.

Proof

Let us prove that $\alpha ({\mathcal {X}})$ is a regular closed set, the same reasoning applies for $\alpha ({\mathcal {Y}})$. The fact that

$$\begin{aligned} \mathrm {Int} (\alpha ({\mathcal {X}}))) \subseteq \alpha ({\mathcal {X}}) \end{aligned}$$

implies:

$$\begin{aligned} \alpha (\mathrm {Int} (\alpha ({\mathcal {X}}))) \subseteq \alpha ({\mathcal {X}}) \end{aligned}$$

by monotonicity of $\alpha$. Conversely, any element $x \in {\mathcal {X}}$ satisfies $\beta (x) = \{x\} \subseteq {\mathcal {X}}$, and so $\mathrm {Int} (\alpha ({\mathcal {X}}))$, which is equal to $\{h \in \alpha ({\mathcal {X}}) \; ; \; \beta (h) \subseteq \alpha ({\mathcal {X}})\}$, contains ${\mathcal {X}}$. This implies that $\alpha (\mathrm {Int} (\alpha ({\mathcal {X}}))) \supseteq \alpha ({\mathcal {X}})$. Thus, $\alpha ({\mathcal {X}})$ is a regular closed set. $\square$

Lemma 9Let ${\mathcal {X}},{\mathcal {Y}}$be two subsets of ${\mathbb {H}}^{n} _n$such that ${\mathcal {X}} \sqcup {\mathcal {Y}} = {\mathbb {H}}^{n} _n$. Then,

$$\begin{aligned} \alpha ({\mathcal {X}}) \sqcup \mathrm {Int} (\alpha ({\mathcal {Y}})) = {\mathbb {H}}^{n}. \end{aligned}$$

Proof

Let us first prove that the union is disjoint. For this aim, let us remark that $h \in \alpha ({\mathcal {X}})$ is equivalent to say that there exists some $x \in {\mathcal {X}}$ such that $h \in \alpha (x)$. In other words, there exists some $x \in {\mathcal {X}}$ such that $x \in \beta (h)$. Now, let us remark that $h \in \mathrm {Int} (\alpha ({\mathcal {Y}}))$ is equivalent to say that $\beta (h) \subseteq \alpha ({\mathcal {Y}})$. If we assume that:

$$\begin{aligned} \alpha ({\mathcal {X}}) \cap \mathrm {Int} (\alpha ({\mathcal {Y}})) \ne \emptyset , \ \ \ (P) \end{aligned}$$

we can deduce that there exists some $h \in {\mathbb {H}}^{n}$ such that there exists $x \in {\mathcal {X}}$ satisfying that:

$$\begin{aligned} x \in \beta (h) \subseteq \alpha ({\mathcal {Y}}), \end{aligned}$$

leading to $x \in \alpha ({\mathcal {Y}})$. Since $x \in {\mathcal {X}} \subset {\mathbb {H}}^{n} _n$, then $x \in {\mathcal {Y}}$, which is a contradiction. Then, (P) is wrong and the intersection of $\alpha ({\mathcal {X}})$ and $\mathrm {Int} (\alpha ({\mathcal {Y}}))$ is empty.

Let us now prove that their union is equal to ${\mathbb {H}}^{n}$. The fact that $\alpha ({\mathcal {X}}) \sqcup \mathrm {Int} (\alpha ({\mathcal {Y}})) \subseteq {\mathbb {H}}^{n}$ is obvious. Now, let us prove the converse inclusion. Let h be a face of ${\mathbb {H}}^{n}$. Two cases are possible:

• either $\beta (h) \subseteq \alpha ({\mathcal {Y}})$, then $h \in \mathrm {Int} (\alpha ({\mathcal {Y}}))$,

• or $\beta (h) \not \subseteq \alpha ({\mathcal {Y}})$, then the fact that

  $$\begin{aligned} \alpha ({\mathcal {X}}) \cup \alpha ({\mathcal {Y}}) = {\mathbb {H}}^{n} \end{aligned}$$

  implies that we have:

  $$\begin{aligned} \beta (h) \cap \alpha ({\mathcal {X}}) \ne \emptyset , \end{aligned}$$

  then there exists $x \in {\mathcal {X}}$ such that:

  $$\begin{aligned} \beta (h) \cap \alpha (x) \ne \emptyset . \end{aligned}$$

  Then, there exists $p \in \beta (h) \cap \alpha (x)$ which means that $h \in \alpha (p)$ and $p \in \alpha (x)$, and then by transitivity, $h \in \alpha (x) \subseteq \alpha ({\mathcal {X}})$.

Then, $\alpha ({\mathcal {X}}) \sqcup \mathrm {Int} (\alpha ({\mathcal {Y}})) \supseteq {\mathbb {H}}^{n}$. The proof is done. $\square$

Proposition 9The hit-transform and the miss-transform of X have the same boundary which is equal to:

$$\begin{aligned} \alpha ({\mathcal {X}}) \cap \alpha ({\mathcal {Y}}). \end{aligned}$$

Proof

The boundary ${\mathfrak {N}} '$ of ${\mathcal {I}}^{\text {hit}} (X)$ is equal to:

$$\begin{aligned} \alpha ({\mathcal {I}}^{\text {hit}} (X)) \cap \alpha ({\mathbb {H}}^{n} \setminus {\mathcal {I}}^{\text {hit}} (X)), \end{aligned}$$

also equal by idempotence of $\alpha$ to:

$$\begin{aligned} \alpha ({\mathcal {X}}) \cap \alpha ({\mathbb {H}}^{n} \setminus \alpha ({\mathcal {X}})), \end{aligned}$$

which is equal by Lemma 9 to:

$$\begin{aligned} \alpha ({\mathcal {X}}) \cap \alpha (\mathrm {Int} (\alpha ({\mathcal {Y}}))), \end{aligned}$$

and by regularity of $\alpha ({\mathcal {Y}})$ (see Proposition 8), then:

$$\begin{aligned} {\mathfrak {N}} ' = \alpha ({\mathcal {X}}) \cap \alpha ({\mathcal {Y}}). \end{aligned}$$

In the same manner, the boundary ${\mathfrak {N}}$ of ${\mathcal {I}}^{\text {miss}} (X)$ is equal to:

$$\begin{aligned} \alpha ({\mathcal {I}}^{\text {miss}} (X)) \cap \alpha ({\mathbb {H}}^{n} \setminus {\mathcal {I}}^{\text {miss}} (X)), \end{aligned}$$

which is equal to:

$$\begin{aligned} \alpha (\mathrm {Int} (\alpha ({\mathcal {X}}))) \cap \alpha ({\mathbb {H}}^{n} \setminus \mathrm {Int} (\alpha ({\mathcal {X}}))), \end{aligned}$$

which is by Proposition 8 equal to:

$$\begin{aligned} \alpha ({\mathcal {X}}) \cap \alpha ({\mathbb {H}}^{n} \setminus \mathrm {Int} (\alpha ({\mathcal {X}}))), \end{aligned}$$

and by Lemma 9, it is equal to:

$$\begin{aligned} \alpha ({\mathcal {X}}) \cap \alpha (\alpha ({\mathcal {Y}})), \end{aligned}$$

equal to:

$$\begin{aligned} \alpha ({\mathcal {X}}) \cap \alpha ({\mathcal {Y}}), \end{aligned}$$

by idempotence of $\alpha$. This way, we have:

$$\begin{aligned} {\mathfrak {N}} = {\mathfrak {N}} ' = \alpha ({\mathcal {X}}) \cap \alpha ({\mathcal {Y}}). \end{aligned}$$

$\square$

Proposition 10For any $z \in {\mathfrak {N}}$, we have the property that $|\alpha ^{\square }_{{\mathfrak {N}}} (z)|$is a $(\textit{dim} (z) - 1)$-surface.

Proof

Since ${\mathfrak {N}}$ is closed, $\forall z \in {\mathfrak {N}}$, $|\alpha ^{\square }_{{\mathfrak {N}}} (z)| = |\alpha ^{\square } (z)|$, which is a $(\rho (z,|{\mathbb {H}}^{n} |) - 1)$-surface by Proposition 3 since ${\mathbb {H}}^{n}$ is an n-surface. Since $\rho (z,|{\mathbb {H}}^{n} |) = \dim (z)$, $|\alpha ^{\square }_{{\mathfrak {N}}} (z)|$ is a $(\textit{dim} (z) - 1)$-surface. $\square$

Lemma 10The immersion ${\mathcal {I}}^{\text {miss}} (X)$of X is AWC iff $\forall z \in {\mathfrak {N}}$, $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$is a $(n-2-\textit{dim} (z))$-surface.

Proof

Let us recall that two disjoint components $C_1$ and $C_2$ of ${\mathfrak {N}}$ are separated: $C_1 \cap \theta (C_2) = \emptyset$.

For this reason, for any $z \in {\mathfrak {N}}$,

$$\begin{aligned} { \begin{array}{rl} \theta ^{\square } _{{\mathfrak {N}}}(z) &{} = \theta ^{\square } (z) \cap \bigcup _{C \in \mathcal {CC} ({\mathfrak {N}})} C,\\ &{} = \bigcup _{C \in \mathcal {CC} ({\mathfrak {N}})} (\theta ^{\square } (z) \cap C),\\ &{} = \theta ^{\square } _{\mathcal {CC} ({\mathfrak {N}},z)}(z). \end{array} } \end{aligned}$$

Since $n \ge 2$, ${\mathcal {I}}^{\text {miss}} (X)$ is AWC iff $\forall C \in \mathcal {CC} ({\mathfrak {N}})$, $C$ is a $(n-1)$-surface, i.e., $\forall C \in \mathcal {CC} ({\mathfrak {N}})$, $\forall z \in C$, $|\theta ^{\square } _{C}(z)|$ is a $(n-2)$-surface, which means that $\forall C \in \mathcal {CC} ({\mathfrak {N}})$, $\forall z \in C$, $|\theta ^{\square }_{{\mathfrak {N}}} (z)|$ is a $(n-2)$-surface, or, in other words, by Proposition 2 and Proposition 10, $\forall z \in {\mathfrak {N}}$, $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ is a $(n-2-\textit{dim} (z))$-surface. $\square$

Proposition 11Let S be a block of dimension $k \in \llbracket 2,n \rrbracket$s.t. $X \cap S = \{p,p'\}$(or s.t. $Y \cap S = \{p,p'\}$) and $p' = \text {antag} _S(p)$, then ${\mathcal {H}} _n \left( \frac{p+p'}{2}\right) \in {\mathfrak {N}}$. In other words, when X contains a primary or secondary critical configuration, the image by ${\mathcal {H}} _n$of the center of the critical configuration belongs to the boundary ${\mathfrak {N}}$of the immersion of X.

Proof

Let v be a 2n-neighbor of p in S, which is possible since $\dim (S) \ge 1$. Then, v and p are 1-antagonists which implies by Lemma 4 that ${\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (v)$ is WD. Hence, by Lemma 2:

$$\begin{aligned} \alpha ({\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (v)) = \alpha \left( {\mathcal {H}} _n (p)\right) \cap \alpha \left( {\mathcal {H}} _n (v)\right) . \end{aligned}$$

If we assume that v belongs to X, it means that $v = p'$, or in other words that $k = 1$, which is wrong since $\textit{dim} (S) \ge 2$. Then, $v \in Y$, and so

$$\begin{aligned} \alpha ({\mathcal {H}} _n (p)) \cap \alpha ({\mathcal {H}} _n (v)) \subseteq {\mathfrak {N}} \end{aligned}$$

by Proposition 9. Now, using Proposition 5,

$$\begin{aligned} {\mathcal {H}} _n \left( \frac{p+p'}{2}\right) = {\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (p'), \end{aligned}$$

which belongs to $\alpha ({\mathcal {H}} _n (p) \wedge {\mathcal {H}} _n (v))$ by Lemma 7, and thus to ${\mathfrak {N}}$. $\square$

Proofs of Section 5

Proposition 12Let p, c be two elements in ${\mathbb {H}}^{n}$. We have the following equivalence:

$$\begin{aligned} \left\{ p \in \beta (c) \right\} \Leftrightarrow \left\{ { \begin{array}{l} \forall i \in \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) ,\\ ~~~~~~ {\mathcal {Z}} (p_i) \in \left\{ {\mathcal {Z}} (c_i) - \frac{1}{2}, {\mathcal {Z}} (c_i) + \frac{1}{2} \right\} ,\\ \forall i \in \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) ,\\ ~~~~~~{\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i),\\ \forall i \in \frac{1}{2} \left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) ,\\ ~~~~~~{\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i),\\ \frac{1}{2} \left( {\mathcal {Z}} _n (p)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) = \emptyset .\\ \end{array}} \right. \end{aligned}$$

Proof

Let us prove first that $p \in \beta (c)$ implies this set of four properties. The relation $p \in \beta (p)$ is equivalent to say that for any $i \in \llbracket 1,n \rrbracket$, $p_i \in \beta (c_i)$. Each term $p_i$ belongs to ${\mathbb {H}}^{1} _1$ or to ${\mathbb {H}}^{1} _0$, and so does $c_i$, which leads to four cases. Then, assuming that for $i \in \llbracket 1,n \rrbracket$, we have $p_i \in \beta (c_i)$, we obtain that:

• $p_i \in {\mathbb {H}}^{1} _1$ and $c_i \in {\mathbb {H}}^{1} _0$, then $p_i \in \beta (c_i)$ implies:

  $$\begin{aligned} {\mathcal {Z}} (p_i) \in \left\{ {\mathcal {Z}} (c_i) - \frac{1}{2}, {\mathcal {Z}} (c_i) + \frac{1}{2} \right\} , \end{aligned}$$

• or $p_i \in {\mathbb {H}}^{1} _1$ and $c_i \in {\mathbb {H}}^{1} _1$, then $p_i \in \beta (c_i)$ implies ${\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i)$,

• or $p_i \in {\mathbb {H}}^{1} _0$ and $c_i \in {\mathbb {H}}^{1} _0$, then $p_i \in \beta (c_i)$ implies ${\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i)$,

• or $p_i \in {\mathbb {H}}^{1} _0$ and $c_i \in {\mathbb {H}}^{1} _1$, then $p_i \in \beta (c_i)$ leads to a contradiction.

This leads to the 4 formulas described above, which concludes the direct implication.

Conversely, it we have these four properties, $\frac{1}{2} \left( p\right) \cap \mathbb {1}\left( c\right) = \emptyset$ shows that:

$$\begin{aligned} \left( \mathbb {1}\left( p\right) \cap \frac{1}{2} \left( c\right) \right) \cup \left( \mathbb {1}\left( p\right) \cap \mathbb {1}\left( c\right) \right) \cup \left( \frac{1}{2} \left( p\right) \cap \frac{1}{2} \left( c\right) \right) = \llbracket 1,n \rrbracket , \end{aligned}$$

and since in these three cases, we obtain that $p_i \in \beta (c_i)$, it is clear that $p \in \beta (c)$. $\square$

Proposition 13Let p, c be two elements of ${\mathbb {H}}^{n}$. Then, $p \succ c$iff there exists $m \in \llbracket 1,n \rrbracket$such that:

$$\begin{aligned} \left\{ \begin{array}{l} \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) = \left\{ m\right\} \\ \text {and}\\ {\mathcal {Z}} _n (p) \in \left\{ {\mathcal {Z}} _n (c) - \frac{1}{2} e^m, {\mathcal {Z}} _n (c) + \frac{1}{2} e^m\right\} . \end{array} \right. \end{aligned}$$

Proof

We can reformulate the fact that we have $p \succ c$ in the following manner (see Proposition 12):

$$\begin{aligned} \left\{ \begin{array}{lr} \forall i \in \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) ,\\ ~~~~~~ {\mathcal {Z}} (p_i) \in \left\{ {\mathcal {Z}} (c_i) - \frac{1}{2}, {\mathcal {Z}} (c_i) + \frac{1}{2} \right\} , &{} (1)\\ \forall i \in \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) ,\\ ~~~~~~ {\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i), &{} (2)\\ \forall i \in \frac{1}{2} \left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) ,\\ ~~~~~~{\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i), &{} (3)\\ \frac{1}{2} \left( {\mathcal {Z}} _n (p)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) = \emptyset , &{} (4)\\ \textit{dim} (p) = \textit{dim} (c) +1. &{} (5)\\ \end{array} \right. \end{aligned}$$

By (4), $\mathbb {1}\left( {\mathcal {Z}} _n (c)\right) \subseteq \mathbb {1}\left( {\mathcal {Z}} _n (p)\right)$, and then (2) can be reformulated:

$$\begin{aligned} \forall i \in \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) , {\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i), \end{aligned}$$

which implies that at least the $\textit{dim} (c)$ integral coordinates of ${\mathcal {Z}} _n (c)$ are integral for ${\mathcal {Z}} _n (p)$. Since $\textit{dim} (p) = \textit{dim} (c) + 1$ by (5), p admits one more integral coordinate than c and it lies into $\mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \setminus \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) = \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right)$, which means that:

$$\begin{aligned} \text {Card} ( \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) ) = 1, \end{aligned}$$

and then there exists one index of coordinate $m \in \llbracket 1,n \rrbracket$ such that $\mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) = \{m\}$. By (1) to (4), we obtain then that for each coordinate $i \in \llbracket 1,n \rrbracket$, ${\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i)$ except for the case $i = m$ where:

$$\begin{aligned} {\mathcal {Z}} (p_i) \in \left\{ {\mathcal {Z}} (c_i) - \frac{1}{2}, {\mathcal {Z}} (c_i) + \frac{1}{2} \right\} , \end{aligned}$$

which concludes the direct sense.

Conversely, if there exists $m \in \llbracket 1,n \rrbracket$ such that :

$$\begin{aligned} \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) = \left\{ m\right\} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {Z}} _n (p) \in \left\{ {\mathcal {Z}} _n (c) - \frac{1}{2} e^m, {\mathcal {Z}} _n (c) + \frac{1}{2} e^m\right\} , \end{aligned}$$

it is clear that (1) is satisfied by hypothesis. Also, for each $i \in \llbracket 1,n \rrbracket \setminus \{m\}$, we have ${\mathcal {Z}} (p_i) = {\mathcal {Z}} (c_i)$, which implies (2) and (3). Now let us assume that (4) is false, it means that there exists some $i \in \frac{1}{2} \left( {\mathcal {Z}} _n (p)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (c)\right)$ such that ${\mathcal {Z}} (c_i)$ is in half and such that ${\mathcal {Z}} (p_i)$ is integral. Then, we obtain that $|{\mathcal {Z}} (c_i) - {\mathcal {Z}} (p_i)| = \frac{1}{2}$, which means that $i = m$ (${\mathcal {Z}} _n (c)$ and ${\mathcal {Z}} _n (p)$ are different only on the $m^{th}$ coordinate). However, i belongs to $\frac{1}{2} \left( {\mathcal {Z}} _n (p)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (c)\right)$ and m belongs to $\mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right)$. We obtain a contradiction:

$$\begin{aligned} \{i\} \subseteq \mathbb {1}\left( {\mathcal {Z}} _n (p)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (p)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) = \emptyset , \end{aligned}$$

then (4) is true. (5) is true because p has one more integral coordinate than c by hypothesis. $\square$

Fig. 53

Assuming that a and b cover c, that the index $i^*$ is defined such that $a_{i^*} \ne c_{i^*}$ and that the index $j^*$ defined such that $b_{j^*} \ne c_{j^*}$. When $i^*$ and $j^*$ are different, we obtain that $\beta (a) \cap \beta (b)$ in light gray is not empty

Lemma 11

Let a, b, c be three elements of ${\mathbb {H}}^{n}$ such that $a = \text {opp}_{c}(b)$, then there exists $m \in \llbracket 1,n \rrbracket$ such that:

• either ${\mathcal {Z}} _n (a) = {\mathcal {Z}} _n (c) - \frac{1}{2} e^m$ and ${\mathcal {Z}} _n (b) = {\mathcal {Z}} _n (c) + \frac{1}{2} e^m$,

• or ${\mathcal {Z}} _n (a) = {\mathcal {Z}} _n (c) + \frac{1}{2} e^m$ and ${\mathcal {Z}} _n (b) = {\mathcal {Z}} _n (c) - \frac{1}{2} e^m$,

which leads in both cases to:

$$\begin{aligned} \frac{{\mathcal {Z}} _n (a) + {\mathcal {Z}} _n (b)}{2} = {\mathcal {Z}} _n (c). \end{aligned}$$

Furthermore,

$$\begin{aligned} { \left\{ \begin{array}{l} \mathbb {1}\left( {\mathcal {Z}} _n (a)\right) = \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) \sqcup \{m\} = \mathbb {1}\left( {\mathcal {Z}} _n (b)\right) ,\\ \frac{1}{2} \left( {\mathcal {Z}} _n (a)\right) \sqcup \{m\} = \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) = \frac{1}{2} \left( {\mathcal {Z}} _n (b)\right) \sqcup \{m\}. \\ \end{array} \right. } \end{aligned}$$

Proof

By Propositions 12 and 13, there exist $i^*,j^* \in \llbracket 1,n \rrbracket$ such that:

$$\begin{aligned} { \left\{ \begin{array}{l} \mathbb {1}\left( {\mathcal {Z}} _n (a)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) = \left\{ i^* \right\} ,\\ {\mathcal {Z}} _n (a) \in \left\{ {\mathcal {Z}} _n (c) - \frac{1}{2} e^{i^*}, {\mathcal {Z}} _n (c) + \frac{1}{2} e^{i^*}\right\} ,\\ \mathbb {1}\left( {\mathcal {Z}} _n (b)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) = \left\{ j^* \right\} ,\\ {\mathcal {Z}} _n (b) \in \left\{ {\mathcal {Z}} _n (c) - \frac{1}{2} e^{j^*}, {\mathcal {Z}} _n (c) + \frac{1}{2} e^{j^*}\right\} .\\ \end{array} \right. } \end{aligned}$$

Also, since a and b cover c, we have that $a \in \beta (c)$ and $b \in \beta (c)$. Applying $\beta$ on these expressions, we obtain that $\beta (a) \subseteq \beta (c)$ and that $\beta (b) \subseteq \beta (c)$ (by transitivity of $\beta$). Due to the fact that for each $i \in \llbracket 1,n \rrbracket$, we have $a_i = c_i$ iff $i \ne i^*$, and $b_j = c_j$ iff $j \ne j^*$, then we have:

$$\begin{aligned} {\left\{ \begin{array}{l} \beta (a_{i^*}) \subset \beta (c_{i^*}),\\ \forall i \in \llbracket 1,n \rrbracket \setminus \{i^* \}, \beta (a_i) = \beta (c_i),\\ \beta (b_{j^*}) \subset \beta (c_{j^*}),\\ \forall i \in \llbracket 1,n \rrbracket \setminus \{j^* \}, \beta (b_i) = \beta (c_i),\\ \end{array}\right. } \end{aligned}$$

If $i^* \ne j^*$ (see Fig. 53), then when $m = i^*$, we have $\beta (a_m) \subset \beta (c_m) = \beta (b_m)$, when $m = j^*$, we have $\beta (b_m) \subset \beta (c_m) = \beta (a_m)$, and when $m \in \llbracket 1,n \rrbracket \setminus \{i^*,j^* \}$, we have $\beta (a_m) = \beta (c_m) = \beta (b_m)$. We obtain that $\beta (a) \cap \beta (b) = \otimes _{i \in \llbracket 1,n \rrbracket } \left( \beta (a_i) \cap \beta (b_i) \right) \ne \emptyset$, which contradicts the hypothesis that a and b are opposites. Then, we have $i^* = j^*$.

Because ${\mathcal {Z}} _n (a),{\mathcal {Z}} _n (b)$ belong to

$$\begin{aligned} \left\{ {\mathcal {Z}} _n (c) - \frac{1}{2} e^{i^*}, {\mathcal {Z}} _n (c) + \frac{1}{2} e^{i^*}\right\} \end{aligned}$$

and because they are different, we obtain that:

• either ${\mathcal {Z}} _n (a) = {\mathcal {Z}} _n (c) - \frac{1}{2} e^{i^*}$ and ${\mathcal {Z}} _n (b) = {\mathcal {Z}} _n (c) + \frac{1}{2} e^{i^*}$,

• or ${\mathcal {Z}} _n (a) = {\mathcal {Z}} _n (c) + \frac{1}{2} e^{i^*}$ and ${\mathcal {Z}} _n (b) = {\mathcal {Z}} _n (c) - \frac{1}{2} e^{i^*}$,

which leads obviously to:

$$\begin{aligned} \frac{{\mathcal {Z}} _n (a) + {\mathcal {Z}} _n (b)}{2} = {\mathcal {Z}} _n (c). \end{aligned}$$

When $m \in \llbracket 1,n \rrbracket \setminus \{i^*\}$, we have then $\textit{dim} (a_m) = \textit{dim} (c_m) = \textit{dim} (b_m)$, and when $m = i^*$, we have $\textit{dim} (a_m) = \textit{dim} (c_m) + 1 = \textit{dim} (b_m)$. We can then conclude that:

$$\begin{aligned} { \left\{ \begin{array}{l} \mathbb {1}\left( {\mathcal {Z}} _n (a)\right) = \mathbb {1}\left( {\mathcal {Z}} _n (c)\right) \sqcup \{i^* \} = \mathbb {1}\left( {\mathcal {Z}} _n (b)\right) ,\\ \frac{1}{2} \left( {\mathcal {Z}} _n (a)\right) \sqcup \{i^* \} = \frac{1}{2} \left( {\mathcal {Z}} _n (c)\right) = \frac{1}{2} \left( {\mathcal {Z}} _n (b)\right) \sqcup \{i^* \}. \\ \end{array} \right. } \end{aligned}$$

$\square$

Proposition 14

Let $t,t',z$ three elements in ${\mathbb {H}}^{n}$ such that t and $t'$ are opposite relatively to z. Now let define ${\mathcal {E}}:= \beta (z) \setminus (\beta (t) \cup \beta (t'))$, and let $m^*$ be the only coordinate in $\llbracket 1,n \rrbracket$ such that $m^* \in \mathbb {1}\left( {\mathcal {Z}} _n (t)\right) \setminus \mathbb {1}\left( {\mathcal {Z}} _n (z)\right)$. Then, the application $\mathfrak {Iso}: {\mathcal {E}} \rightarrow {\mathbb {H}}^{n}$ such that, $\forall u \in {\mathcal {E}},$

$$\begin{aligned} \mathfrak {Iso} (u) := {\mathcal {H}} _n \left( {\mathcal {Z}} _n (u) + ({\mathcal {Z}} (t_{m^*}) - {\mathcal {Z}} (z_{m^*})) \; e ^{m^*} \right) . \end{aligned}$$

is an isomorphism from ${\mathcal {E}}$ to $\beta (t)$. In other words, the order is preserved from ${\mathcal {E}}$ to $\beta (t)$.

Proof

Let $t,t',z$ be three elements of ${\mathbb {H}}^{n}$ such that $t' = \text {opp}_{z}(t)$, then by Proposition 13, we obtain that there exists some value $m \in \llbracket 1,n \rrbracket$ such that $\frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (t)\right) = \{m\}$, ${\mathcal {Z}} _n (t) = {\mathcal {Z}} _n (z) + \frac{1}{2} e^m$, and ${\mathcal {Z}} _n (t') = {\mathcal {Z}} _n (z) - \frac{1}{2} e^m$ (or the converse case ${\mathcal {Z}} _n (t) = {\mathcal {Z}} _n (z) - \frac{1}{2} e^m$, and ${\mathcal {Z}} _n (t') = {\mathcal {Z}} _n (z) + \frac{1}{2} e^m$ but by symmetry, we do not need to treat this case).

We know by Examples 1 that

$$\begin{aligned} \beta (t) = {\mathbb {H}}^{n} _{\left\{ t,\frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) ,\{-\frac{1}{2},0,\frac{1}{2} \}\right\} } \end{aligned}$$

and

$$\begin{aligned} \beta (t') = {\mathbb {H}}^{n} _{\left\{ t',\frac{1}{2} \left( {\mathcal {Z}} _n (t')\right) ,\{-\frac{1}{2},0,\frac{1}{2} \}\right\} }. \end{aligned}$$

Since ${\mathcal {E}}$ is equal to $\beta (z) \setminus (\beta (t) \cup \beta (t'))$, we can reformulate it using Notation 8. Indeed,

$$\begin{aligned}&\beta (z)\\&\quad = {\mathbb {H}}^{n} _{\left\{ z,\frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) ,\left\{ -\frac{1}{2},0,\frac{1}{2} \right\} \right\} },\\&\quad = {\mathcal {H}} _n \left( \left\{ z + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) } \lambda _i e^i \; ; \; \lambda _i \in \left\{ -\frac{1}{2},0,\frac{1}{2} \right\} \right\} \right) ,\\&\quad = {\mathcal {H}} _n \left( \left\{ z + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t) \sqcup \{m\}\right) } \lambda _i e^i \; ; \; \lambda _i \in \left\{ -\frac{1}{2},0,\frac{1}{2} \right\} \right\} \right) ,\\&\quad = {\mathcal {H}} _n \left( \left\{ \begin{array}{l} z + \lambda _m e^m + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i \; ;\\ \; \lambda _m \in \{-\frac{1}{2},0,\frac{1}{2} \},\lambda _i \in \left\{ -\frac{1}{2},0,\frac{1}{2} \right\} \end{array} \right\} \right) ,\\&\quad = {\mathcal {H}} _n \left( \left\{ t + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i \; ; \; \lambda _i \in \left\{ -\frac{1}{2},0,\frac{1}{2} \right\} \right\} \right) \\&\quad \cup \; {\mathcal {H}} _n \left( \left\{ z+ \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i \; ; \; \lambda _i \in \left\{ -\frac{1}{2},0,\frac{1}{2} \right\} \right\} \right) \\&\quad \cup \; {\mathcal {H}} _n \left( \left\{ t' + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i \; ; \; \lambda _i \in \left\{ -\frac{1}{2},0,\frac{1}{2} \right\} \right\} \right) ,\\&\quad = \beta (t) \cup {\mathcal {E}} \cup \beta (t').\\ \end{aligned}$$

Since this is a disjoint union, it is clear that :

$$\begin{aligned} {\mathcal {E}} = {\mathcal {H}} _n \left( \left\{ z+ \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i \; ; \; \lambda _i \in \left\{ -\frac{1}{2},0,\frac{1}{2} \right\} \right\} \right) . \end{aligned}$$

Now that we have this equality, we can prove that their exists an isomorphism between $\beta (t)$, ${\mathcal {E}}$ and $\beta (t')$. By symmetry, it is sufficient to prove that ${\mathcal {E}}$ and $\beta (t)$ are isomorphic. For that, we define the application $\tau ^{+,m} : {\mathbb {H}}^{n} \rightarrow {\mathbb {H}}^{n}$ such that for any $u \in {\mathbb {H}}^{n}$:

$$\begin{aligned} \tau ^{+,m} (u) = {\mathcal {H}} _n \left( {\mathcal {Z}} _n (u) + \frac{1}{2} e^m\right) . \end{aligned}$$

Let us show first that this application maps ${\mathcal {E}}$ to $\beta (t)$. Let u be an element of ${\mathcal {E}}$, then there exists for any $i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right)$ one value $\lambda _i \in \{-\frac{1}{2},0,\frac{1}{2} \}$ such that

$$\begin{aligned} u = {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i\right) . \end{aligned}$$

This way,

$$\begin{aligned} \tau ^{+,m} (u) = {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + \frac{1}{2} e^m + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i\right) . \end{aligned}$$

Since ${\mathcal {Z}} _n (z) + \frac{1}{2} e^m = {\mathcal {Z}} _n (t),$ we obtain that

$$\begin{aligned} \tau ^{+,m} (u) = {\mathcal {H}} _n \left( {\mathcal {Z}} _n (t) + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i\right) , \end{aligned}$$

and then

$$\begin{aligned} \tau ^{+,m} (u) \in {\mathbb {H}}^{n} _{\left\{ t,\frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) ,\{-\frac{1}{2},0,\frac{1}{2} \}\right\} } \end{aligned}$$

which is in fact $\beta (t)$.

Now we want to prove that $\tau ^{+,m}$ is injective, which is immediate because it is a translation. To prove that $\tau ^{+,m}$ is surjective, let us proceed this way: let v be a point in $\beta (t)$, then there exists for any $i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right)$ one value $\lambda _i \in \{-\frac{1}{2},0,\frac{1}{2} \}$ such that $v = {\mathcal {H}} _n ({\mathcal {Z}} _n (t) + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i)$. Its antecedent is simply

$$\begin{aligned} u = {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + \sum _{i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right) } \lambda _i e^i\right) \end{aligned}$$

which obviously belongs to ${\mathcal {E}}$.

This translation is then a bijection from ${\mathcal {E}}$ to $\beta (t)$. Now we need to prove that it preserves the order: let a, b be two elements of ${\mathcal {E}}$ such that $a \succ b$, then, by Proposition 13, there exists a value $i \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right)$ such that:

$$\begin{aligned} { \left\{ \begin{array}{l} {\mathcal {Z}} _n (a) \in \{ {\mathcal {Z}} _n (b) - \frac{1}{2} e^i,{\mathcal {Z}} _n (b) + \frac{1}{2} e^i\},\\ {\mathcal {Z}} (a_i) \in {\mathbb {Z}},\\ {\mathcal {Z}} (b_i) \in ({\mathbb {Z}}/2) \setminus {\mathbb {Z}},\\ \end{array} \right. } \end{aligned}$$

Now let us define $a' = \tau ^{+,m} (a)$ and $b' = \tau ^{+,m} (b)$. We want to prove that $a'$ covers $b'$. In fact, we can write $a' = {\mathcal {H}} _n ({\mathcal {Z}} _n (a) + \frac{1}{2} e^m)$ and $b' = {\mathcal {H}} _n ({\mathcal {Z}} _n (b) + \frac{1}{2} e^m)$. We obtain then that:

$$\begin{aligned} { \begin{array}{lll} {\mathcal {Z}} _n (a') &{} = &{} {\mathcal {Z}} _n (a) + \frac{1}{2} e^m,\\ &{} \in &{} \{{\mathcal {Z}} _n (b) - \frac{1}{2} e^i + \frac{1}{2} e^m,{\mathcal {Z}} _n (b) + \frac{1}{2} e^i + \frac{1}{2} e^m\},\\ &{} \in &{} \{{\mathcal {Z}} _n (b') - \frac{1}{2} e^i, {\mathcal {Z}} _n (b') + \frac{1}{2} e^i\}. \end{array}} \end{aligned}$$

It remains to show that ${\mathcal {Z}} (a'_i)$ belongs to ${\mathbb {Z}}$ and that ${\mathcal {Z}} (b'_i)$ belongs to $({\mathbb {Z}}/2) \setminus {\mathbb {Z}}$. Since we have the three following conditions: (1) ${\mathcal {Z}} (b_i)$ belongs to $({\mathbb {Z}}/2) \setminus {\mathbb {Z}}$, (2) ${\mathcal {Z}} _n (b') = {\mathcal {Z}} _n (b) + \frac{1}{2} e^m$, and (3) $m \not \in \frac{1}{2} \left( {\mathcal {Z}} _n (t)\right)$ (which contains i), then $m \ne i$, which leads to ${\mathcal {Z}} (b'_i) = {\mathcal {Z}} (b_i)$. Then, ${\mathcal {Z}} (b'_i)$ belongs to $({\mathbb {Z}}/2) \setminus {\mathbb {Z}}$. The fact that ${\mathcal {Z}} (a'_i)$ belongs to ${\mathbb {Z}}$ comes from the fact that ${\mathcal {Z}} (a'_i) \in \{{\mathcal {Z}} (b'_i) - \frac{1}{2} e^i, {\mathcal {Z}} (b'_i) + \frac{1}{2} e^i\}.$ This concludes the proof. $\square$

Lemma 12

Assuming $n \ge 2$, let z be an element of ${\mathbb {H}}^{n} \setminus {\mathbb {H}}^{n} _n$ and $t,t'$ be in ${\mathbb {H}}^{n} _{\textit{dim} (z)+1}$ such that they are opposite relatively to z. Then, $\left| \beta ^{\square } (z) \setminus (\beta (t) \cup \beta (t')) \right|$ is a $(n-\textit{dim} (z)-2)$-surface.

Proof

Since t and $t'$ are two opposites,

$$\begin{aligned} \mathbb {1}\left( {\mathcal {Z}} _n (t)\right) = \mathbb {1}\left( {\mathcal {Z}} _n (t')\right) \end{aligned}$$

by Lemma 11, and

$$\begin{aligned} \text {Card} \left( \mathbb {1}\left( {\mathcal {Z}} _n (t)\right) \right) = \text {Card} \left( \mathbb {1}\left( {\mathcal {Z}} _n (z)\right) \right) +1. \end{aligned}$$

Now let $m^*$ be the coordinate in $\llbracket 1,n \rrbracket$ such that $m^* \in \mathbb {1}\left( {\mathcal {Z}} _n (t)\right) \setminus \mathbb {1}\left( {\mathcal {Z}} _n (z)\right)$. We can then write:

$$\begin{aligned} {\mathcal {Z}} _n (t) = {\mathcal {Z}} _n (z) + ({\mathcal {Z}} (t_{m^*}) - {\mathcal {Z}} (z_{m^*})) \; e ^{m^*}. \end{aligned}$$

Now let define ${\mathcal {E}}:= \beta (z) \setminus (\beta (t) \cup \beta (t'))$, and let define the application $\mathfrak {Iso}: {\mathcal {E}} \rightarrow {\mathbb {H}}^{n}$ such that $\forall u \in {\mathcal {E}}$,

$$\begin{aligned} \mathfrak {Iso} (u) := {\mathcal {H}} _n \left( {\mathcal {Z}} _n (u) + ({\mathcal {Z}} (t_{m^*}) - {\mathcal {Z}} (z_{m^*})) \; e ^{m^*} \right) . \end{aligned}$$

Intuitively, this application translates the point u from ${\mathcal {E}}$ to $\beta (t)$ directed by the vector $e ^{m^*}$. More exactly, by Proposition 14, $\mathfrak {Iso}$ is an isomorphism from $|{\mathcal {E}} |$ to $|\beta (t)|$. This way, $|\beta ^{\square } (z) \setminus (\beta (t) \cup \beta (t'))| = |{\mathcal {E}} \setminus \{z\}|$ is isomorphic by Proposition 14 to $|\beta (t) \setminus \{\mathfrak {Iso} (z)\}| = |\beta ^{\square } (t)|$ which is a $(n -\textit{dim} (t) - 1) = (n - 2 - \textit{dim} (z))$-surface. $\square$

Lemma 13

For all $z \in {\mathbb {H}}^{n} \setminus {\mathbb {H}}^{n} _n$ and for all $u \in \beta ^{\square } (z)$,

$$\begin{aligned} {\mathcal {T}}(u) = \left\{ \begin{array}{c} {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + ({\mathcal {Z}} (u_i) - {\mathcal {Z}} (z_i)) . e ^i\right) \; ;\\ i \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \end{array} \right\} \end{aligned}$$

Proof

Let us define:

$$\begin{aligned} {\mathcal {A}}:= \left\{ \begin{array}{c} {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + ({\mathcal {Z}} (u_i) - {\mathcal {Z}} (z_i)) . e ^i\right) \; ;\\ i \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \end{array} \right\} \end{aligned}$$

Let us show first that ${\mathcal {A}} \subseteq {\mathcal {T}}(u)$. Since we have $u \in \beta ^{\square } (z)$, by Proposition 12, $\mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \ne \emptyset$. Then, let t be a face in ${\mathcal {A}}$, t can be written:

$$\begin{aligned} t = {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + ({\mathcal {Z}} (u_{i^*}) - {\mathcal {Z}} (z_{i^*})) . e ^{i^*}\right) , \end{aligned}$$

with $i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$. We recall that

$$\begin{aligned} \frac{1}{2} \left( {\mathcal {Z}} _n (u)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (z)\right) = \emptyset \end{aligned}$$

because $u \in \beta ^{\square } (z)$, we then have the different subcases when $i \in \llbracket 1,n \rrbracket$ :

• 1. i belongs to

  $$\begin{aligned} \left( \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \mathbb {1}\left( {\mathcal {Z}} _n (z)\right) \right) \cup \left( \frac{1}{2} \left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \right) , \end{aligned}$$

  then $u_i = z_i$ and then $t_i = z_i = u_i$ implies that $t_i \in \alpha (u_i) \cap \beta (z_i)$.

• 2. or $i = i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$, then $t_i = u_i$ with ${\mathcal {Z}} (u_i) = {\mathcal {Z}} (z_i) \pm \frac{1}{2}$. Since ${\mathcal {Z}} (t_i) = {\mathcal {Z}} (z_i) \pm \frac{1}{2}$, we have then

  $$\begin{aligned} t_i \in \left\{ \begin{array}{c} \{{\mathcal {Z}} (z_i) - 1/2,{\mathcal {Z}} (z_i)+1/2\},\\ \{{\mathcal {Z}} (z_i) + 1/2,{\mathcal {Z}} (z_i)+3/2\} \end{array} \right\} . \end{aligned}$$

  Also,

  $$\begin{aligned} \beta (z_i) = \left\{ \begin{array}{c} \{{\mathcal {Z}} (z_i) - 1/2,{\mathcal {Z}} (z_i)+1/2\},\\ \{{\mathcal {Z}} (z_i)+1/2\},\\ \{{\mathcal {Z}} (z_i) + 1/2,{\mathcal {Z}} (z_i)+3/2\} \end{array} \right\} \ni t_i. \end{aligned}$$

  This way, $t_i \in \alpha (u_i) \cap \beta (z_i)$.

• 3. or $i \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \setminus \{i^* \}$, then $t_i = z_i$ with ${\mathcal {Z}} (z_i) = {\mathcal {Z}} (u_i) \pm \frac{1}{2}$. Also, we have

  $$\begin{aligned} \alpha (u_i) =\left\{ \begin{array}{c} \{{\mathcal {Z}} (u_i),{\mathcal {Z}} (u_i)+1\},\\ \{{\mathcal {Z}} (u_i)\},\{{\mathcal {Z}} (u_i)+1\} \end{array} \right\} , \end{aligned}$$

  which contains $\left\{ {\mathcal {Z}} (z_i) + \frac{1}{2} \right\} = \{{\mathcal {Z}} (t_i) + \frac{1}{2} \} = t_i$. This way, $t_i \in \alpha (u_i) \cap \beta (z_i)$.

We have finally $t \in \alpha (u) \cap \beta (z)$, and since by construction we have $t \ne z$, $t \in \alpha (u) \cap \beta ^{\square } (z)$. Furthermore, ${\mathcal {Z}} _n (t)$ has the same $\textit{dim} (z)$ integral coordinates as z plus the $i^*$-th one, and then $t \in {\mathbb {H}}^{n} _{\textit{dim} (z)+1}$, then ${\mathcal {A}} \subseteq {\mathcal {T}}(u)$.

Let us show now that ${\mathcal {T}}(u) \subseteq {\mathcal {A}}$. Let t be in ${\mathcal {T}}(u)$. Since $t \in \beta (z)$, we know that:

$$\begin{aligned} \left\{ \forall i \in \llbracket 1,n \rrbracket , i \in \mathbb {1}\left( {\mathcal {Z}} _n (z)\right) \Rightarrow i \in \mathbb {1}\left( {\mathcal {Z}} _n (t)\right) \right\} . \end{aligned}$$

Also, $t \in {\mathbb {H}}^{n} _{\textit{dim} (z)+1}$, then there exists an unique coordinate $i^*$ in $\mathbb {1}\left( {\mathcal {Z}} _n (t)\right) \setminus \mathbb {1}\left( {\mathcal {Z}} _n (z)\right)$. Since $t_i \in \{z_i,u_i\}$ for each $i \in \llbracket 1,n \rrbracket$, then $t_{i^*} = u_{i^*} \ne z_{i^*}$ and for all $i \in \llbracket 1,n \rrbracket \setminus \{i^* \}$, $t_i = z_i$. Because we have $t \in \alpha (u)$, $i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$. Also, we have:

$$\begin{aligned} t = {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + ({\mathcal {Z}} (u_{i^*}) - {\mathcal {Z}} (z_{i^*})) . e ^{i^*}\right) , \end{aligned}$$

then we finally obtain that $t \in {\mathcal {A}}$. $\square$

Lemma 14

Let a, b be two elements of ${\mathbb {H}}^{n}$. Then,

$$\begin{aligned} \left\{ \beta (a) \cap \beta (b) \ne \emptyset \right\} \Leftrightarrow \left\{ a \vee b \text { is WD }\right\} . \end{aligned}$$

Furthermore, when $a \vee b$ is WD, it satisfies the relations:

$$\begin{aligned} { \left\{ \begin{array}{cc} a \vee b = \times _{i \in \llbracket 1,n \rrbracket } (a_i \vee b_i),\\ \beta (a \vee b) = \beta (a) \cap \beta (b).\\ \end{array} \right. } \end{aligned}$$

Proof

Let us treat first the case $a_1,b_1 \in {\mathbb {H}}^{1}$ and let us proceed case by case.

• (1) When $a_1, b_1$ are in ${\mathbb {H}}^{1} _0$, then there exists $i,j \in {\mathbb {Z}}$ such that $a_1 = \{i\}$ et $b_1 = \{j\}$. Then, $\beta (a_1) = \{\{i-1,i\},\{i\},\{i,i+1\}\}$, $\beta (b_1) = \{\{j-1,j\},\{j\},\{j,j+1\}\}$.

  • (A) Then, $i = j$ and $\beta (a_1) \cap \beta (b_1) = \{\{i-1,i\},\{i\},\{i,i+1\}\}$ and $a_1 \vee b_1 = \{i\}$ which implies that $\beta (a_1) \cap \beta (b_1) = \beta (a_1 \vee b_1)$.

  • (B) Or $i = j - 1$, and $\beta (a_1) \cap \beta (b_1) = \{\{i, i+1\}\}$ and $a_1 \vee b_1 = \{i,i+1\}$, which implies $\beta (a_1) \cap \beta (b_1) = \beta (a_1 \vee b_1)$.

  • (C) Or $i = j + 1$, and $\beta (a_1) \cap \beta (b_1) = \{\{j,j+1\}\}$ and $a_1 \vee b_1 = \{j,j+1\}$, which implies $\beta (a_1) \cap \beta (b_1) = \beta (a_1 \vee b_1)$.

  • (D) Or $i \not \in \{j-1,j,j+1\}$, then $\beta (a_1) \cap \beta (b_1) = \emptyset$ and $a_1 \vee b_1$ does not exist.

• (2) When $a_1 \in {\mathbb {H}}^{1} _1$ and $b_1 \in {\mathbb {H}}^{1} _0$, then there exist $i,j \in {\mathbb {Z}}$ such that $a_1 = \{i,i+1\}$ and $b_1 = \{j\}$. Then, $\beta (a_1) = \{\{i,i+1\}\}$, $\beta (b_1) = \{\{j-1,j\},\{j\},\{j,j+1\}\}$.

  • (A) Then, $i = j$, $\beta (a_1) \cap \beta (b_1) = \{\{i,i+1\}\}$, $a_1 \vee b_1 = \{i,i+1\}$ and $\beta (a_1) \cap \beta (b_1) = \beta (a_1 \vee b_1)$.

  • (B) Or $i = j - 1$, $\beta (a_1) \cap \beta (b_1) = \{\{j-1,j\}\}$, $a_1 \vee b_1 = \{j-1,j\}$, and $\beta (a_1) \cap \beta (b_1) = \beta (a_1 \vee b_1)$.

  • (C) Or $i \not \in \{j-1,j\}$, then $\beta (a_1) \cap \beta (b_1) = \emptyset$ and $a_1 \vee b_1$ does not exist.

• (3) When $a_1 \in {\mathbb {H}}^{1} _0$ and $b_1 \in {\mathbb {H}}^{1} _1$, then the reasoning is the same as before.

• (4) When $a_1, b_1 \in {\mathbb {H}}^{1} _1$, then there exists $i,j \in {\mathbb {Z}}$ such that $a_1 = \{i,i+1\}$ and $b_1 = \{j,j+1\}$. We obtain then $\beta (a_1) = \{\{i,i+1\}\}$, $\beta (b_1) = \{\{j,j+1\}\}$.

  • (A) Then, $i = j$, which implies $\beta (a_1) \cap \beta (b_1) = \{\{i,i+1\}\}$, $a_1 \vee b_1 = \{i,i+1\}$ and $\beta (a_1) \cap \beta (b_1) = \beta (a_1 \vee b_1)$.

  • (B) Or $i \ne j$, $\beta (a_1) \cap \beta (b_1) = \emptyset$ and $a_1 \vee b_1$ does not exist.

When a, b belong to ${\mathbb {H}}^{n}$, $n \ge 1$, such that $\beta (a) \cap \beta (b) \ne \emptyset$, we obtain that:

$$\begin{aligned} \begin{array}{ll} \beta (a) \cap \beta (b) &{} = \beta (\times _{i \in \llbracket 1,n \rrbracket } a_i) \cap \beta (\times _{i \in \llbracket 1,n \rrbracket } b_i),\\ &{} = \otimes _{i \in \llbracket 1,n \rrbracket } \beta (a_i) \cap \otimes _{i \in \llbracket 1,n \rrbracket } \beta (b_i),\\ &{} = \otimes _{i \in \llbracket 1,n \rrbracket } \left( \beta (a_i) \cap \beta (b_i)\right) ,\\ &{} \ne \emptyset , \end{array} \end{aligned}$$

then for all $i \in \llbracket 1,n \rrbracket$, $\beta (a_i) \cap \beta (b_i) \ne \emptyset$, which implies that $a_i \vee b_i$ exists and $\beta (a_i) \cap \beta (b_i) = \beta (a_i \vee b_i)$. This way:

$$\begin{aligned} \begin{array}{rl} \beta (a) \cap \beta (b) &{} = \beta (\times _{i \in \llbracket 1,n \rrbracket } a_i) \cap \beta (\times _{i \in \llbracket 1,n \rrbracket } b_i)\\ &{} = \otimes _{i \in \llbracket 1,n \rrbracket } \left( \beta (a_i) \cap \beta (b_i)\right) \\ &{} = \otimes _{i \in \llbracket 1,n \rrbracket } \beta (a_i \vee b_i)\\ &{} = \beta (\times _{i \in \llbracket 1,n \rrbracket } a_i \vee b_i)\\ \end{array} \end{aligned}$$

whose infimum is $\times _{i \in \llbracket 1,n \rrbracket } a_i \vee b_i$ and then $a \vee b = \times _{i \in \llbracket 1,n \rrbracket } a_i \vee b_i$, and $\beta (a \vee b) = \beta (a) \cap \beta (b)$. $\square$

Lemma 15

(Decomposition lemma) Let z be a face in ${\mathbb {H}}^{n} \setminus {\mathbb {H}}^{n} _n$. Each face $u \in \beta ^{\square } (z)$ can be decomposed in the following manner (see Fig. 27):

$$\begin{aligned} u = \bigvee _{v \in {\mathcal {T}}(u)} v. \end{aligned}$$

Proof

We need first to show that $\bigvee _{v \in {\mathcal {T}}(u)} v$ exists. For this aim, it is sufficient to show that $\displaystyle {\bigcap _{t \in {\mathcal {T}}(u)}} \beta (t) \ne \emptyset$. However, $u \in \beta ^{\square } (z)$ implies by Lemma 13 that

$$\begin{aligned} {\mathcal {T}}(u) = \left\{ \begin{array}{c} {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + ({\mathcal {Z}} _n (u_i) - {\mathcal {Z}} _n (z_i)) . e ^i \right) \; ;\\ i \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \end{array} \right\} . \end{aligned}$$

For all $i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$, let us denote:

$$\begin{aligned} t^{i^*} := {\mathcal {H}} _n \left( {\mathcal {Z}} _n (z) + ({\mathcal {Z}} _n (u_{i^*}) - {\mathcal {Z}} _n (z_{i^*})) \; e ^{i^*} \right) , \end{aligned}$$

then ${\mathcal {T}}(u) = \{t^{i^*}\}_{i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }$.

This way,

We want to show that for all $m \in \llbracket 1,n \rrbracket$,

$$\begin{aligned} \displaystyle {\bigcap _{i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }} \beta \left( t^{i^*}_m \right) \ne \emptyset . \end{aligned}$$

Since u belongs to $\beta (z)$, then:

• either $m \in \mathbb {1}\left( {\mathcal {Z}} _n (z)\right) \cup \frac{1}{2} \left( {\mathcal {Z}} _n (u)\right)$, then ${\mathcal {Z}} (u_m) = {\mathcal {Z}} (z_m)$. And because $m \ne i^*$ for all $i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$, $t^{i^*}_m = z_m = u_m$, and:

  $$\begin{aligned} \displaystyle {\bigcap _{i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }} \beta \left( t^{i^*}_m \right) = \beta (u_m) \ne \emptyset . \end{aligned}$$

• or $m \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$, then

  $$\begin{aligned} {\mathcal {Z}} (u_m) \in \left\{ {\mathcal {Z}} (z_m) - \frac{1}{2}, {\mathcal {Z}} (z_m) + \frac{1}{2} \right\} . \end{aligned}$$

  Then, there exists a value $i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$ such that $i^* = m$, for which ${\mathcal {Z}} (t^{i^*}_m) = {\mathcal {Z}} (u_m) \in \{{\mathcal {Z}} (z_m) - \frac{1}{2}, {\mathcal {Z}} (z_m) + \frac{1}{2} \}$, and then:

  $$\begin{aligned}&\displaystyle {\bigcap _{i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }} \beta \left( t^{i^*}_m \right) \\&\quad = \beta (t^m_m) \cap \displaystyle {\bigcap _{i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \setminus \{m\}}} \beta \left( t^{i^*}_m \right) ,\\&\quad = \beta (u_m) \cap \beta (z_m). \end{aligned}$$

  Two cases are possible. Either ${\mathcal {Z}} (u_m) = {\mathcal {Z}} (z_m) - \frac{1}{2}$, then:

  $$\begin{aligned}&\beta (u_m) \cap \beta (z_m)\\&\quad = \beta ({\mathcal {H}} _n ({\mathcal {Z}} (z_m) - \frac{1}{2})) \cap \beta (z_m)\\&\quad = \{\{{\mathcal {Z}} (z_m) - \frac{1}{2},{\mathcal {Z}} (z_m) + \frac{1}{2} \}\}\\&\quad \cap \left\{ \begin{array}{c} \{{\mathcal {Z}} (z_m) - 1/2, {\mathcal {Z}} (z_m) + 1/2\},\\ \{{\mathcal {Z}} (z_m) + 1/2\},\\ \{{\mathcal {Z}} (z_m) + 1/2, {\mathcal {Z}} (z_m) + 3/2\} \end{array} \right\} \\&\quad = \{\{{\mathcal {Z}} (z_m) - \frac{1}{2},{\mathcal {Z}} (z_m) + \frac{1}{2} \}\}\\&\quad = \beta (u_m)\\&\quad \ne \emptyset \end{aligned}$$

  Or ${\mathcal {Z}} (u_m) = {\mathcal {Z}} (z_m) + \frac{1}{2}$, then:

  $$\begin{aligned}&\beta (u_m) \cap \beta (z_m)\\&\quad = \beta ({\mathcal {H}} _n ({\mathcal {Z}} (z_m) + \frac{1}{2})) \cap \beta (z_m)\\&\quad = \{\{{\mathcal {Z}} (z_m) + 1/2,{\mathcal {Z}} (z_m) + 3/2\}\}\\&\quad \cap \left\{ \begin{array}{c} \{{\mathcal {Z}} (z_m) - 1/2, {\mathcal {Z}} (z_m) + 1/2\},\\ \{{\mathcal {Z}} (z_m) + 1/2\},\\ \{{\mathcal {Z}} (z_m) + 1/2, {\mathcal {Z}} (z_m) + 3/2\} \end{array} \right\} \\&\quad = \{\{{\mathcal {Z}} (z_m) + 1/2,{\mathcal {Z}} (z_m) + 3/2\}\}\\&\quad = \beta (u_m)\\&\quad \ne \emptyset \end{aligned}$$

This way, for all $m \in \llbracket 1,n \rrbracket$,

$$\begin{aligned} \displaystyle {\bigcap _{i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }} \beta \left( t^{i^*}_m \right) \ne \emptyset , \end{aligned}$$

and then $\displaystyle {\bigcap _{t \in {\mathcal {T}}(u)}} \beta (t) \ne \emptyset$, which implies that $\bigvee _{t \in {\mathcal {T}}(u)} t$ exists in ${\mathbb {H}}^{n}$.

Let us compute now this term, following the calculus made before:

$$\begin{aligned} \begin{array}{rl} \beta \left( \bigvee _{t \in {\mathcal {T}}(u)} t \right) &{}= \displaystyle {\bigotimes _{m \in \llbracket 1,n \rrbracket }} \; \displaystyle {\bigcap _{i^* \in \mathbb {1}\left( {\mathcal {Z}} _n (u)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }} \beta \left( t^{i^*}_m \right) \\ &{}= \displaystyle {\bigotimes _{m \in \llbracket 1,n \rrbracket }} \; \beta (u_m)\\ &{}= \beta (\times _{m \in \llbracket 1,n \rrbracket } u_m)\\ &{}= \beta (u)\\ \end{array} \end{aligned}$$

and then $u = \bigvee _{t \in {\mathcal {T}}(u)} t$. $\square$

Lemma 17

Let v, f be two faces of ${\mathbb {H}}^{n}$ such that $f \in \beta (v)$. For all $k \in \llbracket \textit{dim} (v), \textit{dim} (f) \rrbracket$:

$$\begin{aligned} \text {Card} \left( \alpha (f) \cap \beta (v) \cap {\mathbb {H}}^{n} _{k}\right) = C^{k - \textit{dim} (v)}_{\textit{dim} (f) - \textit{dim} (v)}. \end{aligned}$$

Proof

This proof is an enumeration problem: we want to find the total number of k-faces of $\alpha (f) \cap \beta (v)$. We recall that by Proposition 12 for all $w \in {\mathbb {H}}^{n}$, we have:

$$\begin{aligned} \begin{array}{c} w \in \alpha (f) \cap \beta (v)\\ \Leftrightarrow \\ \left\{ \begin{array}{l} \forall m \in \mathbb {1}\left( {\mathcal {Z}} _n (f)\right) \cap \frac{1}{2} \left( {\mathcal {Z}} _n (v)\right) , w_m \in \{v_m,f_m\},\\ \text {and}\\ \forall m \in \frac{1}{2} \left( {\mathcal {Z}} _n (f)\right) \cup \mathbb {1}\left( {\mathcal {Z}} _n (v)\right) , w_m = v_m (= f_m).\\ \end{array} \right\} . \end{array} \end{aligned}$$

Then, $\textit{dim} (v)$ coordinates of w in $\alpha (f) \cap \beta (v)$ are fixed and integral, which means that $(k - \textit{dim} (v))$ of its coordinates are free, that is, in half or integral. These last coordinates can be chosen among the $(n - \textit{dim} (v))$ half coordinates of v minus the $(n - \textit{dim} (f))$ half coordinates of f, which explains the $(\textit{dim} (f) - \textit{dim} (v))$ term. This concludes the proof. $\square$

Lemma 18

Let us assume that $n \ge 2$. Let z be in ${\mathbb {H}}^{n}$ such that $\textit{dim} (z) \le n - 2$, and let a, b be in ${\mathbb {H}}^{n} _n \cap \beta ^{\square } (z)$. Then, $\alpha ^{\square } (a) \cap \alpha ^{\square } (b) \cap \beta ^{\square } (z) = \emptyset$ implies that ${\mathcal {Z}} _n (a)$ and ${\mathcal {Z}} _n (b)$ are $(n - \textit{dim} (z))$-antagonist into ${\mathbb {Z}} ^n$.

Proof

The fact that $a, b \in \beta ^{\square } (z)$ implies that $z \in \alpha ^{\square } (a) \cap \alpha ^{\square } (b)$, and then $\alpha (a) \cap \alpha (b) \ne \emptyset$, which implies that $a \wedge b$ exists and $\alpha (a) \cap \alpha (b) = \alpha (a \wedge b)$ by Lemma 2. This way,

$$\begin{aligned} z \in \alpha ^{\square } (a) \cap \alpha ^{\square } (b) \subseteq \alpha (a \wedge b), \end{aligned}$$

and then $a \wedge b \in \beta (z)$. Let us assume that we have $a = b$. Then,

$$\begin{aligned} \alpha ^{\square } (a) \cap \alpha ^{\square } (b) \cap \beta ^{\square } (z) = \alpha ^{\square } (a) \cap \beta ^{\square } (z) \end{aligned}$$

is a $(n-\textit{dim} (z)-2)$-surface by Proposition 7, and then is non-empty (because $(n-\textit{dim} (z)) \ge 2$). This is impossible by hypothesis, and then we have $a \ne b$. Since a and b are different and they are both into ${\mathbb {H}}^{n} _n$, they are not neighbors and this way $\alpha ^{\square } (a) \cap \alpha ^{\square } (b) = \alpha (a) \cap \alpha (b) = \alpha (a \wedge b)$. We obtain that $\alpha (a \wedge b) \cap \beta ^{\square } (z) = \emptyset$. We have seen that $z \in \alpha (a \wedge b)$, then

$$\begin{aligned} \left( \alpha (a \wedge b) \cap \beta ^{\square } (z) \right) \cup \{z\} = \alpha (a \wedge b) \cap \beta (z) = \{z\}, \end{aligned}$$

and then $z = a \wedge b$. By Lemma 4, we deduce that ${\mathcal {Z}} _n (a)$ and ${\mathcal {Z}} _n (b)$ are $(n - \textit{dim} (z))$-antagonists. $\square$

Lemma 19

Let $n \ge 3$ be an integer. Let z be in ${\mathbb {H}}^{n}$ such that $\textit{dim} (z) \le n - 3$, and let $p,p'$ be in S(z) such that $p = \text {antag}_{S(z)}(p')$. Then, $S(z) \setminus \{p,p'\}$ is 2n-connected into ${\mathbb {Z}} ^n$.

Proof

Let x, y be in $S(z) \setminus \{a,b\}$ with $x \ne y$. Then, there exists a value $k \in \llbracket 1, n-\textit{dim} (z) \rrbracket$ such that x and y are k-antagonists (since they belong to the block S(z) of dimension $(n-\textit{dim} (z))$). Let us now proceed by induction on the value of k.

$\underline{\mathrm{Initialization}\,(k = 1):}$ when x and y are 1-antagonists in ${\mathbb {Z}} ^n$, they are 2n-neighbors and then there exists a 2n-path $\pi$ joining them into $S(z) \setminus \{a,b\}$ such that $\pi = (x,y)$.

$\underline{\mathrm{Heredity}\, (k \in \llbracket 2, n - \textit{dim} (z) \rrbracket ):}$ we assume that for all the elements x, y in $S(z) \setminus \{a,b\}$ such that they are $(k-1)$-antagonists into S(z), there exists a 2n-path joining them into $S(z) \setminus \{a,b\}$. Let us show that when x and y are k-antagonists, x and y are 2n-connected into $S(z) \setminus \{a,b\}$. By hypothesis, x and y are k-antagonists with $k \ge 2$ and belong to S(z), then they are antagonists in a block $S \subseteq S(z)$ of dimension k. This way, x admits in S a total number of k 2n-neighbors (which are different from itself) where at most one is into $\{a,b\}$. Indeed, if a and b were neighbors of x, they would be identical or 2-antagonists. These two cases are impossible since a and b are $(n-\textit{dim} (z))$-antagonists with $(n-\textit{dim} (z)) \ge 3$. Then, there exists a 2n-neighbor $v_x$ of x into $S \setminus \{a,b\}$; $v_x$ is then $(k-1)$-antagonist of y and 2n-neighbor of x, then x and y are connected into $S(z) \setminus \{a,b\}$ thanks to the induction hypothesis. $\square$

Proofs of Section 6

Lemma 20

We assume that X is a non-empty finite subset of ${\mathbb {Z}} ^n$, that Y is its complement in ${\mathbb {Z}} ^n$, that ${\mathfrak {N}}$ is the boundary of the hit/miss-transform of X. Then, each component $|F_{i} |$ of $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ is closed in $\beta ^{\square } (z)$.

Proof

Since $\beta ^{\square }_{{\mathfrak {N}}} (z)$ is equal to the intersection ${\mathfrak {N}} \cap \beta ^{\square } (z)$ and ${\mathfrak {N}}$ is closed in ${\mathbb {H}}^{n}$ as intersection of two closed sets of ${\mathbb {H}}^{n}$, then $\beta ^{\square }_{{\mathfrak {N}}} (z)$ is closed in $\beta ^{\square } (z)$. Because when a subset of a topological space is closed, each connected component is closed too, then the connected components of $\beta ^{\square }_{{\mathfrak {N}}} (z)$ are closed in $\beta ^{\square } (z)$, and then for each $i \in {\mathcal {I}}$, $F_{i}$ is closed in $\beta ^{\square } (z)$. $\square$

Lemma 21

We assume that X is a non-empty finite subset of ${\mathbb {Z}} ^n$, that Y is its complement in ${\mathbb {Z}} ^n$, that ${\mathfrak {N}}$ is the boundary of the hit/miss-transform of X. Then, for two different components $F_{i}$ and $F_{j}$ of $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$:

$$\begin{aligned} \beta (F_{i}) \cap F_{j} = \emptyset , \text { and }\alpha (F_{i}) \cap F_{j} = \emptyset . \end{aligned}$$

Proof

Let i, j be two different values of ${\mathcal {I}}$. Now, let us assume that the intersection $\beta (F_{i}) \cap F_{j}$ is not empty, then there exists some $z \in \beta (F_{i}) \cap F_{j}$. It means that we have at the same time $z \in \beta (F_{i})$ and $z \in F_{j}$. It is equivalent to say that there exists some $a \in F_{i}$ such that $z \in \beta (a)$ and $z \in F_{j}$. Then, there exists some $a \in F_{i}$ such that $a \in \alpha (z)$ and $z \in F_{j}$. So, there exists some $a \in F_{i}$ which satisfies $a \in \alpha (F_{j})$. Now, we know that $F_{j}$ is a connected component of the closed set ${\mathfrak {N}}$, and then $\alpha (F_{j}) = F_{j}$. Then, there exists $a \in F_{i} \cap F_{j}$, which would mean that $F_{i} \cap F_{j}$ is non-empty. We have a contradiction. $\square$

Lemma 22

We assume that X is a non-empty finite subset of ${\mathbb {Z}} ^n$, that Y is its complement in ${\mathbb {Z}} ^n$, that ${\mathfrak {N}}$ is the boundary of the hit/miss-transform of X. For each $u \in \beta ^{\square }_{{\mathfrak {N}}} (z)$, there exists one unique index $i^* \in {\mathcal {I}}$ such that $u \in F_{i^*}$ and it satisfies that:

$$\begin{aligned} { \left\{ \begin{array}{l} \alpha ^{\square } _{F_{i^*}}(u) = \alpha ^{\square } _{\beta ^{\square }_{{\mathfrak {N}}} (z)}(u),\\ \beta ^{\square } _{F_{i^*}}(u) = \beta ^{\square } _{\beta ^{\square }_{{\mathfrak {N}}} (z)}(u),\\ \theta ^{\square } _{F_{i^*}}(u) = \theta ^{\square } _{\beta ^{\square }_{{\mathfrak {N}}} (z)}(u). \end{array} \right. } \end{aligned}$$

Proof

Let u be an element of $\beta ^{\square }_{{\mathfrak {N}}} (z)$. Then, there exists one, and only one, value $i^* \in {\mathcal {I}}$ such that u belongs to $F_{i^*}$.

• First, let us remark that:

  $$\begin{aligned}&\alpha ^{\square } _{\beta ^{\square }_{{\mathfrak {N}}} (z)}(u)\\&\quad = \alpha ^{\square } (u) \cap \left( \bigcup _{i \in {\mathcal {I}}} F_{i} \right) ,\\&\quad = (\alpha ^{\square } (u) \cap F_{i^*}) \cup \left( \alpha ^{\square } (u) \cap \left( \bigcup _{i \in {{\mathcal {I}} \setminus \{i^* \}}} F_{i} \right) \right) ,\\&\quad = (\alpha ^{\square } (u) \cap F_{i^*}) \cup \left( \bigcup _{i \in {{\mathcal {I}} \setminus \{i^* \}}} \alpha ^{\square } (u) \cap F_{i} \right) ,\\ \end{aligned}$$

  However, we can remark that if for any $i^*_2 \in {\mathcal {I}} \setminus \{i^* \}$, we have $\alpha ^{\square } (u) \cap F_{i^*_2} \ne \emptyset$, then there exists some $z \in \alpha ^{\square } (u) \cap F_{i^*_2}$, and then there exists $z \in F_{i^*_2}$ such that $u \in \beta ^{\square } (z)$, which implies $u \in \beta (F_{i^*_2})$, and then $F_{i^*}$ and $F_{i^*_2}$ are neighbors. Contradiction.

• By symmetry, we obtain a same reasoning.

• With the two preceding assertions true, we can remark that:

  $$\begin{aligned}&\theta ^{\square } _{\beta ^{\square }_{{\mathfrak {N}}} (z)}(u)\\&\quad = \theta ^{\square } (u) \cap \beta ^{\square }_{{\mathfrak {N}}} (z),\\&\quad = (\alpha ^{\square } (u) \cup \beta ^{\square } (u)) \cap \beta ^{\square }_{{\mathfrak {N}}} (z),\\&\quad = (\alpha ^{\square } (u) \cap \beta ^{\square }_{{\mathfrak {N}}} (z)) \cup (\beta ^{\square } (u) \cap \beta ^{\square }_{{\mathfrak {N}}} (z)),\\&\quad = \alpha ^{\square } _{\beta ^{\square }_{{\mathfrak {N}}} (z)}(u) \cup \beta ^{\square } _{\beta ^{\square }_{{\mathfrak {N}}} (z)}(u),\\&\quad = (\alpha ^{\square } _{F_{i^*}}(u) \cup \beta ^{\square } _{F_{i^*}}(u),\\&\quad = \theta ^{\square } _{F_{i^*}}(u). \end{aligned}$$

$\square$

Property 1

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). Then, $\forall i \in {\mathcal {I}}$, $\forall t \in {\mathbb {H}}^{n} _{\textit{dim} (z)+1},$

$$\begin{aligned} \left\{ t \in F_{i} \Rightarrow \text {opp}_{z}(t) \not \in F_{i} \right\} . \end{aligned}$$

Proof

It is sufficient to show that the hypothesis of non-connectivity of $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ and of presence of two opposite faces in a same connected component of $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ are incompatible. The reasoning of this proof is depicted in Fig. 31. Let i be in ${\mathcal {I}}$ such that there exists $t,t' \in {\mathbb {H}}^{n} _{\textit{dim} (z)+1} \cap F_{i}$ satisfying $t' = \text {opp}_{z}(t)$.

Then, for all $j \in {\mathcal {I}}$, $j \ne i$, we have $\beta (F_{i}) \cap F_{j} = \emptyset$ by Lemma 21, and then $\beta (t) \cap F_{j} = \emptyset$, and $\beta (t') \cap F_{j} = \emptyset$. This way, $F_{j} \subseteq \beta ^{\square } (z) \setminus (\beta (t) \cup \beta (t'))$, which is by Lemma 12 a $(n- \textit{dim} (z)- 2)$-surface (like $F_{j}$). However, when two discrete surfaces of same rank satisfy an inclusion relationship, they are equal (see Proposition 6), then we have :

$$\begin{aligned} F_{j} = \beta ^{\square } (z) \setminus (\beta (t) \cup \beta (t')). \end{aligned}$$

This implies that $F_{i}$ is included into $\beta (t) \cup \beta (t')$ and then $F_{i} = F_{i} \cap (\beta (t) \cup \beta (t'))$. Since t and $t'$ belong to $F_{i}$, we obtain finally that:

$$\begin{aligned} F_{i} = \beta _{F_{i}}(t) \cup \beta _{F_{i}}(t'), \end{aligned}$$

which is a disjoint union of two open non-empty sets, i.e., $F_{i}$ is not connected, which is impossible. $\square$

Property 2

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). For each value i in ${\mathcal {I}}$, $F_{i}$ contains at most $(n-\textit{dim} (z))$$(\textit{dim} (z)+1)$-faces.

Proof

$\beta ^{\square } (z)$ contains exactly $2(n-\textit{dim} (z))$ pairs of opposite $(\textit{dim} (z)+1)$-faces, and then by Property 1, for all i in ${\mathcal {I}}$, $F_{i}$ contains at most $(n-\textit{dim} (z))$$(\textit{dim} (z)+1)$-faces. $\square$

Lemma 23

Let x, y be two elements of ${\mathbb {Z}} ^n$ and S be a block such that $x = \text {antag}_{S}(y)$. Then, for all $z \in S$:

$$\begin{aligned} \left\{ \begin{array}{l} \alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (y)) \subseteq \alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (z)),\\ ~\\ \alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (y)) \subseteq \alpha ({\mathcal {H}} _n (z)) \cap \alpha ({\mathcal {H}} _n (y)). \end{array}\right. \end{aligned}$$

Fig. 54

When z belongs to the block S where x and y are antagonist, we have the relation $\alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (y)) \subseteq \alpha ({\mathcal {H}} _n (z)) \cap \alpha ({\mathcal {H}} _n (y))$.

Proof

By symmetry, it is sufficient to show the first assertion (see Fig. 54). Let ${\mathfrak {I}} = \{i \in \llbracket 1,n \rrbracket \; ; \; x_i \ne y_i\}$. Let z be an element of S, then for all $i \in {\mathfrak {I}}$, $z_i \in \{x_i,y_i\}$ and for all $i \in \llbracket 1,n \rrbracket \setminus {\mathfrak {I}}, x_i = y_i = z_i$.

When $i \in \llbracket 1,n \rrbracket \setminus {\mathfrak {I}}$, $x_i = y_i = z_i$ and then:

$$\begin{aligned}&\alpha ({\mathcal {H}} (x_i)) \cap \alpha ({\mathcal {H}} (y_i))\\&\quad = \alpha ({\mathcal {H}} (x_i)),\\&\quad = \alpha ({\mathcal {H}} (x_i)) \cap \alpha ({\mathcal {H}} (z_i)). \end{aligned}$$

When $i \in {\mathfrak {I}}$, either $z_i = x_i$, and:

$$\begin{aligned}&\alpha ({\mathcal {H}} (x_i)) \cap \alpha ({\mathcal {H}} (y_i))\\&\quad \subseteq \alpha ({\mathcal {H}} (x_i)),\\&\quad \subseteq \alpha ({\mathcal {H}} (x_i)) \cap \alpha ({\mathcal {H}} (z_i)), \end{aligned}$$

or $z_i = y_i$ and we obtain immediately:

$$\begin{aligned} \alpha ({\mathcal {H}} (x_i)) \cap \alpha ({\mathcal {H}} (y_i)) = \alpha ({\mathcal {H}} (x_i)) \cap \alpha ({\mathcal {H}} (z_i)). \end{aligned}$$

A simple application of the Cartesian product is then sufficient to end the proof. $\square$

Lemma 24

Let X be a non-empty finite subset of ${\mathbb {Z}} ^n$, Y its complement in ${\mathbb {Z}} ^n$, and ${\mathfrak {N}}$ be the boundary of the miss-transform of X in ${\mathbb {H}}^{n}$. For each $z \in {\mathfrak {N}}$:

$$\begin{aligned} \beta ^{\square }_{{\mathfrak {N}}} (z) = \bigcup _{f \in {\mathbb {H}}^{n} _{n-1} \cap \beta ^{\square }_{{\mathfrak {N}}} (z)} \alpha (f) \cap \beta ^{\square } (z), \end{aligned}$$

in other words, $\beta ^{\square }_{{\mathfrak {N}}} (z)$ is equal to the union of the closures (into $\beta ^{\square } (z)$) of its $(n-1)$-faces.

Proof

Let us begin with the first inclusion. Since for all $f \in \beta ^{\square }_{{\mathfrak {N}}} (z)$, $f \in {\mathfrak {N}}$, then $\alpha (f) \cap \beta ^{\square } (z) \subseteq \beta ^{\square }_{{\mathfrak {N}}} (z)$ because $\beta ^{\square }_{{\mathfrak {N}}} (z)$ is closed in the subspace $\beta ^{\square } (z)$.

Now let us prove the second inclusion. Let u be an element of $\beta ^{\square }_{{\mathfrak {N}}} (z)$. Let us recall that $S(z) = {\mathcal {Z}} _n (\beta (z) \cap {\mathbb {H}}^{n} _n)$ is the block centered at z (see Proposition 4). Then, by Lemma 8:

$$\begin{aligned} \begin{array}{lll} \beta ^{\square }_{{\mathfrak {N}}} (z) &{} = &{} \alpha ({\mathcal {H}} _n (X)) \cap \alpha ({\mathcal {H}} _n (Y)) \cap \beta ^{\square } (z),\\ &{} = &{} \alpha ({\mathcal {H}} _n (X \cap S(z))) \cap \alpha ({\mathcal {H}} _n (Y \cap S(z))) \cap \beta ^{\square } (z).\\ \end{array} \end{aligned}$$

This way, there exist $x \in X \cap S(z)$ and $y \in Y \cap S(z)$ such that $u \in \alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (y))$. Also, x and y belonging to the same block S(z) and being different, they are k-antagonist, $k \ge 1$.

Now let ${\mathfrak {I}}$ be the set $\{i \in \llbracket 1,n \rrbracket \; ; \; x_i \ne y_i\} = \{j_1, \dots , j_k\}$ where the sequence $(j_i)_i$ is strictly increasing. We can then define the 2n-path $\pi$, i.e., a sequence in ${\mathbb {Z}} ^n$ such as two consecutive elements in the sequence are 2n-neighbors in ${\mathbb {Z}} ^n$, joining x and y into S(z):

$$\begin{aligned} \pi = (p^0 = x, p^1, \dots ,p^{k-1},p^k = y), \end{aligned}$$

satisfying the recursive relation:

$$\begin{aligned} \left\{ \begin{array}{l} p^0 = x,\\ p^{l} = p^{l-1} + (y_{j_l} - x_{j_l}) . e ^{j_l}, \forall l \in \llbracket 1, k \rrbracket , \end{array}\right. \end{aligned}$$

Now, let us define

$$\begin{aligned} l^* := \min \{ l \in \llbracket 1, k \rrbracket \; ; \; p^l \in Y\} - 1, \end{aligned}$$

then we obtain two points $x' := p^{l^*} \in X$ and $y' := p^{l^*+1} \in Y$ which are 2n-neighbors in the block S(z) (see Fig. 55).

Fig. 55

Let x be in X and y be in Y such that they are antagonist in a block $S \subset {\mathbb {Z}} ^n$. They are joined by a 2n-path $\pi \subseteq S$ containing a pair $(x',y') \in X \times Y$ such that $x' \in {\mathcal {N}}^*_{2n} (y')$

Since $y' - x = \displaystyle {\sum _{l \in \llbracket 1, l^*+1 \rrbracket }} (y_{j_l} - x_{j_l}) . e ^{j_l}$, $y'$ and x are antagonist in a block of dimension $(l^*+1)$ that we will call $S'$. Moreover:

$$\begin{aligned} \left\{ \begin{array}{lr} x' = x + \sum _{l \in \llbracket 1, l^* \rrbracket } (y_{j_l} - x_{j_l}) . e ^{j_l}, &{} (1)\\ x' = y' + (x_{j_{l^*+1}} - y_{j_{l^*+1}}) . e ^{j_{l^*+1}}, &{} (2) \end{array}\right. \end{aligned}$$

then $\forall i \in \llbracket 1,n \rrbracket$, $x'_i \in \{y'_i,x_i\}$, which implies $x' \in S'$.

Then, using Lemma 23:

$$\begin{aligned} \alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (y')) \subseteq \alpha ({\mathcal {H}} _n (x')) \cap \alpha ({\mathcal {H}} _n (y')). \end{aligned}$$

Moreover, $y' \in S$ where x and y are antagonist, so one more time using Lemma 23,

$$\begin{aligned} \alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (y)) \subseteq \alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (y')), \end{aligned}$$

and then we obtain by transitivity that:

$$\begin{aligned} \alpha ({\mathcal {H}} _n (x)) \cap \alpha ({\mathcal {H}} _n (y)) \subseteq \alpha ({\mathcal {H}} _n (x')) \cap \alpha ({\mathcal {H}} _n (y')). \end{aligned}$$

This way, $u \in \alpha ({\mathcal {H}} _n (x')) \cap \alpha ({\mathcal {H}} _n (y'))$. Also, $x'$ and $y'$ being 2n-neighbors, they are 1-antagonist and then u belongs to the closure of the $(n-1)$-face $f := {\mathcal {H}} _n (x') \wedge {\mathcal {H}} _n (y')$ by Lemma 2.

Because ${\mathcal {H}} _n (x')$ and ${\mathcal {H}} _n (y')$ belong to $\beta ^{\square } (z)$, we have $z \in \alpha ({\mathcal {H}} _n (x')) \cap \alpha ({\mathcal {H}} _n (y')) = \alpha (f)$ and then $f \in \beta (z)$. Since f is of dimension $n-1$ and z of dimension lower than or equal to $n-3$, f is different from z. Because f is the supremum of $\alpha ({\mathcal {H}} _n (x')) \cap \alpha ({\mathcal {H}} _n (y')) \subseteq {\mathfrak {N}}$, it belongs to ${\mathfrak {N}}$. Finally, we have $u \in \alpha (f) \cap \beta ^{\square } (z)$ with $f \in \beta ^{\square }_{{\mathfrak {N}}} (z) \cap {\mathbb {H}}^{n} _{n-1}$. $\square$

Lemma 25

Let X be a non-empty finite subset of ${\mathbb {Z}} ^n$, Y its complement in ${\mathbb {Z}} ^n$, and ${\mathfrak {N}}$ be the boundary of the miss-transform of X in ${\mathbb {H}}^{n}$. For each $z \in {\mathfrak {N}}$, and for any $i \in {\mathcal {I}}$, the component $F_{i}$ of $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ is the closure in $\beta ^{\square } (z)$ of a set of $(n-1)$-faces, i.e.,

$$\begin{aligned} F_{i} = \bigcup _{f \in {\mathbb {H}}^{n} _{n-1} \cap F_{i}} \alpha (f) \cap \beta ^{\square } (z). \end{aligned}$$

Proof

Using Lemma 24, we have:

$$\begin{aligned} \beta ^{\square }_{{\mathfrak {N}}} (z) = \bigcup _{f \in {\mathbb {H}}^{n} _{n-1} \cap \beta ^{\square }_{{\mathfrak {N}}} (z)} \alpha (f) \cap \beta ^{\square } (z), \end{aligned}$$

where for all $f \in {\mathbb {H}}^{n} _{n-1} \cap \beta ^{\square }_{{\mathfrak {N}}} (z)$, the orders $|\alpha (f) \cap \beta ^{\square } (z)|$ are connected; indeed, any two faces a, b in this poset different from f are connected by the path $\pi = (a,f,b)$.

Now, let us show by a double inclusion that we can prove the result we are looking for.

For any $i \in {\mathcal {I}}$, and for each $f \in {\mathbb {H}}^{n} _{n-1} \cap F_{i}$, $|\alpha (f) \cap \beta ^{\square } (z)|$ is connected, and share f with $F_{i}$. Since they are both subsets of $\beta ^{\square }_{{\mathfrak {N}}} (z)$, by definition of $F_{i}$,

$$\begin{aligned} F_{i} \supseteq \alpha (f) \cap \beta ^{\square } (z). \end{aligned}$$

Hence,

$$\begin{aligned} F_{i} \supseteq \bigcup _{f \in {\mathbb {H}}^{n} _{n-1} \cap F_{i}} \alpha (f) \cap \beta ^{\square } (z). \end{aligned}$$

Conversely, $F_{i}$ is a connected component of $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ which is closed in $\beta ^{\square } (z)$, and then $F_{i}$ is also closed in $\beta ^{\square } (z)$, which means that for $f \in F_{i}$, $\alpha (f) \cap \beta ^{\square } (z) \subseteq F_{i}$, then for any $f \in F_{i} \cap {\mathbb {H}}^{n} _{n-1}$, $\alpha (f) \cap \beta ^{\square } (z) \subseteq F_{i}$, and then:

$$\begin{aligned} F_{i} \subseteq \bigcup _{f \in {\mathbb {H}}^{n} _{n-1} \cap F_{i}} \alpha (f) \cap \beta ^{\square } (z). \end{aligned}$$

That concludes the proof. $\square$

Lemma 26

Let f, z be two elements of ${\mathbb {H}}^{n}$ such that $f \in \beta (z)$, and let be ${\mathfrak {I}} = \{i \in \llbracket 1,n \rrbracket \; ; \; f_i \ne z_i\}$. Then,

$$\begin{aligned} \textit{dim} (f) = \textit{dim} (z) + \text {Card} \left( {\mathfrak {I}} \right) . \end{aligned}$$

Proof

Since $f \in \beta (z)$, then for all $i \in \llbracket 1,n \rrbracket$, $f_i \in \beta (z_i)$ and then three cases are possible:

• $\textit{dim} (z_i) = 1$, and then $f_i = z_i$ (because $\beta (z_i) = \{z_i\}$),

• or $\textit{dim} (z_i) = 0$ and $\textit{dim} (f_i) = 0$, then $f_i = z_i$ (because the only face of dimension 0 in $\beta (z_i)$ is $z_i$),

• or $\textit{dim} (z_i) = 0$ and $\textit{dim} (f_i) = 1$, and then

  $$\begin{aligned} f_i \in \left\{ {\mathcal {H}} \left( {\mathcal {Z}} (z_i) - \frac{1}{2} \right) ,{\mathcal {H}} \left( {\mathcal {Z}} (z_i) + \frac{1}{2} \right) \right\} . \end{aligned}$$

In other words, the number of coordinates where f and z are different is equal to the number of times when the dimension of $f_i$ is strictly greater than the dimension of $z_i$ when i is in $\llbracket 1,n \rrbracket$. $\square$

Property 3

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). Then, $\forall i \in {\mathcal {I}}$, $\forall m \in \llbracket \textit{dim} (z)+1, n-1\rrbracket$:

$$\begin{aligned} F_{i} \cap {\mathbb {H}}^{n} _{m} \ne \emptyset . \end{aligned}$$

Proof

Intuitively, $F_{i}$ being an union of closures of $(n-1)$-faces in $\beta ^{\square } (z)$ by Lemma 25, it contains faces of all dimensions between $(n-1)$ and $(\textit{dim} (z)+1)$.

Formally, since $F_{i}$ is non-empty, there exists one face $f \in {\mathbb {H}}^{n} _{n-1} \cap F_{i} \subseteq \beta ^{\square } (z)$ such that $\alpha (f) \cap \beta ^{\square } (z)$ is included into $F_{i}$. Furthermore, $\alpha (f) \cap \beta ^{\square } (z)$ is not empty because $f \in \beta ^{\square } (z)$, then $\alpha (f) \cap \beta ^{\square } (z)$ is equal to:

$$\begin{aligned} \{{\mathcal {H}} _n (u) \; ; \; u_i \in \{{\mathcal {Z}} (f_i), {\mathcal {Z}} (z_i)\}, \forall i \in \llbracket 1,n \rrbracket \} \setminus \{z\}. \end{aligned}$$

Let us define ${\mathfrak {I}} = \{i \in \llbracket 1,n \rrbracket \; ; \; z_i \ne f_i\} = \{j_1, \dots , j_k\}$ where the sequence $(j_i)_i$ is strictly increasing, and let us define the sequence $(u^l)_{l \in \llbracket 0, k \rrbracket }$ included into ${\mathcal {Z}} _n (\alpha (f) \cap \beta (z))$ defined such that:

$$\begin{aligned} \left\{ \begin{array}{l} u^0 = {\mathcal {Z}} _n (f),\\ u^{l+1} = u^l + ({\mathcal {Z}} (z_{j_l}) - {\mathcal {Z}} (f_{j_l})) \; e ^{j_l}, \forall l \in \llbracket 0, k - 1 \rrbracket . \end{array} \right. \end{aligned}$$

Since f belongs to $\beta (z)$ by hypothesis, $|{\mathcal {Z}} (f_i) - {\mathcal {Z}} (z_i)| = \frac{1}{2}, \forall i \in {\mathfrak {I}}$. In this way, ${\mathcal {H}} _n (u^l)$ is of dimension $(\textit{dim} (f) - l)$ for any $l \in \llbracket 0, k \rrbracket$. By Lemma 26, $k = \textit{dim} (f) - \textit{dim} (z)$, and then $\textit{dim} ({\mathcal {H}} _n (u^l))$ ranges $\llbracket \textit{dim} (z) + 1, n-1 \rrbracket$ when l ranges $\llbracket 0,k-1 \rrbracket$. For these values of l, ${\mathcal {H}} _n (u^l)$ belongs to $\alpha (f) \cap \beta ^{\square } (z)$. This concludes the proof. $\square$

Property 4

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). Now, $\forall i \in {\mathcal {I}}$, $\forall v \in F_{i}$,

$$\begin{aligned} \rho (v, |F_{i} |) = \textit{dim} (v) - \textit{dim} (z) - 1. \end{aligned}$$

Proof

Let u be an element of $F_{i}$ of dimension between $\textit{dim} (z)+1$ and $n-1$; this is possible thanks to Property 3. We want to show by induction that $\rho (u,|F_{i} |) = k$ is equivalent to $\textit{dim} (u) = k + \textit{dim} (z) + 1$.

$\underline{\mathrm{Initialization}\, (k = 0):}$ first, let us assume that u is of dimension $\textit{dim} (u) = \textit{dim} (z) + 1$. Since $F_{i} \subseteq \beta ^{\square } (z)$, $\alpha ^{\square } (u) \cap F_{i} \subseteq \alpha ^{\square } (u) \cap \beta ^{\square } (z) = \emptyset$, then $\alpha ^{\square } _{F_{i}}(u) = \emptyset$, and then $\rho (u,|F_{i} |) = 0$. Now, let us assume that $\rho (u,|F_{i} |) = 0$, then u belongs to $F_{i}$ which is closed in $\beta ^{\square } (z)$, and then $\alpha (u) \cap \beta ^{\square } (z) \subseteq F_{i}$. In this way, the only faces whose rank is 0 in $F_{i}$ are the $(\textit{dim} (z)+1)$-faces of $\beta ^{\square } (z)$. Finally, we have for each $u \in F_{i}$ the equivalence $\rho (u,|F_{i} |) = 0 \Leftrightarrow \textit{dim} (u) = \textit{dim} (z) + 1$.

$\underline{\mathrm{Heredity}\, (k \ge 1):}$ we can assume that for each $l \in \llbracket 0, k - 1 \rrbracket$, $\rho (u,|F_{i} |) = l \Leftrightarrow \textit{dim} (u) = \textit{dim} (z) + 1 + l$. Let us show that for all $v \in F_{i}$ and $k \ge 1$, we have the equivalence $\rho (v,|F_{i} |) = k \Leftrightarrow \textit{dim} (v) = k + \textit{dim} (z) + 1$.

Let v be in $F_{i}$ such that $\textit{dim} (v) = k + \textit{dim} (z) + 1$. Then, using the induction hypothesis, we obtain:

$$\begin{aligned} \begin{array}{ll} &{} \rho (v,|F_{i} |)\\ &{}\quad = \max \left\{ \rho (u,|F_{i} |) \; ; \; u \in \alpha ^{\square } (v) \cap F_{i} \right\} + 1,\\ &{}\quad = \max \left\{ \textit{dim} (u) - \textit{dim} (z) - 1 \; ; \; u \in \alpha ^{\square } _{F_{i}}(v) \right\} + 1,\\ &{}\quad = \max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } _{F_{i}}(v) \right\} - \textit{dim} (z).\\ \end{array} \end{aligned}$$

Since $v \in F_{i}$, $\alpha (v) \cap \beta ^{\square } (z) \subseteq F_{i}$ and then $\alpha ^{\square } (v) \cap \beta ^{\square } (z) \subseteq \alpha ^{\square } _{F_{i}}(v)$, which leads to:

$$\begin{aligned}&\max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } _{F_{i}}(v) \right\} ,\\&\quad \ge \max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } (v) \cap \beta ^{\square } (z)\right\} ,\\&\quad \ge \textit{dim} (v) - 1, \end{aligned}$$

and in the same time,

$$\begin{aligned} \max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } _{F_{i}}(v) \right\} \le \textit{dim} (v) - 1 \end{aligned}$$

because $u \in \alpha ^{\square } (v)$. This way,

$$\begin{aligned} \max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } _{F_{i}}(v) \right\} = \textit{dim} (v) - 1 \end{aligned}$$

and then $\rho (v, |F_{i} |) = \textit{dim} (v) - \textit{dim} (z) - 1 = k$. The direct implication is then proved.

Let us assume now that $v \in F_{i}$ satisfies $\rho (v, |F_{i} |) = k$. By the induction hypothesis, we obtain one more time:

$$\begin{aligned} \rho (v,|F_{i} |) = \max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } _{F_{i}}(v) \right\} - \textit{dim} (z). \end{aligned}$$

In other words, $\max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } _{F_{i}}(v)\right\} = k + \textit{dim} (z)$, and then:

$$\begin{aligned} \max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } (v)\right\} \ge k + \textit{dim} (z). \end{aligned}$$

This way, v is of dimension greater than or equal to $(k+ \textit{dim} (z) + 1)$.

Let us assume now that $\textit{dim} (v) \ge k + \textit{dim} (z) + 2$. Since $v \in F_{i}$, $\alpha (v) \cap \beta ^{\square } (z) \subseteq F_{i}$, and then v covers one or several faces in $F_{i}$ of dimension(s) greater than or equal to $(k+\textit{dim} (z) +1)$, and then

$$\begin{aligned} \max \left\{ \textit{dim} (u) \; ; \; u \in \alpha ^{\square } _{F_{i}}(v)\right\} \ge k + \textit{dim} (z) + 1, \end{aligned}$$

which implies that $\rho (v,|F_{i} |) \ge k + 1$, which is impossible. Then, $\textit{dim} (v) = k + \textit{dim} (z) + 1$. The reciprocal implication is then proved. $\square$

Property 5

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). Then, for all $i \in {\mathcal {I}}$, $F_{i}$ contains at least $(n-\textit{dim} (z))$$(\textit{dim} (z)+1)$-faces.

Proof

Let i be in ${\mathcal {I}}$, then for all $m \in \llbracket \textit{dim} (z)+1, n-1\rrbracket$, $F_{i} \cap {\mathbb {H}}^{n} _{m} \ne \emptyset$ by Property 3. This way, there exists $t \in {\mathbb {H}}^{n} _{n-2} \cap F_{i}$, and because $F_{i}$ is a $(n - 2 - \textit{dim} (z))$-surface, $\left| \beta ^{\square } _{F_{i}}(t)\right|$ is a $((n - \textit{dim} (z) - 2) - \rho (t,|F_{i} |) - 1) = 0$-surface by Property 4. Then, there exists $v,v' \in {\mathbb {H}}^{n} _{n-1} \cap F_{i}$ such that $v \not \in \theta (v')$. However, $\alpha (v) \cap \beta ^{\square } (z)$ and $\alpha (v') \cap \beta ^{\square } (z)$ contain both $(n-\textit{dim} (z)-1)$$(\textit{dim} (z)+1)$-faces (cf. Lemma 17), and $v \ne v'$ implies that ${\mathcal {T}}(v) \ne {\mathcal {T}}(v')$ (cf. Lemma 16), and then there exists at least one face into ${\mathcal {T}}(v')$ which is not among the $(n-\textit{dim} (z)-1)$ faces of ${\mathcal {T}}(v)$. However, ${\mathcal {T}}(v) \cup {\mathcal {T}}(v') \subseteq F_{i}$ (because $F_{i}$ is closed into $\beta ^{\square } (z)$ by Lemma 20). This way, $F_{i} \cap {\mathbb {H}}^{n} _{\textit{dim} (z)+1}$ contains at least $(n-\textit{dim} (z))$ faces. $\square$

Property 7

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). Then,

$$\begin{aligned} \text {Card} \left( {\mathcal {I}} \right) = 2. \end{aligned}$$

Proof

The non-connectivity of $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ implies obviously that $\text {Card} \left( {\mathcal {I}} \right) \ge 2$. Moreover, for each $i \in {\mathcal {I}}$, $F_{i}$ contains $(n-\textit{dim} (z))$$(\textit{dim} (z)+1)$-faces by Property 6, while $\beta ^{\square } (z)$ contains $2(n-\textit{dim} (z))$$(\textit{dim} (z)+1)$-faces, the maximum of components of $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ s then equal to two. $\square$

Property 8

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). For all $i \in {\mathcal {I}}$, ${\mathcal {T}}(F_{i})$ is equal to:

$$\begin{aligned} \left\{ {\mathcal {H}} _n ({\mathcal {Z}} _n (z) + \lambda _m e^m) ; m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \right\} , \end{aligned}$$

with each $\lambda _m$ being exactly one value in $\left\{ - \frac{1}{2}, \frac{1}{2} \right\}$.

Proof

For each $m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$, we have that the two faces ${\mathcal {H}} _n ({\mathcal {Z}} _n (z) - \frac{1}{2} e ^m)$ and ${\mathcal {H}} _n ({\mathcal {Z}} _n (z) + \frac{1}{2} e ^m)$ belong to $\beta ^{\square } (z)$. Then, for all $m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$, we have the possible cases:

• ${\mathcal {H}} _n ({\mathcal {Z}} _n (z) - \frac{1}{2} e ^m) \in {\mathcal {T}}(F_{i})$, and then ${\mathcal {H}} _n ({\mathcal {Z}} _n (z) + \frac{1}{2} e ^m) \not \in {\mathcal {T}}(F_{i})$ (by Property 1),   (P1)

• or ${\mathcal {H}} _n ({\mathcal {Z}} _n (z) + \frac{1}{2} e ^m) \in {\mathcal {T}}(F_{i})$, and then ${\mathcal {H}} _n ({\mathcal {Z}} _n (z) - \frac{1}{2} e ^m) \not \in {\mathcal {T}}(F_{i})$ (for the same reason as before),   (P2)

• or $\{{\mathcal {H}} _n ({\mathcal {Z}} _n (z) - \frac{1}{2} e ^m),{\mathcal {H}} _n ({\mathcal {Z}} _n (z) + \frac{1}{2} e ^m)\} \cap {\mathcal {T}}(F_{i}) = \emptyset$.   (P3)

By (P1) and (P2), we have at most $(n-\textit{dim} (z))$ faces into ${\mathcal {T}}(F_{i}) \subseteq \beta ^{\square } (z)$. If there exists a coordinate $m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$ such that (P3) is true, we will have less than $(n-\textit{dim} (z))$$(\textit{dim} (z)+1)$-faces into ${\mathcal {T}}(F_{i})$, what is impossible by Property 5, then (P3) is never true. We have then either (P1) or (P2) for each $m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$. This way, there exists for each $m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$ exactly one $\lambda _m \in \{-\frac{1}{2},\frac{1}{2} \}$ such that ${\mathcal {H}} _n ({\mathcal {Z}} _n (z) + \lambda _m e ^m) \in {\mathcal {T}}(F_{i})$, then:

$$\begin{aligned} \{{\mathcal {H}} _n ({\mathcal {Z}} _n (z) + \lambda _m e ^m)\}_{m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) } \subseteq {\mathcal {T}}(F_{i}). \end{aligned}$$

Since these two sets have $(n-\textit{dim} (z))$ faces and since they satisfy a inclusion relationship, they are equal, and then:

$$\begin{aligned} \{{\mathcal {H}} _n ({\mathcal {Z}} _n (z) + \lambda _m e^m)\}_{m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) } = {\mathcal {T}}(F_{i}). \end{aligned}$$

$\square$

Property 9

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). For all $i \in {\mathcal {I}}$,

$$\displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t $$

exists in ${\mathbb {H}}^{n} _n \cap \beta ^{\square } (z)$. We will call this face the characteristical face of the component $F_{i}$.

Proof

Let i be a coordinate in ${\mathcal {I}}$. It is sufficient to show that:

$$\begin{aligned} \displaystyle {\bigcap _{t \in {\mathcal {T}}(F_{i})}} \beta (t) \ne \emptyset . \end{aligned}$$

By Property 8, there exists a family of faces

$$\begin{aligned} \{t^m\}_{m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) } = {\mathcal {T}}(F_{i}) \end{aligned}$$

such that for all $m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$, ${\mathcal {Z}} _n (t^m) = {\mathcal {Z}} _n (z) + \lambda _m e ^m$ with $\lambda _m \in \{\frac{1}{2}, - \frac{1}{2} \}$. This way:

When j belongs to $\mathbb {1}\left( {\mathcal {Z}} _n (z)\right)$, we obtain

$$\begin{aligned} t^m_j = z_j \end{aligned}$$

because $t^m$ belongs to $\beta (z)$. Then,

$$\begin{aligned} \displaystyle {\bigcap _{m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }} \beta (t^m_j) = \beta (z_j) \ne \emptyset . \end{aligned}$$

When j belongs to $\frac{1}{2} \left( {\mathcal {Z}} _n (z)\right)$,

$$\begin{aligned}&\displaystyle {\bigcap _{m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }} \beta (t^m_j)\\&\quad = \left( \displaystyle {\bigcap _{m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) \setminus \{j\}}} \beta (t^m_j) \right) \cap \beta (t^j_j),\\&\quad = \beta (z_j) \cap \beta ({\mathcal {H}} ({\mathcal {Z}} (z_j) + \lambda _j)),\\&\quad = \left\{ \begin{array}{c} \{{\mathcal {Z}} (z_j) + \frac{1}{2} \},\\ \{{\mathcal {Z}} (z_j) - \frac{1}{2},{\mathcal {Z}} (z_j) + \frac{1}{2} \},\\ \{{\mathcal {Z}} (z_j) + \frac{1}{2},{\mathcal {Z}} (z_j) + \frac{3}{2}\} \end{array} \right\} ,\\&\quad \cap \; \{\{{\mathcal {Z}} (z_j)+\lambda _j,{\mathcal {Z}} (z_j)+\lambda _j +1\}\},\\&\quad = \{\{{\mathcal {Z}} (z_j)+\lambda _j,{\mathcal {Z}} (z_j)+\lambda _j +1\}\},\\&\quad = \beta (t^j_j),\\&\quad \ne \emptyset . \end{aligned}$$

Then, each term $\displaystyle {\bigcap _{m \in \frac{1}{2} \left( {\mathcal {Z}} _n (z)\right) }} \beta (t^m_j)$ is non-empty, and then $\displaystyle {\bigcap _{t \in {\mathcal {T}}(F_{i})}} \beta (t) \ne \emptyset$.

The fact that $\displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t$ belongs to ${\mathbb {H}}^{n} _n$ is due to the fact that $\text {Card} \left( {\mathcal {T}}(F_{i}) \right) = (n - \textit{dim} (z))$ by Property 6 and to the fact that the faces of ${\mathcal {T}}(F_{i})$ are different two by two. $\square$

Property 10

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). For each $i \in {\mathcal {I}}$:

$$\begin{aligned} F_{i} \subseteq \alpha \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right) . \end{aligned}$$

Proof

Let u be in $F_{i}$, $u \in \beta ^{\square } (z)$, and then by Lemma 15, $u = \displaystyle {\bigvee _{t \in {\mathcal {T}}(u)}} t$. Since $F_{i}$ is closed into $\beta ^{\square } (z)$, $\alpha (u) \cap \beta ^{\square } (z) \subseteq F_{i}$ and then:

$$\begin{aligned} {\mathcal {T}}(u)&= \alpha (u) \cap \beta ^{\square } (z) \cap {\mathbb {H}}^{n} _{\textit{dim} (z)+1},\\&\quad \subseteq F_{i} \cap {\mathbb {H}}^{n} _{\textit{dim} (z)+1},\\&\quad = {\mathcal {T}}(F_{i}). \end{aligned}$$

Now, by Property 1, no face of $F_{i}$ is opposite to each other, and then $\displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t$ exists by Property 9. Since we have ${\mathcal {T}}(u) \subseteq {\mathcal {T}}(F_{i})$, we can then write:

$$\begin{aligned} \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t&= \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(u)}} t \right) \; \bigvee \; \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i}) \setminus {\mathcal {T}}(u)}} t \right) \\&= u \; \bigvee \; \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i}) \setminus {\mathcal {T}}(u)}} t \right) .\\ \end{aligned}$$

By definition of the operator $\vee$:

$$\begin{aligned} \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t = \inf \left( \beta \left( u \right) \cap \beta \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i}) \setminus {\mathcal {T}}(u)}} t \right) \right) , \end{aligned}$$

which implies that $\displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \in \beta (u)$ and then

$$\begin{aligned} u \in \alpha \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right) , \end{aligned}$$

from which we can deduce that

$$\begin{aligned} F_{i} \subseteq \alpha \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right) . \end{aligned}$$

$\square$

Property 11

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). Then,

$$\begin{aligned} F_{i} = \alpha ^{\square } \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right) \cap \beta ^{\square } (z). \end{aligned}$$

Proof

Let i be a coordinate in ${\mathcal {I}}$. Then, $F_{i} \subseteq \beta ^{\square } (z)$, which implies by Property 10 that $F_{i} \subseteq \alpha \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right) \cap \beta ^{\square } (z)$. Nevertheless, $\left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right)$ belongs to ${\mathbb {H}}^{n} _n$ by Property 9. Also, $F_{i} \cap {\mathbb {H}}^{n} _n = \emptyset$, and then we have $F_{i} \subseteq \alpha ^{\square } \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right) \cap \beta ^{\square } (z)$. Since $\displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \in \beta ^{\square } (z)$ (by transitivity of $\beta$), then

$$\begin{aligned} \alpha ^{\square } \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right) \cap \beta ^{\square } (z) \end{aligned}$$

is a $(n-\textit{dim} (z)-2)$-surface by Proposition 7. This is also the case concerning $F_{i}$ by hypothesis. This way,

$$\begin{aligned} F_{i} = \alpha ^{\square } \left( \displaystyle {\bigvee _{t \in {\mathcal {T}}(F_{i})}} t \right) \cap \beta ^{\square } (z) \end{aligned}$$

by Proposition 6. $\square$

Property 12

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). Then, ${\mathcal {Z}} _n ({\mathfrak {a}})$ and ${\mathcal {Z}} _n ({\mathfrak {b}})$ are $(n - \textit{dim} (z))$-antagonist in ${\mathbb {Z}} ^n$.

Proof

Since by Property 11, we have $F_{1} = \alpha ^{\square } ({\mathfrak {a}}) \cap \beta ^{\square } (z)$ and $F_{2} = \alpha ^{\square } ({\mathfrak {b}}) \cap \beta ^{\square } (z)$, and since they are disjoint, then we have:

$$\begin{aligned} \alpha ^{\square } ({\mathfrak {a}}) \cap \alpha ^{\square } ({\mathfrak {b}}) \cap \beta ^{\square } (z) = \emptyset . \end{aligned}$$

By Lemma 18, ${\mathcal {Z}} _n ({\mathfrak {a}})$ and ${\mathcal {Z}} _n ({\mathfrak {b}})$ are then $(n - \textit{dim} (z))$-antagonists. $\square$

Property 13

We assume that $n \ge 3$ and that there exists $z \in {\mathfrak {N}}$ such that $\textit{dim} (z) \le (n-3)$ and that $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is not connected (Hypothesis 1). We assume also that each component of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ is a $(n-\textit{dim} (z)-2)$-surface (Hypothesis 2). Then, X contains a critical configuration of dimension $(n - \textit{dim} (z))$.

Proof

Let assume that Hypothesis 1 is true, that is, $|\beta ^{\square }_{{\mathfrak {N}}} (z)|$ is not connected. At the same time, we assume that the components $|F_{i} |$ of $\left| \beta ^{\square }_{{\mathfrak {N}}} (z)\right|$ are $(n - \textit{dim} (z) - 2)$-surfaces. Then, the two characteristical faces ${\mathfrak {a}}$ and ${\mathfrak {b}}$ exist by Property 12.

Now, let u, v be in $S(z) \setminus \{{\mathcal {Z}} _n ({\mathfrak {a}}),{\mathcal {Z}} _n ({\mathfrak {b}})\}$. Since this set is 2n-connected for $(n-\textit{dim} (z)) \ge 3$ by Lemma 19, there exists a 2n-path $\pi = (p^0 = u, \dots ,p^l = v)$ joining u and v into $S(z) \setminus \{{\mathcal {Z}} _n ({\mathfrak {a}}),{\mathcal {Z}} _n ({\mathfrak {b}})\}$ with $l \ge 1$. We can deduce from it a path $\pi '$ into ${\mathbb {H}}^{n}$ such that:

$$\begin{aligned} \pi ' = \left( \begin{array}{l} {\mathcal {H}} _n (p^0),\\ {\mathcal {H}} _n (p^0) \wedge {\mathcal {H}} _n (p^1),\\ {\mathcal {H}} _n (p^1),\\ \dots ,\\ {\mathcal {H}} _n (p^{l-1}),\\ {\mathcal {H}} _n (p^{l-1}) \wedge {\mathcal {H}} _n (p^l),\\ {\mathcal {H}} _n (p^l) \end{array} \right) . \end{aligned}$$

For all m into $\llbracket 0, l-1 \rrbracket$, we have

$$\begin{aligned} {\mathcal {H}} _n (p^{m-1}) \wedge {\mathcal {H}} _n (p^m) \in {\mathbb {H}}^{n} _{n-1} \end{aligned}$$

since $p^{m-1}$ and $p^m$ are 2n-neighbors into ${\mathbb {Z}} ^n$.

Let us assume now that there exists a value $m \in \llbracket 0, l-1 \rrbracket$ such that ${\mathcal {H}} _n (p^{m-1}) \wedge {\mathcal {H}} _n (p^m) \in {\mathfrak {N}}$, then ${\mathcal {H}} _n (p^{m-1}) \wedge {\mathcal {H}} _n (p^m) \in \beta ^{\square }_{{\mathfrak {N}}} (z)$ and then:

• either ${\mathcal {H}} _n (p^{m-1}) \wedge {\mathcal {H}} _n (p^m) \in \alpha ^{\square } ({\mathfrak {a}}) \cap \beta ^{\square } (z)$ and then

  $$\begin{aligned} \beta ^{\square } ({\mathcal {H}} _n (p^{m-1}) \wedge {\mathcal {H}} _n (p^m)) = \{{\mathcal {H}} _n (p^{m-1}),{\mathcal {H}} _n (p^m)\} \end{aligned}$$

  contains ${\mathfrak {a}}$, which is impossible by definition of $\pi$,

• or ${\mathcal {H}} _n (p^{m-1}) \wedge {\mathcal {H}} _n (p^m) \in \alpha ^{\square } ({\mathfrak {b}}) \cap \beta ^{\square } (z)$ and then

  $$\begin{aligned} \beta ^{\square } ({\mathcal {H}} _n (p^{m-1}) \wedge {\mathcal {H}} _n (p^m)) = \{{\mathcal {H}} _n (p^{m-1}),{\mathcal {H}} _n (p^m)\} \end{aligned}$$

  contains ${\mathfrak {b}}$, which is impossible for the same reason.

This way, ${\mathcal {H}} _n (p^{m-1}) \wedge {\mathcal {H}} _n (p^m) \not \in {\mathfrak {N}}$, and then either all the points of $\pi$ belong to X or they all belong to Y.

In other words, either $S(z) \setminus \{{\mathcal {Z}} _n ({\mathfrak {a}}),{\mathcal {Z}} _n ({\mathfrak {b}})\} \subseteq X$ or $S(z) \setminus \{{\mathcal {Z}} _n ({\mathfrak {a}}),{\mathcal {Z}} _n ({\mathfrak {b}})\} \subseteq Y$.

Now, let $v^{{\mathfrak {a}}}$ be a 2n-neighbor of ${\mathcal {Z}} _n ({\mathfrak {a}})$ in the set $S(z) \setminus \{{\mathcal {Z}} _n ({\mathfrak {a}}),{\mathcal {Z}} _n ({\mathfrak {b}})\}$ and let $v^{{\mathfrak {b}}}$ be a 2n-neighbor of ${\mathcal {Z}} _n ({\mathfrak {b}})$ into $S(z) \setminus \{{\mathcal {Z}} _n ({\mathfrak {a}}),{\mathcal {Z}} _n ({\mathfrak {b}})\}$. Then,

$$\begin{aligned} {\mathcal {H}} _n (v^{{\mathfrak {a}}}) \wedge {\mathfrak {a}} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {H}} _n (v^{{\mathfrak {b}}}) \wedge {\mathfrak {b}} \end{aligned}$$

belong to ${\mathfrak {N}}$ (because they belong to $\beta ^{\square }_{{\mathfrak {N}}} (z)$) and then we have the two possible configurations:

• either ${\mathcal {Z}} _n ({\mathfrak {a}}) \in X$, then $v^{{\mathfrak {a}}} \in Y$, from which we deduce that $v^{{\mathfrak {b}}} \in Y$, and then ${\mathcal {Z}} _n ({\mathfrak {b}}) \in X$ (and X contains a primary critical configuration),

• or ${\mathcal {Z}} _n ({\mathfrak {a}}) \in Y$, then $v^{{\mathfrak {a}}} \in X$, from which we deduce that $v^{{\mathfrak {b}}} \in X$, and then ${\mathcal {Z}} _n ({\mathfrak {b}}) \in Y$ (and X contains a secondary critical configuration).

By Property 12, this critical configuration is of dimension $(n-\textit{dim} (z))$.

In brief, either we have $X \cap S(z) = \{{\mathcal {Z}} _n ({\mathfrak {a}}),{\mathcal {Z}} _n ({\mathfrak {b}})\}$ or we have $X \cap S(z) = S(z) \setminus \{{\mathcal {Z}} _n ({\mathfrak {a}}),{\mathcal {Z}} _n ({\mathfrak {b}})\}$ where both are critical configurations of dimension $(n-\textit{dim} (z))$. $\square$