Equivalence Classes of Full-Dimensional $0/1$-Polytopes with Many Vertices

Abstract

Let $Q_n$ denote the $n$-dimensional hypercube with vertex set $V_n=\{0,1\}^n$. A $0/1$-polytope of $Q_n$ is the convex hull of a subset of $V_n$. This paper is concerned with the enumeration of equivalence classes of full-dimensional $0/1$-polytopes under the symmetries of the hypercube. With the aid of a computer program, Aichholzer obtained the number of equivalence classes of full-dimensional $0/1$-polytopes of $Q_4$ and $Q_5$ with any given number of vertices and those of $Q_6$ up to 12 vertices. Let $F_n(k)$ denote the number of equivalence classes of full-dimensional 0/1-polytopes of $Q_n$ with $k$ vertices. We present a method to compute $F_n(k)$ for $k>2^{n-2}$. Let $A_n(k)$ denote the number of equivalence classes of $0/1$-polytopes of $Q_n$ with $k$ vertices, and let $H_n(k)$ denote the number of equivalence classes of $0/1$-polytopes of $Q_n$ with $k$ vertices that are not full-dimensional. So we have $A_n(k)= F_n(k)+ H_n(k)$. It is known that $A_n(k)$ can be computed by using the cycle index of the hyperoctahedral group. We show that for $k>2^{n-2}$, $H_n(k)$ can be determined by the number of equivalence classes of $0/1$-polytopes with $k$ vertices that are contained in every hyperplane spanned by a subset of $V_n$. We also find a way to compute $H_n(k)$ when $k$ is close to $2^{n-2}$. For the case of $Q_6$, we can compute $F_6(k)$ for $k>12$. Together with the computation of Aichholzer, we reach a complete solution to the enumeration of equivalence classes of full-dimensional 0/1-polytopes of $Q_6$.

1 Introduction

Let $Q_n$ denote the $n$-dimensional hypercube with vertex set $V_n=\{0,1\}^n$. A $0/1$-polytope of $Q_n$ is defined to be the convex hull of a subset of $V_n$. The study of $0/1$-polytopes has received much attention, see, for example [6, 7, 11–13, 15, 18, 19].

In this paper, we are concerned with the problem of determining the number of equivalence classes of $n$-dimensional $0/1$-polytopes of $Q_n$ under the symmetries of $Q_n$, which has been considered as a difficult problem, see Ziegler [18]. It is also listed by Zong [19, Problem 5.1] as one of the fundamental problems concerning $0/1$-polytopes.

An $n$-dimensional $0/1$-polytope of $Q_n$ is also called a full-dimensional $0/1$-polytope of $Q_n$. Two $0/1$-polytopes are said to be equivalent if one can be transformed to the other by a symmetry of $Q_n$. Such an equivalence relation is called the $0/1$-equivalence relation. For example, Fig. 1 gives the representatives of $0/1$-equivalence classes of $Q_2$, among which (d) and (e) are full-dimensional.

Fig. 1

Representatives of 0/1-equivalence classes of $Q_2$

As the first nontrivial case, full-dimensional 0/1-equivalence classes of $Q_4$ were counted by Below, see Ziegler [18]. With the aid of a computer program, Aichholzer [1] completed the enumeration of full-dimensional $0/1$-equivalence classes of $Q_5$, and those of $Q_6$ up to $12$ vertices, see also Aichholzer [3] and Ziegler [18]. The 5-dimensional hypercube $Q_5$ has been considered as the last case that one can hope for a complete solution to the enumeration of full-dimensional $0/1$-equivalence classes.

Let $F_n(k)$ denote the number of full-dimensional $0/1$-equivalence classes of $Q_n$. The objective of this paper is to present a method to compute $F_n(k)$ for $k>2^{n-2}$. We also find a way to compute $F_n(k)$ when $k$ is close to $2^{n-2}$. Using our approach, we can determine $F_6(k)$ for $k> 12$. Combining the computation of Aichholzer [1], we reach a complete solution for the case of $Q_6$.

To describe our approach, let $A_n(k)$ denote the number of $0/1$-equivalence classes of $Q_n$ with $k$ vertices, and let $H_n(k)$ denote the number of $0/1$-equivalence classes of $Q_n$ with $k$ vertices that are not full-dimensional. So we have

$$\begin{aligned} A_n(k)=F_n(k)+H_n(k). \end{aligned}$$

(1.1)

It is clear that $F_n(k)=0$ for $1\le k\le n$ since any full-dimensional $0/1$-polytope of $Q_n$ has at least $n+1$ vertices. As will be seen in Sect. 2, the values $A_n(k)$ for any $k$ can be computed from the cycle index of the hyperoctahedral group $B_n$. Hence $F_n(k)$ can be determined by $H_n(k)$.

To compute $H_n(k)$, we need a relation between the dimension of a $0/1$-polytope and the number of vertices. Let $P$ be a $0/1$-polytope of $Q_n$, and let ${\mathrm {dim}}(P)$ denote the dimension of $P$. It is known that $P$ is affinely equivalent to a full-dimensional $0/1$-polytope of $Q_{d}$ for some $d\le n$, see Ziegler [18]. Thus we have the following consequence.

Theorem 1.1

Let $P$ be a $0/1$-polytope of $Q_n$ with more than $2^{m}$ vertices, where $1\le m<n$. Then we have

$$\begin{aligned} \mathrm {dim}(P)\ge m+1. \end{aligned}$$

From Theorem 1.1, we see that if a $0/1$-polytope $P$ of $Q_n$ has more than $2^{n-1}$ vertices, then $P$ has dimension $n$. Thus, for $k>2^{n-1}$, we have $F_n(k)=A_n(k)$.

Based on Theorem 1.1, we show that the computation of $H_n(k)$ for $2^{n-2}<k\le 2^{n-1}$ can be carried out by determining the number of equivalence classes of $0/1$-polytopes with $k$ vertices that are contained in every hyperplane spanned by vertices of $Q_n$. When $2^{n-2}<k\le 2^{n-1}$, we can apply Pólya’s theorem to count equivalence classes of $0/1$-polytopes with $k$ vertices that are contained in a hyperplane spanned by vertices of $Q_n$. In particular, when $n=6$, we obtain $F_6(k)$ for $16<k\le 32$.

We also find a way to compute $H_n(k)$ when $k$ is close to $2^{n-2}$. In particular, when $n=6$, we obtain $F_6(k)$ for $13\le k \le 16$.

This paper is organized as follows. In Sect. 2, we recall a method introduced by Chen [9] to determine the cycle structure of a symmetry $w$ in the hyperoctahedral group $B_n$ in terms of the number of vertices of $Q_n$ fixed by $w$. Sections 3–6 are devoted to the computation of $H_n(k)$ for $2^{n-2}<k\le 2^{n-1}$. In Sect. 7, we provide a way to compute $H_n(k)$ when $k$ is close to $2^{n-2}$. This enables us to determine $H_n(k)$ for $n=6$ and $13\le k\le 16$.

2 The Cycle Index of the Hyperoctahedral Group

The group of symmetries of $Q_n$ is known as the hyperoctahedral group $B_n$. In this section, we give an overview of a method introduced by Chen [9] to compute the cycle index of $B_n$, which will be used in the determination of the cycle index of the subgroup consisting of symmetries that fix a hyperplane spanned by vertices of $Q_n$.

We proceed with a brief review of the cycle index of a finite group acting on a finite set, see, for example, Brualdi [8]. Let $G$ be a finite group that acts on a finite set $X$. Then each element $g\in G$ induces a permutation on $X$. The cycle type of a permutation is defined to be a multiset $\{1^{k_1},2^{k_2},\ldots \}$, where $k_i$ is the number of cycles of length $i$ that appear in the cycle decomposition of the permutation. For $g\in G$, let $c(g)$ denote the cycle type of the permutation on $X$ induced by $g$. Let $z=(z_1,z_2,\ldots )$ be a sequence of indeterminants, and let

$$\begin{aligned} z^{c(g)}=z_1^{k_1}z_2^{k_2}\cdots . \end{aligned}$$

The cycle index of $G$ is defined as

$$\begin{aligned} Z_G(z)=Z_G(z_1,z_2,\ldots )=\frac{1}{|G|}\sum _{g\in G}z^{c(g)}. \end{aligned}$$

(2.1)

Pólya’s enumeration theorem shows that the cycle index in (2.1) can be applied to count nonisomorphic colorings of $X$ by using a given number of colors. To be more specific, let us color the elements of $X$ by using $m$ colors, say $c_1,c_2,\ldots ,c_m$. Let $C_G(u_1,\ldots ,u_m)$ be the polynomial obtained from the cycle index $Z_G(z)$ by substituting $z_i$ with $u_1^i+\cdots +u_m^i$. Pólya’s enumeration theorem states that the number of nonisomorphic colorings of $X$ by using the $m$ colors $c_1, \ldots ,c_m$ such that $a_i$ elements of $X$ receive the color $c_i$ equals

$$\begin{aligned} \left[ u_1^{a_1}\cdots u_m^{a_m}\right] C_G(u_1,\ldots ,u_m), \end{aligned}$$

where $\left[ u_1^{a_1}\cdots u_m^{a_m}\right] C_G(u_1,\ldots ,u_m)$ is the coefficient of $u_1^{a_1}\cdots u_m^{a_m}$ in $C_G(u_1,\ldots ,u_m)$.

For a coloring of $Q_n$ with two colors, say, black and white, the black vertices can be considered as vertices of a $0/1$-polytope of $Q_n$. This establishes a one-to-one correspondence between equivalence classes of colorings and $0/1$-equivalence classes of $Q_n$. Let $Z_n(z)$ denote the cycle index of $B_n$ acting on the vertex set $V_n$, and let $C_n(u_1,u_2)$ be the polynomial obtained from $Z_n(z)$ by substituting $z_i$ with $u_1^i+u_2^i$. By Pólya’s theorem, we have

$$\begin{aligned} A_n(k)=\big [u_1^ku_2^{2^n-k}\big ]C_n(u_1,u_2). \end{aligned}$$

(2.2)

The computation of $Z_n(z)$ has been studied by Pólya [16] and Harrison and High [14]. Explicit expressions of $Z_n(z)$ for $n\le 6$ are given by Aguila [5], which are listed below.

$$\begin{aligned} Z_1(z)\!&= \!z_1,\\ Z_2(z)\!&= \!\frac{1}{8} \Big (z_1^4+2z_1^2z_2+3z_2^2+2z_4\Big ),\\ Z_3(z)\!&= \!\frac{1}{48} \Big (z_1^8+6z_1^4z_2^2+13z_2^4+8z_1^2z_3^2+12z_4^2+8z_2z_6\Big ),\\ Z_4(z)\!&= \!\frac{1}{384} \left( \begin{array}{l} z_1^{16}+12z_1^8z_2^4+12z_1^4z_2^6+51z_2^8+48z_8^2\\ +\,48z_1^2z_2z_4^3+84z_4^4+96z_2^2z_6^2+32z_1^4z_3^4 \end{array}\right) ,\\ Z_5(z)\!&= \!\frac{1}{3840} \left( \begin{array}{l} z_1^{32}+20z_1^{16}z_2^8+ 60z_1^8z_2^{12}+231z_2^{16}+ 80z_1^8z_3^8+240z_1^4z_2^2z_4^6\\ +\,240z_2^4z_4^6+520z_4^8+384z_1^2z_5^6+160z_1^4z_2^2z_3^4z_6^2+ 720z_2^4z_6^4\\ +\,480z_8^4+384z_2z_{10}^3+320z_4^2z_{12}^2 \end{array}\right) ,\\ Z_6(z)\!&= \!\frac{1}{46080} \left( \begin{array}{l} z_1^{64}\!+\!30z_1^{32}z_2^{16}\!+\!180z_1^{16}z_2^{24}\!+\!120z_1^8z_2^{28}\!+\! 1053z_2^{32}\!+\!160z_1^{16}z_3^{16}\\ \!+\,640z_1^4z_3^{20}\!+\!720z_1^8z_2^4z_4^{12}\!+\! 1440z_1^4z_2^6z_4^{12}\!+\!2160z_2^8z_4^{12}\!+\!4920z_4^{16}\\ \!+\,2304z_1^4z_5^{12}\!+\! 960z_1^8z_2^4z_3^8z_6^4\!+\!5280z_2^8z_6^8\!+\!3840z_1^2z_2z_3^2z_6^9\!+\!5760z_8^8\\ \!+\,1920z_2^2z_6^{10}+6912z_2^2z_{10}^6+3840z_4^4z_{12}^4+ 3840z_4z_{12}^5 \end{array}\right) \!. \end{aligned}$$

For $k>2^{n-1}$, we have shown that $F_n(k)=A_n(k)$. Thus, by (2.2) we obtain that for $k>2^{n-1}$,

$$\begin{aligned} F_n(k)=\big [u_1^ku_2^{2^n-k}\big ]C_n(u_1,u_2). \end{aligned}$$

For $n=4,5$ and $6$, the values of $F_n(k)$ for $k>2^{n-1}$ are given in Tables 1, 2 and 3.

Table 1 $F_4(k)$ for $k>8$

Table 2 $F_5(k)$ for $k>16$

Table 3 $F_6(k)$ for $k>32$

We next recall the method of Chen [9] for computing the cycle index of $B_n$. A symmetry of $Q_n$ can be represented as a signed permutation on $\{1,2,\ldots ,n\}$, which is a permutation on $\{1,2,\ldots ,n\}$ with a plus or a minus sign attached to each element. Following the notation in Chen and Stanley [10] or Chen [9], we may write a signed permutation as the form of the cycle decomposition and ignore the plus signs. For example, $(\overline{2}4\overline{5})(3)(1\overline{6})$ represents a signed permutation, where $(245)(3)(16)$ is its underlying permutation. The action of a signed permutation $w\in B_n$ on the vertices of $Q_n$ is defined as follows. For a vertex $(x_1, x_2,\ldots , x_n)$ of $Q_n$, we define $w(x_1,x_2,\ldots ,x_n)$ to be the vertex $(y_1,y_2,\ldots ,y_n)$ of $Q_n$ as given by

$$\begin{aligned} y_i= \left\{ \begin{array}{ll} x_{\pi (i)} \qquad &{}\hbox {if}\; i\hbox { is associated with a plus sign},\\ 1-x_{\pi (i)} \qquad &{}\hbox {if}\; i\hbox { is associated with a minus sign}, \end{array}\right. \end{aligned}$$

(2.3)

where $\pi$ is the underlying permutation of $w$.

We end this section with the following formula of Chen [9], which will be used in Sect. 5 to compute the cycle structure of a symmetry that fixes a hyperplane spanned by vertices of $Q_n$.

Let $n$ be a positive integer, and let $p_1^{n_1}\ldots p_r^{n_r}$ be the prime factorization of $n$. Let $\mu (n)$ be the classical number-theoretic Möbius function, that is, $\mu (n)=(-1)^r$ if $n_1=\cdots =n_r=1$, and $\mu (n)=0$ otherwise.

Theorem 2.1

Let $G$ be a group that acts on a finite set $X$. For any $g\in G$, the number of $i$-cycles of the permutation on $X$ induced by $g$ is given by

$$\begin{aligned} \frac{1}{i}\sum _{j|i}\mu (i/j)\psi (g^j), \end{aligned}$$

where $\psi (g^j)$ is the number of fixed points of $g^j$ acting on $X$.

3 $H_n(k)$ for $2^{n-2}<k\le 2^{n-1}$

Recall that $H_n(k)$ is the number of $0/1$-equivalence classes of $Q_n$ with $k$ vertices that are not full-dimensional. In this section, we show that for $2^{n-2}<k\le 2^{n-1}$, the number $H_n(k)$ is determined by the number of equivalence classes of $0/1$-polytopes with $k$ vertices that are contained in every hyperplane spanned by vertices of $Q_n$. For this reason, it is necessary to consider all possible hyperplanes spanned by vertices of $Q_n$.

A hyperplane spanned by vertices of $Q_n$ is also called a spanned hyperplane of $Q_n$. In other words, a spanned hyperplane of $Q_n$ is a hyperplane in $\mathbb {R}^n$ such that the affine space spanned by the vertices of $Q_n$ contained in this hyperplane is of dimension $n-1$. Let

$$\begin{aligned} H:a_1x_1+a_2x_2+\cdots +a_nx_n=b \end{aligned}$$

be a spanned hyperplane of $Q_n$, where $a_1,\ldots ,a_n$ and $b$ are integers. For $n\le 8$, all spanned hyperplanes of $Q_n$ have been found by Aichholzer and Aurenhammer [4].

As will be seen, in order to compute $H_n(k)$ for $2^{n-2}<k\le 2^{n-1}$, we need to consider equivalence classes of spanned hyperplanes of $Q_n$ under the symmetries of $Q_n$. Note that the symmetries of $Q_n$ can be expressed by permuting the coordinates and changing $x_i$ to $1-x_i$ for some indices $i$. Therefore, for each equivalence class of spanned hyperplanes of $Q_n$, we can choose a representative of the form

$$\begin{aligned} a_1x_1+a_2x_2+\cdots +a_tx_t=b, \end{aligned}$$

(3.1)

where $t\le n$ and $0<a_1\le a_2\le \cdots \le a_t$.

A complete list of spanned hyperplanes of $Q_n$ for $n\le 6$ can be found in Aichholzer [2]. The following hyperplanes are representatives of equivalence classes of spanned hyperplanes of $Q_4$:

$$\begin{aligned}&x_1=0,\\&x_1+x_2=1,\\&x_1+x_2+x_3=1,\\&x_1+x_2+x_3+x_4=\hbox {1 or 2},\\&x_1+x_2+x_3+2x_4=2. \end{aligned}$$

In addition to the above hyperplanes, which can also be viewed as spanned hyperplanes of $Q_5$, we have the following representatives of equivalence classes of spanned hyperplanes of $Q_5$:

$$\begin{aligned}&x_1+x_2+x_3+x_4+x_5=\hbox {1 or 2},\\&x_1+x_2+x_3+x_4+2x_5=\hbox {2 or 3},\\&x_1+x_2+x_3+2x_4+2x_5=\hbox {2 or 3},\\&x_1+x_2+2x_3+2x_4+2x_5=\hbox {3 or 4},\\&x_1+x_2+x_3+x_4+3x_5=3,\\&x_1+x_2+x_3+2x_4+3x_5=3,\\&x_1+x_2+2x_3+2x_4+3x_5=4. \end{aligned}$$

When $n=6$, for the purpose of computing $F_6(k)$ for $16<k\le 32$, we need the representatives of equivalence classes of spanned hyperplanes of $Q_6$ containing more than 16 vertices. There are 6 such representatives:

$$\begin{aligned}&x_1=0,\\&x_1+x_2=1,\\&x_1+x_2+x_3=1,\\&x_1+x_2+x_3+x_4=2,\\&x_1+x_2+x_3+x_4+x_5=2,\\&x_1+x_2+x_3+x_4+x_5+x_6=3. \end{aligned}$$

Note that two equivalent spanned hyperplanes of $Q_n$ contain the same number of vertices of $Q_n$ because the symmetry of $Q_n$ preserves the number of vertices. So we may say that an equivalence class of spanned hyperplanes of $Q_n$ contains $k$ vertices, by which we mean that every spanned hyperplane in this class contains $k$ vertices of $Q_n$.

To state the main result of this section, we need to define an equivalence relation on $0/1$-polytopes contained in a set of points in $\mathbb {R}^n$. Given a set $S\subset \mathbb {R}^n$, consider the set of $0/1$-polytopes of $Q_n$ that are contained in $S$. Restricting the $0/1$-equivalence relation to this set induces an equivalence relation. More precisely, two $0/1$-polytopes in the set of $0/1$-polytopes of $Q_n$ contained in $S$ are equivalent if one can be transformed to the other by a symmetry of $Q_n$. Such an equivalence class is called a partial $0/1$-equivalence class of $S$. Denote by $\mathcal {P}(S,k)$ the set of partial $0/1$-equivalence classes of $S$ with $k$ vertices. The cardinality of $\mathcal {P}(S,k)$ is denoted by $N_S(k)$.

Let $h(n,k)$ denote the number of equivalence classes of spanned hyperplanes of $Q_n$ that contain at least $k$ vertices. Assume that $H_1,H_1,\ldots ,H_{h(n,k)}$ are the representatives of equivalence classes of spanned hyperplanes of $Q_n$ containing at least $k$ vertices. We use $\mathcal {H}_n(k)$ to denote the set of $0/1$-equivalence classes of $Q_n$ with $k$ vertices that are not full-dimensional. We shall define a map, denoted $\Phi$, from the (disjoint) union of $\mathcal {P}(H_i,k)$, where $1\le i\le h(n,k)$, to $\mathcal {H}_n(k)$. Given a partial $0/1$-equivalence class $\mathcal {P}\in \mathcal {P}(H_i,k)$, we define $\varPhi (\mathcal {P})$ to be the unique $0/1$-equivalence class in $\mathcal {H}_n(k)$ containing $\mathcal {P}$. Then we have the following theorem.

Theorem 3.1

For $2^{n-2}<k\le 2^{n-1}$, the map $\varPhi$ is a bijection.

Proof

We first show that $\varPhi$ is injective. Let $\mathcal {P}_1$ and $\mathcal {P}_2$ be two distinct partial $0/1$-equivalence classes with $k$ vertices, which are contained in the spanned hyperplanes $H_i$ and $H_j$ of $Q_n$, respectively. Let $P_1$ be a $0/1$-polytope in $\mathcal {P}_1$, and $P_2$ be a $0/1$-polytope in $\mathcal {P}_2$. To prove that $\varPhi$ is an injection, it suffices to show that $P_1$ and $P_2$ are not equivalent. This is clear when $i=j$. We now consider the case $i\ne j$. Suppose to the contrary that $P_1$ and $P_2$ are equivalent. So there exists a symmetry $w\in B_n$ such that $w(P_1)=P_2$. Since $2^{n-2}<k\le 2^{n-1}$, by Theorem 1.1 we see that $P_1$ and $P_2$ are of dimension $n-1$. For a spanned hyperplane $H$ of $Q_n$, we use $w(H)$ to denote the hyperplane obtained from $H$ under the action of $w$. So we have $w(H_i)=H_j$, contradicting the fact that the spanned hyperplanes $H_i$ and $H_j$ are not equivalent. Consequently, the 0/1-polytopes $P_1$ and $P_2$ are not equivalent.

It remains to show that $\varPhi$ is surjective. For any $\mathcal {C}\in \mathcal {H}_n(k)$, we aim to find a partial $0/1$-equivalence class $\mathcal {P}$ such that $\varPhi (\mathcal {P})=\mathcal {C}$. Let $P$ be any $0/1$-polytope in $\mathcal {C}$. Since $P$ is not full-dimensional, there exists a spanned hyperplane $H$ of $Q_n$ such that $P$ is contained in $H$. It follows that $H$ contains at leat $k$ vertices. Thus there exists a representative $H_j$ ($1\le j\le h(n,k)$) such that $H$ is in the equivalence class of $H_j$. Assume that $w(H)=H_j$ for some $w\in B_n$. So $w(P)$ is contained in $H_j$. Let $\mathcal {P}$ be the partial $0/1$-equivalence class of $H_j$ containing $w(P)$. Clearly, we have $\varPhi (\mathcal {P})=\mathcal {C}$. This completes the proof. $\square$

It should also be noted that in the proof of Theorem 3.1, the condition $2^{n-2}<k\le 2^{n-1}$ is required. When $k\le 2^{n-2}$, the map $\varPhi$ is not necessarily an injection while is always a surjection. For a $0/1$-polytope $P$ with $k\le 2^{n-2}$ vertices contained in a spanned hyperplane of $Q_n$, it is not always true that $\mathrm {dim}(P)=n-1$. So there may exist equivalent $0/1$-polytopes $P$ and $P'$ with $k$ vertices and nonequivalent spanned hyperplanes $H$ and $H'$ such that $P$ is contained in $H$ and $P'$ is contained in $H'$. If this is the case, then $\varPhi$ maps these two partial 0/1-equivalence classes containing $P$ and $P'$ to the same 0/1-equivalence class in $\mathcal {H}_n(k)$.

As a consequence of Theorem 3.1, we obtain the following formula.

Corollary 3.2

For $2^{n-2}< k\le 2^{n-1}$,

$$\begin{aligned} H_n(k)=\sum _{i=1}^{h(n,k)}N_{H_i}(k). \end{aligned}$$

(3.2)

By Corollary 3.2, the computation of $H_n(k)$ for $2^{n-2}<k\le 2^{n-1}$ is carried out by determining the number of partial $0/1$-equivalence classes of every spanned hyperplane of $Q_n$. We shall explain how to compute the latter in the rest of this section.

For $2^{n-2}<k\le 2^{n-1}$, let $H$ be a spanned hyperplane of $Q_n$ containing at least $k$ vertices. Let $P$ and $P'$ be two distinct 0/1-polytopes of $Q_n$ with $k$ vertices that are contained in $H$. Assume that $P$ and $P'$ belong to the same partial $0/1$-equivalence class of $H$. Then there exists a symmetry $w\in B_n$ such that $w(P)=P'$. By Theorem 1.1, both $P$ and $P'$ have dimension $n-1$. Hence we have $w(H)=H$.

Let

$$\begin{aligned} F(H)=\{w\in B_n\,|\, w(H)=H\} \end{aligned}$$

be the stabilizer subgroup of $H$, namely, the subgroup of $B_n$ that fixes $H$. By the above argument, we see that $P$ and $P'$ belong to the same partial $0/1$-equivalence class of $H$ if and only if one can be transformed to the other by a symmetry in $F(H)$. So, for $2^{n-2}<k\le 2^{n-1}$, we can use Pólya’s theorem to compute the number $N_H(k)$ of partial $0/1$-equivalence classes of $H$ with $k$ vertices.

Denote by $V_n(H)$ the set of vertices of $Q_n$ that are contained in $H$. Consider the action of $F(H)$ on $V_n(H)$. Assume that each vertex in $V_n(H)$ is assigned one of the two colors, say, black and white. For such a coloring of the vertices in $V_n(H)$, assume that the black vertices are vertices of a $0/1$-polytope contained in $H$. Clearly, for $2^{n-2}< k\le 2^{n-1}$, this leads to a one-to-one correspondence between partial $0/1$-equivalence classes of $H$ with $k$ vertices and equivalence classes of colorings of the vertices in $V_n(H)$ with $k$ black vertices.

Write $Z_H(z)$ for the cycle index of $F(H)$, and let $C_H(u_1,u_2)$ denote the polynomial obtained from $Z_H(z)$ by substituting $z_i$ with $u_1^i+u_2^i$.

Theorem 3.3

Assume that $2^{n-2}<k\le 2^{n-1}$, and let $H$ be a spanned hyperplane of $Q_n$ containing at least $k$ vertices of $Q_n$. Then we have

$$\begin{aligned} N_H(k)=\big [u_1^ku_2^{|V_n(H)|-k}\big ]C_H(u_1,u_2). \end{aligned}$$

We shall compute the cycle index $Z_H(z)$ in Sects. 4 and 5. Section 4 is devoted to a characterization of the stabilizer group $F(H)$. In Sect. 5, we will give an explicit expression for $Z_H(z)$.

4 The Structure of the Stabilizer $F(H)$

In this section, we aim to characterize the stabilizer $F(H)$ for a given spanned hyperplane $H$ of $Q_n$.

As mentioned in Sect. 3, for every equivalence class of spanned hyperplanes of $Q_n$, we can choose a representative of the form

$$\begin{aligned} H:\, a_1x_1+a_2x_2+\cdots +a_tx_t=b, \end{aligned}$$

(4.1)

where the coefficients $a_i$ are positive integers with $a_1\le a_2\le \cdots \le a_t$, and $b$ is a nonnegative integer.

From now on, we shall restrict our attention only to spanned hyperplanes of $Q_n$ in the form of (4.1). We define the type of the spanned hyperplane $H$ in (4.1) to be a vector $\alpha =(\alpha _1,\alpha _2,\ldots ,\alpha _\ell )$, where $\alpha _i$ is the multiplicity of $i$ occurring in the set $\{a_1,a_2,\ldots ,a_t\}$. For example, let

$$\begin{aligned} H:x_1+x_2+2x_3+2x_4+3x_5=4 \end{aligned}$$

be a spanned hyperplane of $Q_5$. Then the type of $H$ is $\alpha =(\alpha _1,\alpha _2,\alpha _3)=(2,2,1)$.

For positive integers $i$ and $j$ with $i\le j$, let $[i,j]$ denote the interval $\{i,i+1,\ldots ,j\}$. Let $\alpha =(\alpha _1,\alpha _2,\ldots ,\alpha _\ell )$ be the type of a spanned hyperplane. For $i=1,2,\ldots ,\ell$, let $S_{\alpha _i}$ be the group of permutations on the interval

$$\begin{aligned} \big [\alpha _1+\cdots +\alpha _{i-1}+1,\alpha _1+\cdots + \alpha _{i-1}+\alpha _i\big ], \end{aligned}$$

(4.2)

where we assume that $\alpha _0=0$. We define

$$\begin{aligned} S_\alpha =S_{\alpha _1}\times S_{\alpha _2}\times \cdots \times S_{\alpha _\ell }, \end{aligned}$$

(4.3)

where $\times$ denotes the direct product of groups. We also define

$$\begin{aligned} \overline{S}_\alpha =\overline{S}_{\alpha _1}\times \overline{S}_{\alpha _2}\times \cdots \times \overline{S}_{\alpha _\ell }, \end{aligned}$$

(4.4)

where $\overline{S}_{\alpha _i}$ is the set of signed permutations on the interval (4.2) for which every element is associated with the minus sign.

Let

$$\begin{aligned} P(H)=\left\{ \begin{array}{lll} S_\alpha &{} \ \ \mathrm {if} \ \sum _{i=1}^t a_i\ne 2b,\\ S_\alpha \cup \overline{S}_\alpha &{} \ \ \mathrm {if}\ \sum _{i=1}^t a_i=2b. \end{array}\right. \end{aligned}$$

(4.5)

We have the following characterization of the stabilizer of a spanned hyperplane of $Q_n$.

Theorem 4.1

Let $H:a_1x_1+a_2x_2+\cdots +a_tx_t=b$ be a spanned hyperplane of $Q_n$. Then

$$\begin{aligned} F(H)=P(H)\times B_{n,t}, \end{aligned}$$

where $B_{n,t}$ is the group of signed permutations on the interval $[t+1,n]$.

To give a proof of Theorem 4.1, we need to describe the action of a symmetry of $Q_n$ on a hyperplane in $\mathbb {R}^n$. Let $H:a_1x_1+a_2x_2+\cdots +a_nx_n=b$ be a hyperplane in $\mathbb {R}^n$, and $w$ be a symmetry in $B_n$. Recall that $w(H)$ is the hyperplane obtained from $H$ under the action of $w$. Let $s(w)$ be the set of entries of $w$ that are assigned the minus sign. In view of (2.3), we see that $w(H)$ is of the form

$$\begin{aligned} \sum _{i\,\notin s(w)}a_{\pi (i)}x_i+ \sum _{j\,\in s(w)}a_{\pi (j)}(1-x_j)=b, \end{aligned}$$

(4.6)

where $\pi$ is the underlying permutation of $w$. For $1\le j\le n$, let

$$\begin{aligned} s(w,j)= \left\{ \begin{array}{l@{\quad }l} -1 \qquad &{}\text{ if } j\in s(w),\\ 1 \qquad &{}\text{ otherwise }.\end{array}\right. \end{aligned}$$

Then (4.6) can be rewritten as

$$\begin{aligned} s(w,1)\cdot a_{\pi (1)}x_1+s(w,2)\cdot a_{\pi (2)} x_2+\cdots +s(w,n)\cdot a_{\pi (n)}x_n=b-\sum _{j\in s(w)}a_{\pi (j)}. \end{aligned}$$

(4.7)

For example, let

$$\begin{aligned} H:x_1-x_2-x_3+2x_4=1 \end{aligned}$$

be a hyperplane in $\mathbb {R}^4$, and let $w=(1)(\bar{2}\bar{3})(4)\in B_4$. Then $w(H)$ is the following hyperplane:

$$\begin{aligned} x_1+x_2+x_3+2x_4=3. \end{aligned}$$

We are now in a position to prove Theorem 4.1.

Proof

Assume that $w\in F(H)$ and $\pi$ is the underlying permutation of $w$. We aim to show that $w\in P(H)\times B_{n,t}$. Notice that $w(H)$ can be expressed in the form of (4.7). Since $H=w(H)$, it follows that for $1\le j\le t$, $s(w,j)$ are either all positive or all negative. So we have the following two cases.

Case 1: $s(w,j)$ is positive for $1\le j\le t$. In this case, it is clear that $w(H)$ is of the following form:

$$\begin{aligned} a_{\pi (1)}x_1+a_{\pi (2)}x_2+\cdots +a_{\pi (t)}x_t=b, \end{aligned}$$

where $a_{\pi (j)}=a_j$ for $1\le j\le t$. So we deduce that, for any $1\le j\le t$, $\pi (j)$ is in the interval $\big [\alpha _1+\cdots +\alpha _{i-1}+1,\alpha _1+\cdots + \alpha _{i-1}+\alpha _i\big ]$ that contains the element $j$. This implies that $w\in S_\alpha \times B_{n,t}$.

Case 2: $s(w,j)$ is negative for $1\le j\le t$. Then $w(H)$ is of the following form:

$$\begin{aligned} -a_{\pi (1)}x_1-a_{\pi (2)}x_2-\cdots -a_{\pi (t)}x_t=b-(a_1+\cdots +a_t). \end{aligned}$$

Since $w(H)=H$, we have $a_{\pi (j)}=a_j$ for $1\le j\le t$ and $b-(a_1+\cdots +a_t)=-b$. This yields that $w\in \overline{S}_\alpha \times B_{n,t}$.

Combining the above two cases, we deduce that $w\in P(H)\times B_{n,t}$. It remains to show that if $w$ belongs to $P(H)\times B_{n,t}$, then $w$ fixes $H$. Write $w=\pi \, \sigma$, where $\pi \in P(H)$ and $\sigma \in B_{n,t}$. We have the following two cases.

Case 1: $\pi \in S_{\alpha }$. By (4.7), the hyperplane $w(H)$ is of the following form:

$$\begin{aligned} a_{\pi (1)}x_1+\cdots +a_{\pi (t)}x_t=b. \end{aligned}$$

By the definition of $S_\alpha$, we see that $a_{\pi (i)}=a_i$ for $1\le i\le t$. So we have $w(H)=H$.

Case 2: $\pi \in \overline{S}_{\alpha }$. Let $\pi _0$ be the underlying permutation of $\pi$. By (4.7), the hyperplane $w(H)$ can be expressed as

$$\begin{aligned} -a_{\pi _0(1)}x_1-\cdots -a_{\pi _0(t)}x_t=b-(a_1+\cdots +a_t). \end{aligned}$$

By the definition of $\overline{S}_\alpha$, we see that $a_{\pi _0(i)}=a_i$ for $1\le i\le t$, which, together with the following relation:

$$\begin{aligned} 2b=a_1+\cdots +a_t, \end{aligned}$$

implies that $w(H)=H$. This completes the proof. $\square$

We conclude this section with a sufficient condition to determine whether two elements in the subgroup $P(H)$ are in the same conjugacy class. Recall that for a group $G$, two elements $g_1$ and $g_2$ are in the same conjugacy class of $G$ if there exists an element $g\in G$ such that $g_1=gg_2 g^{-1}$. This condition will be used in Sect. 5 for the purpose of computing the cycle index of the stabilizer group of a spanned hyperplane $H$.

Let $H$ be a spanned hyperplane of $Q_n$ of type $\alpha =(\alpha _1,\ldots ,\alpha _\ell )$. Recall that each element $\pi$ in the subgroup $P(H)$ is either in $S_\alpha$ or in $\overline{S}_{\alpha }$. Hence $\pi$ can be expressed as a product $\pi =\pi _1\pi _2\cdots \pi _\ell$, where, for $1\le i\le \ell$, $\pi _i$ belongs to $S_{\alpha _i}$ if $\pi \in S_\alpha$, and $\pi _i$ belongs to $\overline{S}_{\alpha _i}$ if $\pi \in \overline{S}_\alpha$.

Theorem 4.2

Let $\pi =\pi _1\pi _2\cdots \pi _\ell$ and $\pi '=\pi _1'\pi _2'\cdots \pi _\ell '$ be two elements in $P(H)$ such that $\pi$ and $\pi '$ are both in $S_{\alpha }$, or $\pi$ and $\pi '$ are both in $\overline{S}_{\alpha }$. If the underlying permutations of $\pi _i$ and $\pi _i'$ have the same cycle type for any $1\le i\le \ell$, then $\pi$ and $\pi '$ are in the same conjugacy class of $P(H)$.

Proof

We first consider the case when both $\pi$ and $\pi '$ are in $S_\alpha$. Since $\pi _i$ and $\pi _i'$ are permutations of the same cycle type, they are in the same conjugacy class. So there is a permutation $w_i\in S_{\alpha _i}$ such that $\pi _i=w_i\pi _i'w_i^{-1}$. It follows that $\pi =(w_1\pi _1'w_1^{-1})\cdots (w_\ell \pi _\ell 'w_\ell ^{-1})=w\pi 'w^{-1}$, where $w=w_1\cdots w_\ell \in S_\alpha$. This shows that $\pi$ and $\pi '$ are in the same conjugacy class.

It remains to consider the case when both $\pi$ and $\pi '$ are in $\overline{S}_\alpha$. Let $\pi _0$ (resp., $\pi '_0$) be the underlying permutation of $\pi$ (resp., $\pi '$). Then there is a symmetry $w\in S_\alpha$ such that $\pi _0=w\pi '_0w^{-1}$. We claim that $\pi =w\pi 'w^{-1}$. Indeed, it is enough to show that $\pi (x_1,x_2,\ldots ,x_t) =w\pi 'w^{-1}(x_1,x_2,\ldots ,x_t)$ for any point $(x_1,x_2,\ldots ,x_t)$ in $\mathbb {R}^t$. Assume that $\pi (x_1,x_2,\ldots ,x_t)=(y_1,y_2,\ldots ,y_t)$ and $w\pi 'w^{-1}(x_1,x_2,\ldots ,x_t)=(z_1,z_2,\ldots ,z_t)$. Since every element of $\pi$ is associated with the minus sign, by (2.3) we find that $y_i=1-x_{\pi _0(i)}$ for $1\le i\le t$. On the other hand, using (2.3), it is easy to check that $z_i=1-x_{w^{-1}\pi '_0w(i)}$ for $1\le i\le t$. Since $\pi _0=w\pi '_0w^{-1}$, we deduce that $\pi _0(i)=w^{-1}\pi '_0w(i)$. Therefore, we have $y_i=z_i$ for $1\le i\le t$. So the claim is justified. This completes the proof. $\square$

5 The Computation of $Z_H(z)$

In this section, we obtain a formula for the cycle index $Z_H(z)$ of the stabilizer group $F(H)$ of a spanned hyperplane $H$ of $Q_n$.

Let

$$\begin{aligned} H:\, a_1x_1+a_2x_2+\cdots +a_tx_t=b \end{aligned}$$

(5.1)

be a spanned hyperplane of $Q_n$. Recall that $V_n(H)$ is the set of vertices of $Q_n$ contained in $H$. To compute the cycle index $Z_H(z)$, we need to determine the cycle structures of permutations on $V_n(H)$ induced by the symmetries in $F(H)$. By Theorem 4.1, each symmetry in $F(H)$ can be written uniquely as a product $\pi w$, where $\pi \in P(H)$ and $w\in B_{n,t}$. We shall define two group actions for the subgroups $P(H)$ and $B_{n,t}$, and we derive an expression for the cycle type of the permutation on $V_n(H)$ induced by $\pi w$ in terms of the cycle types of the permutations induced by $\pi$ and $w$.

Let $H$ be a spanned hyperplane of $Q_n$ as given in (5.1). To define the action of $P(H)$, we should consider $H$ as a hyperplane in $\mathbb {R}^t$. Clearly, if $H$ is regarded as a hyperplane in $\mathbb {R}^t$, it is a spanned hyperplane of $Q_t$. Denote by $V_t(H)$ the set of vertices of $Q_t$ that are contained in $H$, namely,

$$\begin{aligned} V_t(H)=\{(x_1,x_2,\ldots ,x_t)\in V_t\,|\, a_1x_1+a_2x_2+\cdots +a_tx_t=b\}. \end{aligned}$$

Since $P(H)$ stabilizes the set $V_t(H)$, we get an action of the group $P(H)$ on $V_t(H)$.

We also need to describe the action of a symmetry in the group $B_{n,t}$ on the set of vertices of $Q_{n-t}$. Assume that $w\in B_{n,t}$, namely, $w$ is a signed permutation on the interval $[t+1,n]$. Subtracting each element of $w$ by $t$, we get a signed permutation on $[1,n-t]$. In this way, each signed permutation in $B_{n,t}$ corresponds to a symmetry of $Q_{n-t}$. Hence, $B_{n,t}$ is isomorphic to the group $B_{n-t}$ of symmetries of $Q_{n-t}$. This leads to an action on $V_{n-t}$.

Let $\pi w$ be a symmetry in $F(H)$, where $\pi \in P(H)$ and $w\in B_{n,t}$. The following lemma shows that the cycle type of the permutation on $V_n(H)$ induced by $\pi w$ is determined by the cycle types of the permutations on $V_t(H)$ and $V_{n-t}$ induced by $\pi$ and $w$. For an element $g$ in a group $G$ acting on a finite set $X$, we use $c(g)$ to denote the cycle type of the permutation on $X$ induced by $g$, which is written as a multiset $\{1^{c_1}, 2^{c_2}, \ldots \}$.

Lemma 5.1

Let $H:\, a_1x_1+a_2x_2+\cdots +a_tx_t=b$ be a spanned hyperplane of $Q_n$, and $\pi w$ be a symmetry in $F(H)$, where $\pi \in P(H)$ and $w\in B_{n,t}$. Assume that $c(\pi )=\{1^{m_1},2^{m_2},\ldots \}$ and $c(w)=\{1^{k_1},2^{k_2},\ldots \}$. Then we have

$$\begin{aligned} c(\pi w)=\bigcup _{i\ge 1}\bigcup _{j\ge 1} \left\{ \Big (\mathrm {lcm}(i,j)\right) ^{\frac{ijm_ik_j}{\mathrm {lcm}(i,j)}}\Big \}, \end{aligned}$$

(5.2)

where $\bigcup$ denotes the disjoint union of multisets, and $\mathrm {lcm}(i,j)$ denotes the least common multiple of $i$ and $j$.

Proof

Clearly, each vertex in $V_n(H)$ can be expressed as a vector of the following form

$$\begin{aligned} (x_1,\ldots ,x_{t},y_1,\ldots ,y_{n-t}), \end{aligned}$$

where $(x_1,\ldots ,x_{t})$ is a vertex in $V_t(H)$ and $(y_1,\ldots ,y_{n-t})$ is a vertex of $Q_{n-t}$. Assume that $|V_t(H)|=m$. Let $V_t(H)=\{u_1,u_2, \ldots ,u_{m}\}$ and $V_{n-t}=\{v_1,v_2,\ldots ,v_{2^{n-t}}\}$. Then each vertex in $V_n(H)$ can be expressed as an ordered pair $(u_i,v_j)$, where $1\le i\le m$ and $1\le j\le 2^{n-t}$.

Let $C_i=({s_1},\ldots ,{s_i})$ be an $i$-cycle of the permutation on $V_t(H)$ induced by $\pi$, that is, $C_i$ maps the vertex $u_{s_p}$ to the vertex $u_{s_{p+1}}$ for $1\le p\le i-1$, and to the vertex $u_{s_1}$ for $p=i$. Similarly, let $C_j=({t_1},\ldots ,t_j)$ be a $j$-cycle of the permutation on $V_{n-t}$ induced by $w$, that is, $C_j$ maps the vertex $v_{t_q}$ to the vertex $v_{t_{q+1}}$ for $1\le q \le j-1$, and to the vertex $v_{t_1}$ for $q=j$. Define $C_{i,j}$ to be the permutation on the subset $\{(u_{s_p},v_{t_q})\,|\, 1\le p\le i,\ 1\le q\le j\}$ of $V_n(H)$ such that

$$\begin{aligned} C_{i,j}(u_{s_p},v_{t_q})=(C_i(u_{s_p}),C_j(v_{t_q})). \end{aligned}$$

It is easily seen that the induced permutation of $\pi w$ on $V_n(H)$ is the direct product of $C_{i,j}$, where $C_i$ (resp., $C_j$) runs over the cycles of the permutation on $V_t(H)$ (resp., $V_{n-t}$) induced by $\pi$ (resp., $w$).

It can be verified that the cycle type of $C_{i,j}$ is

$$\begin{aligned} \Big \{\left( \mathrm {lcm}(i,j)\right) ^{\frac{ij}{\mathrm {lcm} (i,j)}}\Big \}. \end{aligned}$$

Thus the cycle type of the induced permutation of $\pi w$ on $V_n(H)$ is given by (5.2). This completes the proof. $\square$

For convenience, we introduce the following notation. Let $\pi$ be a symmetry in $P(H)$. Assume that the cycle type of the permutation on $V_t(H)$ induced by $\pi$ is

$$\begin{aligned} c(\pi )=\{1^{m_1},2^{m_2}, \ldots \}. \end{aligned}$$

For $j\ge 1$, we define

$$\begin{aligned} f_{\pi ,j}(z)=\prod _{i\ge 1}(z_{\mathrm {lcm}(i, j)})^{\frac{ijm_i}{\mathrm {lcm}(i,j)}}. \end{aligned}$$

(5.3)

We have the following proposition.

Proposition 5.2

Let $H$ be a spanned hyperplane of $Q_n$ of type $\alpha$. Assume that $\pi =\pi _1\pi _2\cdots \pi _\ell$ and $\pi '=\pi _1'\pi _2' \cdots \pi _\ell '$ are two symmetries in $P(H)$ such that $\pi$ and $\pi '$ are both in $S_{\alpha }$, or $\pi$ and $\pi '$ are both in $\overline{S}_{\alpha }$. If the underlying permutations of $\pi _i$ and $\pi _i'$ have the same cycle type for $1\le i\le \ell$, then, for $j\ge 1$,

$$\begin{aligned} f_{\pi ,j}(z)=f_{\pi ',j}(z). \end{aligned}$$

(5.4)

Proof

It follows from Theorem 4.2 that $\pi$ and $\pi '$ are in the same conjugacy class of $P(H)$. Hence the permutations on $V_t(H)$ induced by $\pi$ and $\pi '$ are in the same conjugacy class, that is, $c(\pi )=c(\pi ')$. Since $f_{\pi ,j}(z)$ depends only on the cycle type $c(\pi )$, we deduce that $f_{\pi ,j}(z)=f_{\pi ',j}(z)$. This completes the proof. $\square$

To compute the cycle index $Z_H(z)$, we recall some notation and terminology on integer partitions. A partition $\lambda$ of a positive integer $n$, denoted $\lambda \vdash n$, will be expressed in the multiset form, that is, $\lambda =\{1^{m_1},2^{m_2},\ldots \}$, where $m_i$ is the number of occurrences of $i$ in $\lambda$. Denote by $\ell (\lambda )$ the number of parts of $\lambda$, that is, $\ell (\lambda )=m_1+m_2+\cdots$. For a partition $\lambda =\{1^{m_1},2^{m_2},\ldots \}$, let

$$\begin{aligned} m_\lambda =1^{m_1} m_1! 2^{m_2}m_2!\cdots . \end{aligned}$$

For two partitions $\lambda$ and $\mu$, define $\lambda \cup \mu$ to be the partition obtained by putting the parts of $\lambda$ and $\mu$ together. For example, for $\lambda =\{1,2\}$ and $\mu =\{1^2,3\}$, we have $\lambda \cup \mu =\{1^3,2,3\}$.

Let $H$ be a spanned hyperplane of $Q_n$ of type $\alpha =(\alpha _1,\alpha _2,\ldots ,\alpha _\ell )$. For $1\le i\le \ell$, let $\mu ^i$ be a partition of $\alpha _i$, and let $\mu =\mu ^1\cup \cdots \cup \mu ^\ell$. Assume that $\pi =\pi _1\pi _2\cdots \pi _\ell$ (resp., $\pi '=\pi _1'\pi _2'\cdots \pi _\ell '$) is a symmetry in $S_\alpha$ (resp., $\overline{S}_\alpha$) such that the underlying permutation of $\pi _i$ (resp., $\pi _i'$) has cycle type $\mu ^i$ for $1\le i\le \ell$. For $j\ge 1$, define

$$\begin{aligned} g_{\mu ,j}(z)=f_{\pi ,j}(z) \end{aligned}$$

and

$$\begin{aligned} \overline{g}_{\mu ,j}(z)=f_{\pi ',j}(z). \end{aligned}$$

By Proposition 5.2, the functions $g_{\mu ,j}(z)$ and $\overline{g}_{\mu ,j}(z)$ are well defined.

Let

$$\begin{aligned} g_\mu (z)=(g_{\mu ,1}(z),g_{\mu ,2}(z),\ldots ) \end{aligned}$$

and

$$\begin{aligned} \overline{g}_\mu (z)=(\overline{g}_{\mu ,1}(z),\overline{g}_{\mu , 2}(z),\ldots ). \end{aligned}$$

In the above notation, we obtain the following formula for the cycle index $Z_H(z)$.

Theorem 5.3

Let $H:a_1x_1+a_2x_2+\cdots +a_tx_t=b$ be a spanned hyperplane of $Q_n$. Assume that $H$ is of type $\alpha =(\alpha _1,\alpha _2,\ldots ,\alpha _\ell )$. Then we have

$$\begin{aligned} Z_H(z)=\frac{1}{2^{\delta (H)}}\sum _{(\mu ^1,\ldots ,\mu ^\ell )} \prod _{i=1}^\ell m_{\mu ^i}^{-1}\left( Z_{n-t}(g_{\mu }(z))+ \delta (H)Z_{n-t}(\overline{g}_\mu (z))\right) , \end{aligned}$$

(5.5)

where $\mu ^i\vdash \alpha _i$, $\mu =\mu ^1\cup \cdots \cup \mu ^\ell$, $\delta (H)=1$ if $\sum _{i=1}^t a_i= 2b$ and $\delta (H)=0$ otherwise.

Proof

Let $\pi \in P(H)$ and $w\in B_{n,t}$, and let

$$\begin{aligned} c(w)=\{1^{k_1},2^{k_2},\ldots \} \end{aligned}$$

be the cycle type of the permutation on $V_{n-t}$ induced by $w$. In view of Lemma 5.1, we have

$$\begin{aligned} {z}^{c(\pi w)}=f_{\pi ,1}(z)^{k_1}f_{\pi ,2}(z)^{k_2}\cdots \end{aligned}$$

(5.6)

Summing over signed permutations $w$ in $B_{n,t}$ and using (2.1) and (5.6), we deduce that

$$\begin{aligned} \sum _{\pi w}{z}^{c(\pi w)}&=\sum _{w}f_{\pi ,1}(z)^{k_1}f_{\pi ,2}(z)^{k_2}\cdots \\&=(n-t)!2^{n-t}Z_{n-t}(f_{\pi ,1}(z),f_{\pi ,2}(z),\ldots )\\&=(n-t)!2^{n-t}Z_{n-t}(f_\pi ({z})), \end{aligned}$$

where

$$\begin{aligned} f_{\pi }(z)=(f_{\pi ,1}(z),f_{\pi ,1}(z),\ldots ). \end{aligned}$$

Thus,

$$\begin{aligned} Z_H({z})&=\frac{1}{|F(H)|}\sum _{\pi w\in F(H)}{z}^{c(\pi w)}\nonumber \\&=\frac{1}{|F(H)|}\sum _{\pi \in P(H)}(n-t)!2^{n-t}Z_{n-t}(f_\pi ({z})) \nonumber \\&=\frac{(n-t)!2^{n-t}}{|F(H)|}\Bigg (\sum _{\pi \in S_\alpha }Z_{n-t}(f_\pi ({z}))+ \delta (H)\sum _{\pi '\in \overline{S}_\alpha }Z_{n-t}(f_{\pi '}({z}))\Bigg ), \end{aligned}$$

(5.7)

where $\delta (H)=1$ if $\sum _{i=1}^t a_i= 2b$ and $\delta (H)=0$ otherwise.

For a partition $\lambda \vdash n$, there are $\frac{n!}{m_\lambda }$ permutations on $\{1,2,\ldots ,n\}$ that are of type $\lambda$, see, for example, Stanley [17, Proposition 1.3.2]. So the number of symmetries $\pi =\pi _1\pi _2\ldots \pi _\ell$ in $S_\alpha$ (or, $\overline{S}_\alpha$) such that for $i=1,2,\ldots ,\ell$, the underlying permutation of $\pi _i$ is of type $\mu ^i$ equals

$$\begin{aligned} \prod _{i=1}^\ell \frac{\alpha _i!}{m_{\mu ^i}}. \end{aligned}$$

(5.8)

Combining (5.7), (5.8) and Proposition 5.2, we obtain that

$$\begin{aligned} Z_H({z}) =\frac{(n-t)!2^{n-t}}{|F(H)|}\sum _{(\mu ^1,\ldots ,\mu ^\ell )} \prod _{i=1}^\ell \frac{\alpha _i!}{m_{\mu ^i}}\big (Z_{n-t}(g_\mu ({z}))+ \delta (H)Z_{n-t}(\overline{g}_\mu ({z}))\big ),\nonumber \\ \end{aligned}$$

(5.9)

where $\mu ^i\vdash \alpha _i$ and $\mu =\mu ^1\cup \cdots \cup \mu ^\ell$.

It is easily seen that

$$\begin{aligned} |F(H)|=(n-t)!2^{n-t+\delta (H)}\prod _{i=1}^\ell \alpha _i!. \end{aligned}$$

(5.10)

Substituting (5.10) into (5.9), we are led to (5.5). $\square$

By Theorem 5.3, to compute the cycle index $Z_H({z})$, it suffices to determine the cycle type $c(\pi )$ of the permutation on $V_t(H)$ induced by $\pi \in P(H)$. Let $c(\pi )=\{1^{m_1},2^{m_2},\ldots \}$. By Theorem 2.1, we have

$$\begin{aligned} m_i=\frac{1}{i}\sum _{j|i}\mu (i/j)\psi (\pi ^j), \end{aligned}$$

(5.11)

where $\psi (\pi ^j)$ is the number of vertices in $V_t(H)$ that are fixed by $\pi ^j$. The following theorem gives a formula for $\psi (\pi )$, leading to a formula for $\psi (\pi ^j)$ .

Theorem 5.4

Let $H:a_1x_1+a_2x_2+\cdots +a_tx_t=b$ be a spanned hyperplane of $Q_n$. Assume that $\pi =\pi _1\pi _2\cdots \pi _\ell$ is a symmetry in $P(H)$ such that the underlying permutation of $\pi _i$ is of type $\mu ^i=\{1^{m_{i1}},2^{m_{i2}},\ldots \}$ for $i=1,2,\ldots ,\ell$. Then

$$\begin{aligned} \psi (\pi )= \left\{ \begin{array}{lll} \left[ x^b\right] \prod _{i=1}^\ell \prod _{j\ge 1}(1+x^{ij})^{m_{ij}} &{} \ \ \mathrm {if} \ \ \pi \in S_\alpha ,\\ \chi (\mu )2^{\ell (\mu )} &{} \ \ \mathrm {if}\ \ \pi \in \overline{S}_\alpha , \end{array}\right. \end{aligned}$$

(5.12)

where $\mu =\mu ^1\cup \cdots \cup \mu ^\ell$, $\chi (\mu )=1$ if $\mu$ has no odd parts and $\chi (\mu )=0$ otherwise.

Proof

We first consider the case when $\pi$ is in $S_\alpha$. Observe that, a vertex $v=(x_1,x_2,\ldots ,x_t)$ of $Q_t$ is both fixed by $\pi$ and contained in $V_t(H)$ if and only if

• (1) For $1\le i\le \ell$ and each $k$-cycle $({j_1},{j_2},\ldots ,{j_k})$ of $\pi _i$, we have

  $$\begin{aligned} x_{{j_1}}=x_{{j_2}}=\cdots =x_{{j_k}}. \end{aligned}$$

• (2) $a_1x_1+a_2x_2+\cdots +a_tx_t=b$, or equivalently,

  $$\begin{aligned} b_1+2b_2+\cdots +\ell b_\ell =b, \end{aligned}$$

  where $b_i$ ($1\le i\le \ell$) is the sum of the entries of $v$ equal to 1.

It can be easily deduced that the number of vertices of $Q_t$ satisfying the above conditions is given by

$$\begin{aligned} \big [x^b\big ]\prod _{i=1}^\ell \prod _{j\ge 1}\big (1+ x^{ij}\big )^{m_{ij}}. \end{aligned}$$

This proves (5.12) for the case when $\pi \in S_\alpha$.

We now consider the case when $\pi$ is in $\overline{S}_\alpha$. Notice that a vertex $v=(x_1,x_2,\ldots ,x_t)$ of $Q_t$ is fixed by $\pi$ if and only if, for any $k$-cycle $(\overline{j_1},\overline{j_2},\ldots ,\overline{j_k})$ of $\pi$, we have

$$\begin{aligned} (x_{{j_1}},x_{{j_2}},\ldots ,x_{{j_k}}) =(1-x_{{j_2}},1-x_{{j_3}},\ldots ,1-x_{{j_1}}). \end{aligned}$$

(5.13)

Consequently, if a vertex $v=(x_1,x_2,\ldots ,x_t)$ of $Q_t$ is fixed by $\pi$, then, for any $k$-cycle $(\overline{j_1},\overline{j_2},\ldots ,\overline{j_k})$ of $\pi$, the vector $(x_{j_1},x_{j_2},\ldots , x_{j_k})$ is either $(0,1,\ldots ,0,1)$ or $(1,0,\ldots ,1,0)$. This implies that $k$ is even. Thus $\pi$ does not have any fixed points if $\pi$ contains an odd cycle.

We now assume that $\pi$ has only even cycles. In this case, the number of vertices of $Q_t$ fixed by $\pi$ equals $2^{\ell (\mu )}$. To prove $\psi (\pi )=2^{\ell (\mu )}$, we need to demonstrate that any vertex of $Q_t$ fixed by $\pi$ is in $V_t(H)$. Let $v=(x_1,x_2,\ldots ,x_t)$ be a vertex of $Q_t$ fixed by $\pi$. Since, for each cycle $(\overline{j_1},\overline{j_2},\ldots ,\overline{j_k})$ of $\pi$, the vector $(x_{{j_1}},x_{{j_2}},\ldots ,x_{{j_k}})$ is either $(0,1,\ldots ,0,1)$ or $(1,0,\ldots ,1,0)$, we deduce that $a_1x_1+a_2x_2+\cdots +a_tx_t=b$ by applying the relation $a_1+\cdots +a_t=2b.$ Hence the vertex $v$ is in $V_t(H)$. This completes the proof. $\square$

Based on Theorem 5.4, we can compute $\psi (\pi ^j)$ since the cycle structure of $\pi ^j$ is easily determined by the cycle structure of $\pi$. Let $\pi =\pi _1\pi _2\ldots \pi _\ell$ be a symmetry in $P(H)$ such that for $1\le i \le \ell$, the underlying permutation of $\pi _i$ is of type $\mu ^i=\{1^{m_{i1}},2^{m_{i2}},\ldots \}$. Clearly, we have $\pi ^j=\pi _1^j\pi _2^j\ldots \pi _\ell ^j$. Moreover, we see that $\pi ^j$ belongs to $S_\alpha$ if $\pi$ is in $S_\alpha$ or $\pi$ is in $\overline{S}_\alpha$ and $j$ is even, and $\pi ^j$ belongs to $\overline{S}_\alpha$ otherwise. Let $\mathrm {gcd}(i,j)$ denote the greatest common divisor of $i$ and $j$. Then the cycle type of the underlying permutation of $\pi _i^j$ is given by

$$\begin{aligned} \Big \{1^{m_{i1}}, \mathrm {gcd}(2,j)^{\frac{2m_{i2}}{\mathrm {gcd} (2,j)}}, \mathrm {gcd}(3,j)^{\frac{3m_{i3}}{\mathrm {gcd} (3,j)}}, \ldots \Big \}. \end{aligned}$$

6 $F_n(k)$ for $n=4,5,6$ and $2^{n-2}< k\le 2^{n-1}$

This section is devoted to the computation of $F_n(k)$ for $n=4,5,6$ and $2^{n-2}< k\le 2^{n-1}$. This requires the cycle index $Z_H(z)$ for every spanned hyperplane $H$ of $Q_n$ for $n=4,5,6$ that contains more than $2^{n-2}$ vertices of $Q_n$.

Recall that $h(n,k)$ denotes the number of equivalence classes of spanned hyperplanes of $Q_n$ containing at least $k$ vertices. Let $H_1,H_2,\ldots ,H_{h(n,k)}$ be the representatives of these equivalence classes. When $2^{n-2}< k\le 2^{n-1}$, combining relation (1.1), Corollary 3.2 and Theorem 3.3, we deduce that

$$\begin{aligned} F_n(k)&=A_n(k)-H_n(k) \nonumber \\&=A_n(k)-\sum _{i=1}^{h(n,k)}N_{H_i}(k)\nonumber \\&=A_n(k)-\sum _{i=1}^{h(n,k)}\big [u_1^{k}u_2^{|V_n(H_i)| -k}\big ]C_{H_i}(z_1,z_2). \end{aligned}$$

(6.1)

Using formula (6.1), we proceed to compute $F_n(k)$ for $n=4,5,6$ and $2^{n-2}< k\le 2^{n-1}$. We start with the computation of $F_4(k)$ for $4<k\le 8$. For $t\le n$, we use $H_n^t$ to denote the following hyperplane in $\mathbb {R}^n$

$$\begin{aligned} x_1+x_2+\cdots +x_t=\left\lfloor t/2\right\rfloor . \end{aligned}$$

In this notation, representatives of equivalence classes of spanned hyperplanes of $Q_4$ containing more than $4$ vertices are as follows:

$$\begin{aligned}&H_4^1:x_1=0,\\&H_4^2:x_1+x_2=1,\\&H_4^3:x_1+x_2+x_3=1,\\&H_4^4:x_1+x_2+x_3+x_4=2. \end{aligned}$$

Employing the techniques in Sect. 5, we obtain the cycle indices $Z_{H_4^1}(z)$ and $Z_{H_4^2}(z)$ as given below.

$$\begin{aligned} Z_{H_4^1}(z)&=Z_3(z),\\ Z_{H_4^2}(z)&= \frac{1}{16} \big (9z_2^4+4z_4^2+2z_1^4z_2^2+z_1^8\big ). \end{aligned}$$

For the remaining two hyperplanes $H=H_4^3$ and $H_4^4$, it is easily checked that $N_H(k)=1$ for $k=5,6$, and $N_H(k)=0$ for $k=7,8$. Thus, applying (6.1) we can determine $F_4(k)$ for $k=5,6,7,8$. These values are given in Table 4, which agree with the computation of Aichholzer [1].

Table 4 $F_4(k)$ for $k=5,6,7,8$

Observing that $F_4(k)=0$ for $k\le 4$, thus we have completed the enumeration of full-dimensional $0/1$-equivalence classes of $Q_4$.

We now compute $F_5(k)$ for $8< k\le 16$. Representatives of equivalence classes of spanned hyperplanes of $Q_5$ containing more than $8$ vertices are $H_5^1,H_5^2,H_5^3,H_5^4,H_5^5$. By utilizing the techniques in Sect. 5, we obtain that

$$\begin{aligned} Z_{H_5^1}({z})&=Z_4({z}),\\ Z_{H_5^2}({z})&=\frac{1}{96} \big (z_1^{16}+6z_1^8z_2^4+33z_2^8+8z_1^4z_3^4+24z_4^4+ 24z_2^2z_6^2\big ),\\ Z_{H_5^3}({z})&=\frac{1}{48} \big (12z_2^6+8z_4^3+2z_1^6z_2^3+ z_1^{12}+6z_1^2z_2^5+3z_1^4z_2^4+ 6z_6^2+4z_{12}+4z_3^2z_6\\&\qquad \qquad +2z_3^4\big ),\\ Z_{H_5^4}({z})&=\frac{1}{96}\big (z_1^{12}+27z_2^6+9z_1^4z_2^4+8z_3^4 +24z_6^2+18z_2^2z_4^2+6z_1^4z_4^2+3z_1^8z_2^2\big ),\\ Z_{H_5^5}({z})&=\frac{1}{120} \big (24z_5^2+30z_2z_4^2+20z_1 z_3z_6+20z_1z_3^3+15z_1^2z_2^4+10z_1^4z_2^3+z_1^{10}\big ). \end{aligned}$$

Consequently, the values $F_5(k)$ for $8<k\le 16$ can be derived from (6.1), and they agree with the computation of Aichholzer [1], see Table 5.

Table 5 $F_5(k)$ for $8< k\le 16$

The main objective of this section is to compute $F_6(k)$ for $16<k\le 32$. As mentioned in Sect. 4, there are 6 representatives of equivalence classes of spanned hyperplanes of $Q_6$ containing more than 16 vertices, namely, $H_6^1, H_6^2, H_6^3, H_6^4, H_6^5, H_6^6$. Again, by applying the techniques in Sect. 5, we obtain that

$$\begin{aligned} Z_{H_6^1}({z})&=Z_5({z}),\\ Z_{H_6^2}({z})&=\frac{1}{768} \left( \begin{array}{l} z_1^{32}+12z_1^{16}z_2^8+12z_1^8z_2^{12}+127z_2^{16}+32z_1^8z_3^8\\ +\,\, 48z_1^4z_2^2z_4^6+168z_4^8+224z_2^4z_6^4+96z_8^4+48z_2^4z_4^6 \end{array}\right) ,\\ Z_{H_6^3}({z})&=\frac{1}{288} \left( \begin{array}{l} z_1^{24}+6z_1^{12}z_2^6+52z_2^{12}+18z_3^8+ 48z_4^6+32z_2^3z_6^3+3z_1^8z_2^8\\ +\,\, 18z_1^4z_2^{10}+24z_1^2z_3^2z_2^2z_6^2+8z_1^6z_3^6+12z_3^4 z_6^2+42z_6^4+24z_{12}^2 \end{array}\right) ,\\ Z_{H_6^4}({z})&=\frac{1}{384} \left( \begin{array}{l} z_1^{24}+81z_2^{12}++2z_1^{12}z_2^6+18z_1^4z_2^{10}+ 15z_1^8z_2^8+72z_6^4+32z_{12}^2\\ +\,\, 64z_4^6+16z_3^4z_6^2+8z_3^8 +54z_2^4z_4^4+12z_1^4z_2^2z_4^4+6z_1^8z_4^4+3z_1^{16}z_2^4 \end{array}\right) ,\\ Z_{H_6^5}({z})&=\frac{1}{240} \left( \begin{array}{l} z_1^{20}+24z_{10}^2+60z_2^2z_4^4+26z_2^{10}+20z_1^2z_3^2z_6^2\\ +\,\, 20z_1^2z_3^6+15z_1^4z_2^8+10z_1^8z_2^6+40z_2z_6^3+24z_5^4 \end{array}\right) ,\\ Z_{H_6^6}({z})&=\frac{1}{1440}\left( \begin{array}{l} z_1^{20}+144z_5^4+144z_{10}^2+320z_2z_6^3+270z_2^2z_4^4 +76z_2^{10}\\ +\,\, 90z_1^4z_4^4+30z_1^8z_2^6+45z_1^4z_2^8+240z_1^2z_3^2z_6^2+80 z_1^2z_3^6 \end{array}\right) . \end{aligned}$$

Using (6.1), we can compute $F_6(k)$ for $16<k\le 32$. These values are listed in Table 6.

Table 6 $F_6(k)$ for $16<k\le 32$

7 $H_6(k)$ for $k=13,14, 15,16$

In this section, we compute $H_6(k)$ for $k=13,14, 15,16$. Together with the computation of Aichholzer for $n=6$ and $k\le 12$, we complete the enumeration of full-dimensional $0/1$-equivalence classes of the 6-dimensional hypercube. In fact, we can compute $H_n(k)$ when $n>4$ and $k$ is close to $2^{n-2}$.

Let us recall the map $\varPhi$ defined in Sect. 3. Let $H_1,H_2,\ldots ,H_{h(n,k)}$ be the representatives of equivalence classes of spanned hyperplanes of $Q_n$ containing at least $k$ vertices. As before, we use $\mathcal {P}(H_i,k)$ to denote the set of partial $0/1$-equivalence classes of $H_i$ with $k$ vertices, and use $N_{H_i}(k)$ to denote the cardinality of $\mathcal {P}(H_i,k)$. Let $\mathcal {P}$ be a partial $0/1$-equivalence class in the (disjoint) union of $\mathcal {P}(H_i,k)$ where $1\le i\le h(n,k)$. Then $\varPhi$ maps $\mathcal {P}$ to the unique $0/1$-equivalence class in $\mathcal {H}_n(k)$ that contains $\mathcal {P}$.

When $k\le 2^{n-2}$, it is possible that there exist equivalent $0/1$-polytopes $P$ and $P'$ that are contained respectively in $H_i$ and $H_j$, where $1\le i\ne j\le h_{n,k}$. Let $\mathcal {P}$ and $\mathcal {P'}$ be the partial $0/1$-equivalence classes of $H_i$ and $H_j$ that contain $P$ and $P'$ respectively. Then we have $\varPhi (\mathcal {P})=\varPhi (\mathcal {P'})$. So $\varPhi$ is not necessarily an injection when $k\le 2^{n-2}$. Note that when restricted to $\mathcal {P}(H_i,k)$, $\varPhi$ is always an injection. Thus, in order to compute $H_n(k)$ for $k\le 2^{n-2}$, we need to compute the number $N_{H_i}(k)$ of partial 0/1-equivalence classes of each spanned hyperplane $H_i$ as well as the number of partial $0/1$-equivalence classes with $k$ vertices that are contained in the intersection of distinct spanned hyperplanes.

The objective of this section is to find a way to compute $N_{H_i}(k)$ when $k$ is close to $2^{n-2}$. As will be seen, when $2^{n-3}<k\le 2^{n-2}$, to compute $N_{H_i}(k)$ we need to consider all possible symmetries $w\in B_n$ such that the intersections of $H_i$ and $w(H_i)$ contain at least $k$ vertices. To be more specific, we need to determine the number of partial $0/1$-equivalence classes with $k$ vertices that are contained in the intersection $H_i\cap w(H_i)$. Moreover, when $k$ is close to $2^{n-2}$, there are only a few symmetries $w$ such that the intersection $H_i\cap w(H_i)$ contains at least $k$ vertices. This makes it possible to compute $N_{H_i}(k)$ when $k$ is close to $2^{n-2}$.

When $k$ is close to $2^{n-2}$, the same technique can be applied to determine the number of partial $0/1$-equivalence classes with $k$ vertices that are contained in the intersection of distinct spanned hyperplanes.

Notice that

$$\begin{aligned} \mathcal {H}_n(k)=A_1\cup A_2\cup \cdots \cup A_{h(n,k)}, \end{aligned}$$

where

$$\begin{aligned} A_i=\varPhi (\mathcal {P}(H_i,k)). \end{aligned}$$

By the principle of inclusion–exclusion, we have the following expression for $H_n(k)$.

Lemma 7.1

Let $H$ be a spanned hyperplane of $Q_n$. Then we have

$$\begin{aligned} H_n(k)=&\sum _{1\le i\le h(n,k)}|A_i|- \sum _{1\le i_1<i_2\le h(n,k)}|A_{i_1}\cap A_{i_2}|\nonumber \\&+\sum _{1\le i_1<i_2<i_3 \le h(n,k)}|A_{i_1}\cap A_{i_2}\cap A_{i_3}|-\cdots \end{aligned}$$

(7.1)

By Lemma 7.1, the computation of $H_n(k)$ reduces to the evaluation of the cardinalities of $A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}$, where $1\le i_1<\cdots <i_m \le h(n,k)$. Since $\varPhi$ is an injection when restricted to $\mathcal {P}(H_i,k)$, we have $|A_i|=N_{H_i}(k)$. Moreover, as will be shown, when $2^{n-3}<k\le 2^{n-2}$ and $m\ge 2$, the computation of $|A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}|$ can be transformed to the determination of partial $0/1$-equivalence classes contained in the intersection of distinct spanned hyperplanes.

We now focus on the computation of $N_{H}(k)$, where $H$ is a spanned hyperplane of $Q_n$ and $k$ is close to $2^{n-2}$. Let $S\subseteq H$ be a subset of $H$. In Sect. 3, we have defined the partial $0/1$-equivalence relation on the set of $0/1$-polytopes of $Q_n$ contained in $S$. Here we need another equivalence relation on this set, that is, two $0/1$-polytopes are said to be equivalent if one can be transformed to the other by a symmetry in the stabilizer $F(H)$ of $H$. The associated equivalence classes are called local $0/1$-equivalence classes of $S$. Since $F(H)$ is a subgroup of $B_n$, each local $0/1$-equivalence class of $S$ is contained in a unique partial $0/1$-equivalence class of $S$.

Denote by $\mathcal {L}(S,k)$ the set of local $0/1$-equivalence classes of $S$ with $k$ vertices. To compute $N_{H}(k)$ when $k$ is close to $2^{n-2}$, we need to compute the cardinality of $\mathcal {L}(H,k)$ and the cardinality of $\mathcal {L}(S, k)$, where $S$ can be expressed as $S=H\cap w(H)$ for a symmetry $w$ in $B_n$ satisfying certain conditions. The cardinality of $\mathcal {L}(H,k)$ can be obtained from the cycle index $Z_H(z)$ of the stabilizer $F(H)$. In the following formula, $C_H(u_1,u_2)$ denotes the polynomial obtained from $Z_H(z)$ by substituting $z_i$ with $u_1^i+u_2^i$, as defined in Sect. 3.

Lemma 7.2

For any $1\le k\le 2^{n-1}$, we have

$$\begin{aligned} |\mathcal {L}(H,k)|=\big [u_1^ku_2^{|V_n(H)-k|}\big ]C_H(u_1,u_2). \end{aligned}$$

(7.2)

In the remaining of this section, we assume that $2^{n-3}<k\le 2^{n-2}$. Keep in mind that $N_H(k)$ is the cardinality of the set $\mathcal {P}(H,k)$ of partial 0/1-equivalence classes of $H$. To compute $|\mathcal {P}(H,k)|$, we shall define a subset $\mathcal {L}_1(H,k)$ of $\mathcal {L}(H,k)$ and a subset $\mathcal {P}_1(H,k)$ of $\mathcal {P}(H,k)$, which satisfy the following relation:

$$\begin{aligned} |\mathcal {P}(H,k)|=|\mathcal {L}(H,k)|-|\mathcal {L}_1(H,k)|+ |\mathcal {P}_{1}(H,k)|. \end{aligned}$$

We first define the subset $\mathcal {L}_1(H,k)$, which depends on a map $\Psi$ from the set of local 0/1-equivalence classes of certain intersections $H\cap w(H)$ to the set $\mathcal {L}(H,k)$. To define $\Psi$, let $E(H,k)$ denote the set of affine subspaces $H\cap w(H)$, where $w$ ranges over symmetries in $B_n$ such that

• (1) $H\ne w(H)$, that is, the symmetry $w$ of $Q_n$ does not fix $H$;

• (2) $H\cap w(H)$ contains at least $k$ vertices of $Q_n$.

Consider the equivalence classes of $E(H,k)$ under the symmetries in $F(H)$. This means that two elements $H\cap w(H)$ and $H\cap w'(H)$ in $E(H,k)$ are equivalent if there exists a symmetry $\sigma \in F(H)$ such that

$$\begin{aligned} H\cap w(H)=\sigma (H\cap w'(H)). \end{aligned}$$

Denote by $h_1(H,k)$ the number of equivalence classes of $E(H,k)$ under the symmetries in $F(H)$. Let

$$\begin{aligned} E_1(H,k)=\{H\cap w_{i}(H)\,|\, 1\le i\le h_1(H,k)\} \end{aligned}$$

be the set of representatives of these equivalence classes of $E(H,k)$.

The map $\Psi$ is defined from the (disjoint) union of $\mathcal {L}(H\cap w_i(H),k)$, where $1\le i\le h_1(H,k)$, to $\mathcal {L}(H,k)$. Let $\mathcal {L}$ be a local $0/1$-equivalence class in $\mathcal {L}(H\cap w_i(H),k)$. Define $\Psi (\mathcal {L})$ to be the unique local $0/1$-equivalence class in $\mathcal {L}(H,k)$ containing $\mathcal {L}$. We have the following property.

Theorem 7.3

For $n>4$ and $2^{n-3}<k\le 2^{n-2}$, the map $\Psi$ is an injection.

Proof

Let $\mathcal {L}$ and $\mathcal {L}'$ be two distinct local $0/1$-equivalence classes with $k$ vertices. Assume that $\mathcal {L}$ is contained in $\mathcal {L}(H\cap w_i(H),k)$ and $\mathcal {L}'$ is contained in $\mathcal {L}(H\cap w_j(H),k)$), where $1\le i, j\le h_1(H,k)$. To prove that $\Psi$ is an injection, we need to show that $\Psi (\mathcal {L})\ne \Psi (\mathcal {L}')$. If $i=j$, from the definition of the local 0/1-equivalence relation, it is clear that $\Psi (\mathcal {L})\ne \Psi (\mathcal {L}')$.

We now consider the case $i\ne j$. Let $P$ and $P'$ be two $0/1$-polytopes contained in $\mathcal {L}$ and $\mathcal {L}'$, respectively. We claim that $\mathrm {dim}(P)=\mathrm {dim}(P')= n-2$. We only give a proof of the assertion that $\mathrm {dim}(P)= n-2$. The relation $\mathrm {dim}(P')= n-2$ can be justified by the same argument.

Since $P$ has more than $2^{n-3}$ vertices, it follows from Theorem 1.1 that $\mathrm {dim}(P)\ge n-2$. On the other hand, since $P$ is contained in the intersection $H\cap w_i(H)$, we see that $\mathrm {dim}(P)\le n-2$. Hence we have $\mathrm {dim}(P)= n-2$.

Based on the above claim, it can be shown that $\Psi (\mathcal {L})\ne \Psi (\mathcal {L}')$. Suppose to the contrary that $\Psi (\mathcal {L})= \Psi (\mathcal {L}')$. Then there is a symmetry $w\in F(H)$ such that $P=w(P')$. Since $\mathrm {dim}(P)=\mathrm {dim}(P')= n-2$, we deduce that $H\cap w_i(H)= w(H\cap w_j(H))$, which contradicts the fact that $H\cap w_i(H)$ and $H\cap w_j(H)$ are not equivalent under the symmetries in $F(H)$. This completes the proof. $\square$

We can now give the definition of the subset $\mathcal {L}_1(H,k)$ of $\mathcal {L}(H,k)$. Notice that for each $1\le i\le h_1(H,k)$, $\Psi (\mathcal {L}(H\cap w_i(H),k))$ is a subset of $\mathcal {L}(H,k)$. By Theorem 7.3, these subsets are disjoint. We define $\mathcal {L}_1(H,k)$ to be the union of $\Psi (\mathcal {L}(H\cap w_i(H),k))$, where $1\le i\le h_1(H,k)$.

We proceed to define the subset $\mathcal {P}_1(H,k)$ of $\mathcal {P}(H,k)$. Let $\overline{\mathcal {L}}_1(H,k)$ be the complement of $\mathcal {L}_1(H,k)$, that is,

$$\begin{aligned} \overline{\mathcal {L}}_1(H,k)=\mathcal {L}(H,k)\backslash \mathcal {L}_1(H,k). \end{aligned}$$

(7.3)

In the above notation, for any local $0/1$-equivalence class $\mathcal {L}\in \overline{\mathcal {L}}_1(H,k)$ and any $0/1$-polytope $P\in \mathcal {L}$, if $w\in B_n$ is a symmetry such that $w(P)$ is contained in $H$, then $w(H)=H$. This yields that $\mathcal {L}$ is also a partial $0/1$-equivalence class of $H$. Consequently, when $2^{n-3}<k\le 2^{n-2}$, $\overline{\mathcal {L}}_1(H,k)$ is a subset of $\mathcal {P}(H,k)$. Define

$$\begin{aligned} \mathcal {P}_1(H,k)=\mathcal {P}(H,k)\backslash \overline{\mathcal {L}}_1(H,k). \end{aligned}$$

(7.4)

From (7.3) and (7.4), we see that $N_H(k)$ can be expressed in terms of the cardinalities of $\mathcal {L}(H,k)$, $\mathcal {L}_1(H,k)$ and $\mathcal {P}_1(H,k)$. More precisely,

$$\begin{aligned} N_H(k)&=|\mathcal {P}(H,k)|\nonumber \\&=|\overline{\mathcal {L}}_1(H,k)|+|\mathcal {P}_1(H,k)|\nonumber \\&=|\mathcal {L}(H,k)|-|\mathcal {L}_1(H,k)|+|\mathcal {P}_1(H,k)|. \end{aligned}$$

(7.5)

By Lemma 7.2, $|{\mathcal {L}}(H,k)|$ can be computed from the cycle index $Z_H(z)$. From Theorem 7.3, $|\mathcal {L}_1(H,k)|$ can be derived from the cardinalities of $\mathcal {L}(H\cap w(H),k)$, where $H\cap w(H)\in E_1(H,k)$. To compute $|\mathcal {P}_{1}(H,k)|$, we need a map $\Gamma$ defined as follows.

Let $h_2(H,k)$ denote the number of equivalence classes of $E(H,k)$ under the symmetries in $B_n$, and let

$$\begin{aligned} E_2(H,k)=\{H\cap w_{i}(H)\,|\, 1\le i\le h_2(H,k)\} \end{aligned}$$

be the set of representatives of these equivalence classes of $E(H,k)$. We define a map $\Gamma$ from the (disjoint) union of $\mathcal {P}(H\cap w_i(H),k)$, where $1\le i\le h_2(H,k)$, to $\mathcal {P}_{1}(H,k)$. Let $\mathcal {P}$ be a partial $0/1$-equivalence class in $\mathcal {P}(H\cap w_i(H),k)$. Then $\Gamma$ maps $\mathcal {P}$ to the unique partial $0/1$-equivalence class in $\mathcal {P}_1(H,k)$ that contains $\mathcal {P}$.

When $2^{n-3}<k\le 2^{n-2}$, it has been shown that each 0/1-polytope with $k$ vertices contained in the intersection $H\cap w_i(H)$ has dimension $n-2$. This enables us to use the same argument as in the proof Theorem 7.3 to reach the following assertion.

Theorem 7.4

For $n>4$ and $2^{n-3}<k\le 2^{n-2}$, the map $\Gamma$ is a bijection.

Combining Lemma 7.2, Theorem 7.3 and Theorem 7.4, formula (7.5) can be rewritten as

$$\begin{aligned} N_H(k)=\,\big [u_1^{k}u_2^{|V_n(H)|-k}\big ]C_H(u_1,u_2)&-\sum _{H\cap w(H)\in E_1(H,k)}|\mathcal {L}(H\cap w(H),k)|\nonumber \\&+\sum _{H\cap w(H)\in E_2(H,k)}|\mathcal {P}(H\cap w(H),k)|.\quad \end{aligned}$$

(7.6)

So, to compute $N_H(k)$, it is enough to determine $|\mathcal {L}(H\cap w(H),k)|$ and $|\mathcal {P}(H\cap w(H),k)|$. We can compute $|\mathcal {L}(H\cap w(H),k)|$ and $|\mathcal {P}(H\cap w(H),k)|$ by applying Pólya’s theorem.

We first consider $|\mathcal {L}(H\cap w(H),k)|$. Let $P$ and $P'$ be any two $0/1$-polytopes belonging to the same local $0/1$-equivalence class in $\mathcal {L}(H\cap w(H),k)$. Then there exists a symmetry $\sigma$ in $F(H)$ such that $\sigma (P)=P'$. It is clear from Theorem 1.1 that both $P$ and $P'$ have dimension $n-2$. So we deduce that $w'(H\cap w(H))=H\cap w(H)$.

Let $F_1(H,w)$ be the subgroup of $F(H)$ that stabilizes $H\cap w(H)$, that is,

$$\begin{aligned} F_1(H,w)=\left\{ \sigma \in F(H)\,|\, \sigma \left( H\cap w(H)\right) =H\cap w(H)\right\} . \end{aligned}$$

Denote by $V_n(H\cap w(H))$ the set of vertices of $Q_n$ contained in $H\cap w(H)$. Consider the action of $F_1(H,w)$ on $V_n(H\cap w(H))$. Assume that each vertex in $V_n(H\cap w(H))$ is assigned one of the two colors, say, black and white. Clearly, when $2^{n-3}< k\le 2^{n-2}$, this leads to a one-to-one correspondence between local 0/1-equivalence classes in $\mathcal {L}(H\cap w(H),k)$ and equivalence classes of colorings of the vertices in $V_n(H\cap w(H))$ with $k$ black vertices.

Denote by $Z_{(H,w)}({z})$ the cycle index of $F_1\left( H,w\right)$ acting on $V_n(H\cap w(H))$. Write $C_{(H,w)}(u_1,u_2)$ for the polynomial obtained from $Z_{(H,w)}({z})$ by substituting $z_i$ with $u_1^i+u_2^i$. For $2^{n-3}<k\le 2^{n-2}$, we obtain that

$$\begin{aligned} |\mathcal {L}(H\cap w(H),k)|=\big [u_1^{k}u_2^{|V_n(H\cap w(H))|-k}\big ]\, C_{(H,w)}(u_1,u_2). \end{aligned}$$

(7.7)

Similarly, we can use Pólya’s theorem to compute $|\mathcal {P}(H\cap w(H),k)|$. Let $F_2\left( H,w \right)$ be the subgroup of $B_n$ that stabilizes $H\cap w(H)$, that is,

$$\begin{aligned} F_2\left( H,w \right) =\big \{\sigma \in B_n\,|\, \sigma \left( H\cap w(H)\right) =H\cap w(H)\big \}. \end{aligned}$$

Denote by $Z_{H\cap w(H)}({z})$ the cycle index of $F_2\left( H,w\right)$ acting on $V_n(H\cap w(H))$. Write $C_{H\cap w(H)}(u_1,u_2)$ for the polynomial obtained from $Z_{H\cap w(H)}(z)$ by substituting $z_i$ with $u_1^i+u_2^i$. For $2^{n-3}<k\le 2^{n-2}$, we have

$$\begin{aligned} |\mathcal {P}(H\cap w(H),k)|=\big [u_1^{k}u_2^{|V_n(H\cap w(H))|-k}\big ]C_{H\cap w(H)}(u_1,u_2). \end{aligned}$$

(7.8)

Now, plugging (7.7) and (7.8) into (7.6), we arrive at the following formula for $N_H(k)$.

Theorem 7.5

Assume that $n>4$ and $2^{n-3}<k\le 2^{n-2}$. Let $H$ be a spanned hyperplane of $Q_n$ containing at least $k$ vertices of $Q_n$. Let $q(w)=|V_n(H\cap w(H))|$. Then we have

$$\begin{aligned} N_H(k)=\,\big [u_1^{k}u_2^{|V_n(H)|-k}\big ]C_H(u_1,u_2)&-\sum _{H\cap w(H)\in E_1(H,k)}\big [u_1^{k}u_2^{q(w)-k}\big ]C_{(H,w)}(u_1,u_2)\nonumber \\&+\sum _{H\cap w(H)\in E_2(H,k)}\big [u_1^{k}u_2^{q(w)- k}\big ]C_{H\cap w(H)}(u_1,u_2).\nonumber \\ \end{aligned}$$

(7.9)

For $n=6$ and $k=13,14,15,16$, we can use Theorem 7.5 to compute $N_H(k)$, where $H$ is a spanned hyperplane of $Q_6$ containing more than $12$ vertices. By the computation of Aichholzer [2], in addition to the spanned hyperplanes $H_6^1, H_6^2, H_6^3, H_6^4, H_6^5, H_6^6$, there are $8$ representatives of equivalence classes of spanned hyperplanes of $Q_6$ containing more than $12$ vertices, namely,

$$\begin{aligned}&H_1:x_1+x_2+x_3+2x_4=2,\\&H_2:x_1+x_2+x_3+x_4=1,\\&H_3:x_1+x_2+x_3+x_4+2x_5=3,\\&H_{4}:x_1+x_2+x_3+x_4+x_5+2x_6=3,\\&H_5:x_1+x_2+x_3+x_4+x_5+x_6=2,\\&H_6:x_1+x_2+x_3+x_4+2x_5=2,\\&H_7:x_1+x_2+x_3+2x_4+2x_5=3,\\&H_8:x_1+x_2+x_3+x_4+2x_5+2x_6=4. \end{aligned}$$

Using a Maple program, when $k=13,14,15,16$, it is routine to check that $E(H,k)=\emptyset$ for $H=H_6^3,H_6^4,H_6^5,H_6^6$ and $H=H_1,H_2,\ldots ,H_8$. Therefore, for these spanned hyperplanes, by Theorem 7.5 we obtain that

$$\begin{aligned} N_H(k)=\big [u_1^{k}u_2^{|V_n(H)|-k}\big ]\,C_H(u_1,u_2). \end{aligned}$$

(7.10)

The cycle indices $Z_H(z)$ for $H=H_6^3,H_6^4,H_6^5,H_6^6$ are given in Sect. 6. Using the techniques in Sect. 5, we can derive the cycle indices for $H_1,H_2,\ldots , H_5$, which are given below.

$$\begin{aligned} Z_{H_1}({z})&=\frac{1}{48} \left( \begin{array}{l} z_1^{16}+4z_{12}z_4+4z_3^2z_6z_1^2z_2+2z_3^4z_1^4\\ +12z_2^8+8z_4^4+6z_1^4z_2^6+5z_1^8z_2^4+6z_6^2z_2^2 \end{array}\right) ,\\ Z_{H_2}({z})&=\frac{1}{192} \left( \begin{array}{l} z_1^{16}+68z_4^4+24z_6^2z_2^2+16z_{12}z_4+8z_3^4z_1^4\\ +\,\, 39z_2^8+12z_1^4z_2^6+8z_1^8z_2^4+16z_3^2z_6z_1^2z_2 \end{array}\right) ,\\ Z_{H_3}({z})&=\frac{1}{96} \left( \begin{array}{l} z_1^{16}+24z_6^2z_2^2+8z_3^4z_1^4+33z_2^8+ 6z_1^8z_2^4+24z_4^4 \end{array}\right) ,\\ Z_{H_4}({z})&=\frac{1}{120} \left( \begin{array}{l} z_1^{15}\!+\!24z_5^3\!+\!30z_2z_4^3z_1\!+\!20z_1z_3^2z_6z_2\!+\! 20z_1^3z_3^4\!+\!15z_1^3z_2^6\!+\!10z_1^7z_2^4 \end{array}\right) ,\\ Z_{H_5}({z})&=\frac{1}{720} \left( \begin{array}{l} z_1^{15}+120z_3z_6^2+144z_5^3+40z_3^5+180z_1z_2z_4^3\\ +\,\, 40z_1^3z_3^4+60z_1^3z_2^6+ 15z_1^7z_2^4+ 120z_1z_2z_3^2z_6 \end{array}\right) . \end{aligned}$$

For $H=H_6$, $H_7$, $H_8$, we obtain that $N_{H}(13)=2$, $N_{H}(14)=1$, and $N_{H}(15)=N_{H}(16)=0$ without computing the cycle index $Z_H(z)$. For example, for $H=H_6$, since $H_6$ contains $14$ vertices of $Q_6$, we have $N_{H}(14)=1$ and $N_{H}(15)=N_{H}(16)=0$. On the other hand, there are 14 0/1-polytopes with 13 vertices contained in $H_6$. It is easy to check that these 14 0/1-polytopes form two partial 0/1-equivalence classes. So we have $N_{H}(13)=2$. Similarly, we get $N_{H}(13)=2$, $N_{H}(14)=1$, and $N_{H}(15)=N_{H}(16)=0$ for $H=H_7, H_8$.

It remains to compute $N_H(k)$ for $H=H_6^1, H_6^2$ and $k=13,14,15,16$. We first consider $H_6^1$. Keep in mind that $H_6^1$ is the spanned hyperplane $x_1=0$. Thus, for $H_6^1$ and $k=13,14,15,16$, it is easily seen that the intersections $H_6^1\cap w(H_6^1)$ in $E(H_6^1,k)$ form only one equivalence class under the symmetries in $F(H_6^1)$ or $B_n$. A representative of this equivalence class can be chosen as $H_6^1\cap w(H_6^1)$, where $w=(1,2)(3)(4)(5)(6)$. So we have

$$\begin{aligned} E_1(H_6^1,k)=E_2(H_6^1,k) =\big \{(x_1,x_2,\ldots ,x_6)\in \mathbb {R}^6\,|\, x_1=x_2=0\big \}. \end{aligned}$$

Moreover, for $k=13,14,15,16$, it is easy to check that if two $0/1$-polytopes in $H_6^1\cap w(H_6^1)$ with $k$ vertices are equivalent under the symmetries in $B_n$, then they are equivalent under the symmetries in $F(H_6^1)$. This implies that each local 0/1-equivalence class of $H_6^1\cap w(H_6^1)$ is also a partial 0/1-equivalence class of $H_6^1\cap w(H_6^1)$ and vice versa. Hence we obtain

$$\begin{aligned} \mathcal {L}(H_6^1\cap w(H_6^1),k)=\mathcal {P}(H_6^1\cap w(H_6^1),k). \end{aligned}$$

Therefore, for $k=13,14,15,16$, by formula (7.6) we have

$$\begin{aligned} N_{H_6^1}(k)=\big [u_1^{k}u_2^{32-k}\big ]C_{H_6^1}(u_1,u_2). \end{aligned}$$

(7.11)

We now compute $N_{H_6^2}(k)$ for $k=13,14,15,16$. Recall that $H_6^2$ is the spanned hyperplane $x_1+x_2=1$. It is not hard to check that the intersections $H_6^2\cap w(H_6^2)$ in $E(H_6^2,k)$ form two equivalence classes under the symmetries in $F(H_6^2)$ or $B_n$. Moreover, each equivalence class in $E(H_6^2,k)$ under the symmetries in $F(H_6^2)$ is an equivalence class in $E(H_6^2,k)$ under the symmetries in $B_n$ and vice versa. The representatives of these two equivalence classes can be chosen as $H_6^2\cap w_1(H_6^2)$ and $H_6^2\cap w_2(H_6^2)$, where $w_1=(1,3,2)(4)(5)(6)$ and $w_2=(1,3)(2,4)(5)(6)$. Notice that the intersections $H_6^2\cap w_1(H_6^2)$ and $H_6^2\cap w_2(H_6^2)$ are of the following form:

$$\begin{aligned} H_6^2\cap w_1(H_6^2)=&\, \big \{(x_1,x_2,\ldots ,x_6)\in \mathbb {R}^6\,|\, x_1+x_2=1\ \ \text {and}\ \ x_2+x_3=1 \big \}, \\ H_6^2\cap w_2(H_6^2)=&\,\big \{(x_1,x_2,\ldots ,x_6)\in \mathbb {R}^6\,|\, x_1+x_2=1\ \ \text {and}\ \ x_3+x_4=1 \big \}. \end{aligned}$$

Since the set of vertices contained in $H_6^2\cap w_1(H_6^2)$ is

$$\begin{aligned}&\{(1,0,1,x_4,x_5,x_6),\, (0,1,0,x_4,x_5,x_6)\,|\, x_i=0\;\hbox {or 1 for}\; i=4,5,6\}, \end{aligned}$$

it is easy to check that for $k=13,14,15,16$, if two 0/1-polytopes contained in $H_6^2\cap w_1(H_6^2)$ with $k$ vertices are equivalent under the symmetries in $B_n$, then they are equivalent under the symmetries in $F(H_6^2)$. This means that each local 0/1-equivalence class of $H_6^2\cap w_1(H_6^2)$ is also a partial 0/1-equivalence class of $H_6^2\cap w_1(H_6^2)$ and vice versa. So, we have

$$\begin{aligned} \mathcal {L}(H_6^2\cap w_1(H_6^2),k)=\mathcal {P}(H_6^2\cap w_1(H_6^2),k). \end{aligned}$$

Therefore, by formula (7.6) we obtain that for $k=13,14,15,16$,

$$\begin{aligned} N_{H_6^2}(k)=\big [u_1^{k}u_2^{32-k}\big ]C_{H_6^2}(u_1,u_2)+ |\mathcal {P}(H_6^2\cap w(H_6^2),k)-|\mathcal {L}(H_6^2\cap w(H_6^2),k),\nonumber \\ \end{aligned}$$

(7.12)

where $w=(1,3)(2,4)(5)(6)$.

Combining (7.10), (7.11) and (7.12) , for $n=6$ and $k=13,14,15,16$, we obtain that

$$\begin{aligned} \sum _{i=1}^{h(6,k)}|A_i|=&\sum _{i=1}^6\big [u_1^{k}u_2^{|V_6(H_6^i)| -k}\big ]C_{H_6^i}(u_1,u_2)+\sum _{i=1}^8\big [u_1^{k}u_2^{|V_6(H_i) |-k}\big ]C_{H_i}(u_1,u_2)\nonumber \\&+|\mathcal {P}(H_6^2\cap w(H_6^2),k)-|\mathcal {L}(H_6^2\cap w(H_6^2),k), \end{aligned}$$

(7.13)

where $w=(1,3)(2,4)(5)(6)$.

By Lemma 7.1, to determine $H_6(k)$ for $k=13,14,15,16$, we still need to compute $|A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}|$ for $m\ge 2$. We first consider the case $m=2$. The computation of the general case can be carried out in the same way.

We now demonstrate how to compute $|A_i\cap A_j|$ for $1\le i<j\le h(n,k)$. Let $E(H_i,H_j,k)$ be the set of affine subspaces $H_i\cap w(H_j)$ that contain at least $k$ vertices of $Q_n$. Denote by $h(H_i,H_j,k)$ the number of equivalence classes in $E(H_i,H_j,k)$ under the symmetries in $B_n$, and let

$$\begin{aligned} E_1(H_i,H_j,k)=\left\{ H_i\cap w_t(H_j)\,|\, 1\le t \le h(H_i,H_j,k)\right\} \end{aligned}$$

be the set of representatives of equivalence classes in $E(H_i,H_j,k)$.

We consider the union of the sets $\mathcal {P}(H_i\cap w_t(H_j),k)$ of partial 0/1-equivalence classes of $H_i\cap w_t(H_j)$ with $k$ vertices, where $1\le t\le h(H_i,H_j,k)$, and we define a map $\Upsilon$ from this set of partial 0/1-equivalence classes to $A_i\cap A_j$. Let $\mathcal {P}$ be a partial $0/1$-equivalence class in $\mathcal {P}(H_i\cap w_t(H_j),k)$. Then there is a unique $0/1$-equivalence class $\mathcal {P}'$ in $A_i\cap A_j$ that contains $\mathcal {P}$. Define $\Upsilon (\mathcal {P})=\mathcal {P}'$. We have the following property. The proof is omitted since it is similar to that of Theorem 7.3.

Theorem 7.6

For $n>4$ and $2^{n-3}<k\le 2^{n-2}$, the map $\Upsilon$ is a bijection.

As a consequence of Theorem 7.6, for $n>4$ and $2^{n-3}<k\le 2^{n-2}$, we have

$$\begin{aligned} |A_i\cap A_j|=\sum _{t=1}^{h(H_i,H_j,k)}|\mathcal {P}(H_i\cap w_t(H_j),k)|. \end{aligned}$$

(7.14)

The above approach can be used to determine $|A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}|$ for $m\ge 3$. Let

$$\begin{aligned} E(H_{i_1},\ldots ,H_{i_m},k) \end{aligned}$$

be the set of affine subspaces $H_{i_1}\cap w_2(H_{i_2})\cap \cdots \cap w_m(H_{i_m})$, where $w_2,\ldots , w_m$ are symmetries in $B_n$ such that $H_{i_1}\cap w_2(H_{i_2})\cap \cdots \cap w_m(H_{i_m})$ contains at least $k$ vertices of $Q_n$. Denote by $E_1(H_{i_1},\ldots ,H_{i_m},k)$ the set of representatives of equivalence classes of $E(H_{i_1},\ldots ,H_{i_m},k)$ under the symmetries in $B_n$.

Consider the union of the sets $\mathcal {P}(H_{i_1}\cap w_2(H_{i_2})\cap \cdots \cap w_m(H_{i_m}),k)$ of partial 0/1-equivalence classes, where

$$\begin{aligned} H_{i_1}\cap w_2(H_{i_2})\cap \cdots \cap w_m(H_{i_m})\in E_1(H_{i_1},\ldots ,H_{i_m},k). \end{aligned}$$

We define a map $\Omega$ from this set of partial 0/1-equivalence classes to $A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}$. Let $\mathcal {P}$ be a partial 0/1-equivalence of $H_{i_1}\cap w_2(H_{i_2})\cap \cdots \cap w_m(H_{i_m})$. Then $\Omega$ maps $\mathcal {P}$ to the unique $0/1$-equivalence class in $A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}$ that contains $\mathcal {P}$. Using the same argument as in the proof of Theorem 7.3, we obtain the following property.

Theorem 7.7

For $n>4$ and $2^{n-3}<k\le 2^{n-2}$, the map $\Omega$ is a bijection.

As a consequence of Theorem 7.7, we see that for $n> 4$ and $2^{n-3}<k\le 2^{n-2}$,

$$\begin{aligned} |A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}|=\sum |\mathcal {P}(H_{i_1} \cap w_2(H_{i_2})\cap \cdots \cap w_m(H_{i_m}),k)|, \end{aligned}$$

(7.15)

where the sum ranges over the representatives $H_{i_1}\cap w_2(H_{i_2})\cap \cdots \cap w_m(H_{i_m})$ of equivalence classes in $E(H_{i_1},\ldots ,H_{i_m},k)$.

The following theorem shows that for $m\ge 3$, the set $E(H_{i_1},\ldots ,H_{i_m},k)$ is empty under certain conditions. When $n=6$ and $k=13,14,15,16$, this property allows us to deduce that for any $m\ge 4$ and any spanned hyperplanes $H_{i_1},\ldots ,H_{i_m}$, the set $E(H_{i_1},\ldots ,H_{i_m},k)$ is empty.

Theorem 7.8

Let $n>4$ and $2^{n-3}<k\le 2^{n-2}$. If there exist $1\le p< q\le m$ such that $E(H_{i_{p}},H_{i_{q}},k)$ is empty, then $E(H_{i_1},\ldots ,H_{i_m},k)$ is empty.

Proof

Assume that there exist $1\le p< q\le m$ such that $E(H_{i_{p}},H_{i_{q}},k)$ is empty. Suppose to the contrary that $E(H_{i_1},\ldots ,H_{i_m},k)$ is nonempty. Let

$$\begin{aligned} S=H_{i_1}\cap w_2(H_{i_2})\cap \cdots \cap w_m(H_{i_m}) \end{aligned}$$

be an affine space belonging to $E(H_{i_1},\ldots ,H_{i_m},k)$. Let $w_1$ be the identity element $e$ in $B_n$. We claim that

$$\begin{aligned} S=w_p(H_{i_p})\cap w_q(H_{i_q}). \end{aligned}$$

(7.16)

Clearly, $S\subseteq w_p(H_{i_p})\cap w_q(H_{i_q})$. Since $\mathrm {dim}(w_p(H_{i_p})\cap w_q(H_{i_q}))=n-2$, to prove (7.16), it suffices to show that $\mathrm {dim}(S)=n-2$. Since $S$ contains more than $2^{n-3}$ vertices of $Q_n$, by Theorem 1.1, we deduce that $\mathrm {dim}(S)\ge n-2$. But $S\subseteq w_p(H_{i_p})\cap w_q(H_{i_q})$, so we have $\mathrm {dim}(S)= n-2$. This proves the claim.

Let $w=(w_{p})^{-1}$. By (7.16), we see that $w(S)$ is an affine space in $E(H_{i_{p}},H_{i_{q}},k)$, contradicting the assumption that $E(H_{i_{p}},H_{i_{q}},k)$ is empty. This completes the proof. $\square$

Using formulas (7.14) and (7.15), we can compute $|A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}|$ for $n=6$, $k=13,14,15,16$ and $m\ge 2$. We first consider the case when $m=2$. Using a Maple program, it can be checked that there are only four pairs for which $E(H_i,H_j,k)$ is nonempty. Recall that for $t\le n$, $H_n^t$ denotes the hyperplane $x_1+\cdots +x_t= \left\lfloor t/2\right\rfloor$ in $\mathbb {R}^n$.

Case 1: $(H_6^1, H_6^2)$. In this case, it can be easily checked that the affine subspaces in $E(H_6^1, H_6^2,k)$ form two equivalence classes under the symmetries in $B_n$. The representatives can be chosen as $H_6^1\cap H_6^2$ and $H_6^1\cap w(H_6^2)$, where $w=(1,3,2)(4)(5)(6)$. Notice that $w(H_6^2)$ is the hyperplane $x_2+x_3=1$. So we have

$$\begin{aligned} E_1(H_6^1, H_6^2,k)=\big \{H_6^1\cap H_6^2,\ H_6^1\cap H_6^3\big \}. \end{aligned}$$

(7.17)

Case 2: $(H_6^1,H_6^3)$. In this case, the affine subspaces in $E(H_6^1, H_6^3,k)$ form only one equivalence class under the symmetries in $B_n$. The representative can be chosen as $H_6^1\cap H_6^3$, and hence

$$\begin{aligned} E_1(H_6^1, H_6^3,k)=\big \{H_6^1\cap H_6^3\big \}. \end{aligned}$$

(7.18)

Case 3: $(H_6^2,H_6^3)$. This case is similar to Case 2. We have

$$\begin{aligned} E_1(H_6^2, H_6^3,k)=\big \{H_6^1\cap H_6^3\big \}. \end{aligned}$$

(7.19)

Case 4: $(H_6^2, H_6^4)$. In this case, it can be verified that

$$\begin{aligned} E_1(H_6^2, H_6^4,k)=\big \{H_6^2\cap H_6^4\big \}. \end{aligned}$$

(7.20)

By (7.17)–(7.20), we obtain that for $n=6$ and $k=13,14,15,16$,

$$\begin{aligned} \sum _{1\le i<j\le h(6,k)}|A_i\cap A_j|=|\mathcal {P}(H_6^1\cap H_6^2,k)|+3|\mathcal {P}(H_6^1\cap H_6^3,k)|+|\mathcal {P}(H_6^2\cap H_6^4,k)|. \end{aligned}$$

(7.21)

Finally, we compute $|A_{i_1}\cap A_{i_2}\cap \cdots \cap A_{i_m}|$ for $n=6$, $k=13,14,15,16$ and $m\ge 3$. We claim that $E(H_{i_1},\ldots ,H_{i_m},k)$ is empty for any $m\ge 4$. If this is not the case, then, by Theorem 7.8, for any $1\le p< q\le m$, $E(H_{i_p},H_{i_q},k)$ is nonempty. Since $m\ge 4$, there are at least six pairs $(H_i,H_j)$ with $1\le i<j\le h(6,k)$ for which $E(H_i,H_j,k)$ is nonempty. However, as shown before, there are only four pairs $(H_i,H_j)$ with $1\le i<j\le h(6,k)$ for which $E(H_i,H_j,k)$ is nonempty, leading to a contradiction. So the claim is proved.

When $m=3$, it is easy to check that $E(H_{i_1},H_{i_2},H_{i_3},k)$ is nonempty if and only if

$$\begin{aligned} (H_{i_1},H_{i_2},H_{i_3})=(H_6^1, H_6^2,H_6^3). \end{aligned}$$

Moreover, we have

$$\begin{aligned} E_1(H_6^1,H_6^2,H_6^3,k)=\big \{H_6^1\cap H_6^3\big \}. \end{aligned}$$

Thus, for $n=6$, $k=13,14,15,16$ and $m\ge 3$, we have

$$\begin{aligned} \sum _{1\le i_1<\cdots <i_m \le h(6,k)}|A_{i_1}\cap \cdots \cap A_{i_m}|= \Big \{\begin{array}{l@{\quad }l} |\mathcal {P}(H_6^1\cap H_6^3,k)| &{} \text{ if }\; m=3,\\ 0 &{} \text{ if }\; m>3.\end{array} \end{aligned}$$

(7.22)

By Lemma 7.1 and formulas (7.13), (7.21) and (7.22), we deduce that for $n=6$ and $k=13,14,15,16$,

$$\begin{aligned} H_6(k)=&\sum _{i=1}^6\big [u_1^{k}u_2^{|V_6(H_6^i)|-k}\big ]C_{H_6^i} (u_1,u_2)+\sum _{i=1}^8\big [u_1^{k}u_2^{|V_6(H_i)|-k}\big ]C_{H_i} (u_1,u_2)\nonumber \\&+|\mathcal {P}(H_6^2\cap w(H_6^2),k)|-|\mathcal {P}(H_6^1\cap H_6^2,k)|-2|\mathcal {P}(H_6^1\cap H_6^3,k)|\nonumber \\&-|\mathcal {P}(H_6^2\cap H_6^4,k)| -|\mathcal {L}(H_6^2\cap w(H_6^2),k)|, \end{aligned}$$

(7.23)

where $w=(1,3)(2,4)(5)(6)$. Notice that for $w=(1,3)(2,4)(5)(6)$,

$$\begin{aligned} H_6^2\cap w(H_6^2)\!=\!H_6^2\cap H_6^4\!=\!\big \{(x_1,x_2,\ldots ,x_6)\in \mathbb {R}^6\,|\, x_1 \!+\! x_2\!=\!1\ \ \text {and}\ \ x_3 \!+\! x_4 \!=\! 1 \big \}. \end{aligned}$$

Thus, (7.23) can be rewritten as

$$\begin{aligned} H_6(k)=&\sum _{i=1}^6\big [u_1^{k}u_2^{|V_6(H_6^i)|-k}\big ] C_{H_6^i}(u_1,u_2)+\sum _{i=1}^8\big [u_1^{k}u_2^{|V_6(H_i)| -k}\big ]C_{H_i}(u_1,u_2)\nonumber \\&-|\mathcal {P}(H_6^1\cap H_6^2,k)|-2|\mathcal {P}(H_6^1\cap H_6^3,k)| -|\mathcal {L}(H_6^2\cap w(H_6^2),k)|, \end{aligned}$$

(7.24)

where $w=(1,3)(2,4)(5)(6)$.

As for $|\mathcal {P}(H_6^1\cap H_6^2,k)|$, we notice that

$$\begin{aligned} H_6^1\cap H_6^2=\{(0,1,x_3,x_4,x_5,x_6)\,|\, x_i=0\;\text{ or } \text{1 } \text{ for }\; i=3,4,5,6\}. \end{aligned}$$

Thus the vertices of $Q_6$ contained in $H_6^1\cap H_6^2$ are in one-to-one correspondence with the vertices of $Q_4$. To be more specific, given a vertex $(0,1,x_3,x_4,x_5,x_6)$ of $Q_6$ contained in $H_6^1\cap H_6^2$, we get a vertex $(x_3,x_4,x_5,x_6)$ of $Q_4$ and vice versa. Moreover, the partial 0/1-equivalence classes of $H_6^1\cap H_6^2$ are in one-to-one correspondence with the 0/1-equivalence classes of $Q_4$. Hence, for $n=6$ and $k=13,14,15,16$, we have

$$\begin{aligned} |\mathcal {P}(H_6^1\cap H_6^2,k)|=\big [u_1^{k} u_2^{16-k}\big ]C_4(u_1,u_2). \end{aligned}$$

(7.25)

We now compute $|\mathcal {P}(H_6^1\cap H_6^3,k)|$. Since

$$\begin{aligned} H_6^1\cap H_6^3=\{(0,x_2,x_3,x_4,x_5,x_6)\,|\, x_2+x_3=1\}, \end{aligned}$$

we see that each vertex $(0,x_2,x_3,x_4,x_5,x_6)$ of $Q_6$ contained in $H_6^1\cap H_6^3$ corresponds to a vertex $(x_2,x_3,x_4,x_5,x_6)$ of $Q_5$ contained in the spanned hyperplane $H_5^2$ of $Q_5$ and vice versa. Hence the partial 0/1-equivalence classes of $H_6^1\cap H_6^3$ are in one-to-one correspondence with the partial 0/1-equivalence classes of the spanned hyperplane $H_5^2$ of $Q_5$. Therefore, for $n=6$ and $k=13,14,15,16$, we have

$$\begin{aligned} |\mathcal {P}(H_6^1\cap H_6^3,k)|=\big [u_1^{k}u_2^{16 -k}\big ]C_{H_5^2}(u_1,u_2). \end{aligned}$$

(7.26)

Finally, we determine $|\mathcal {L}(H_6^2\cap w(H_6^2),k)|$ for $w=(1,3)(2,4)(5)(6)$. By (7.7), we see that $|\mathcal {L}(H_6^2\cap w(H_6^2),k)|$ can be obtained from the cycle index $Z_{(H_6^2,w)}({z})$. Using the technique in Sect. 5, we obtain that

$$\begin{aligned} Z_{(H_6^2,w)}(z)=\frac{1}{32} \left( z_1^{16}+21z_2^8 +8z_4^4+2z_1^8z_2^4\right) . \end{aligned}$$

Hence

$$\begin{aligned} |\mathcal {L}(H_6^2\cap H_6^4,k)|=\big [u_1^{k}u_2^{16 -k}\big ]C_{(H_6^2,w)}(u_1,u_2), \end{aligned}$$

(7.27)

where $C_{(H_6^2,w)}(u_1,u_2)$ is the polynomial obtained from $Z_{(H_6^2,w)}({z})$ by substituting $z_i$ with $u_1^i+u_2^i$.

Using (7.24)–(7.27), we can compute $H_6(k)$ for $k=13,14,15,16$. Since $F_6(k)=A_6(k)-H_6(k)$, we obtain $F_6(k)$ for $k=13,14,15,16$ as given in Table 7.

Table 7 $F_6(k)$ for $k=13,14,15,16$