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This article is cited in 2 scientific papers (total in 2 papers) Semiregular Gosset polytopes V. N. Berestovskiia, Yu. G. Nikonorovb a Sobolev Institute of Mathematics, Siberian Branch of the Russian Academy of Sciences, Novosibirsk b Southern Mathematical Institute of the Vladikavkaz Scientific Center of the Russian Academy of Sciences, Vladikavkaz
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2022, Volume 86, Issue 4, DOI: https://doi.org/10.4213/im9169 
Bibliographic databases:
    § 1. Introduction and main resultIn the paper [1], the class of finite homogeneous metric spaces and its subclasses of normal, generalised normal, strongly generalised normal homogeneous spaces, and Clifford–Wolf homogeneous spaces and relations between these classes were introduced and studied. Similar classes have been studied for Riemannian manifolds, see the book [2], which contains an extensive bibliography on the topic. Special cases of finite homogeneous metric spaces are sets of vertices of compact convex polytopes (including regular and semiregular ones [3]–[5]) in Euclidean spaces with a group of isometries which is transitive on the set of vertices. By a semiregular polytope we mean a compact convex polytope in Euclidean space with regular facets and a homogeneous set of vertices. In [1], the sets of vertices of regular and semiregular polytopes in three-dimensional Euclidean space, as well as of hypercubes, their dual polytopes (hyperoctahedra), and regular simplices in Euclidean spaces of arbitrary dimension were examined for belonging to the corresponding subclasses. In [6], the metric properties of all regular and Archimedean polytopes are investigated. In particular, special properties of Archimedean solids and their explicit representations as convex hulls of the set their vertices are studied in [6]. However, the main result of this paper is the proof of the fact that the set of vertices of every regular polytope of dimension at least $4$, different from a 120-cell in $\mathbb{R}^4$, is Clifford–Wolf homogeneous. The classification of regular polytopes of arbitrary dimension was obtained for the first time by L. Schläfli and presented in his book [7], see also the book of H. S. M. Coxeter [4] and his papers [8]–[10]. The list of semiregular polytopes of arbitrary dimension was first given without proof in the paper of Th. Gosset [11]. Later, this list occurred in the paper of E. L. Elte [12]. The proof of the completeness of this list was obtained much later by G. Blind and R. Blind, see the paper [13] and the works of the authors of the paper cited in the bibliography. In the present paper, we are interested only in semiregular (non-regular) polytopes in $\mathbb{R}^n$ for $n\geqslant 4$, which we call Gosset polytopes. Much more information can be obtained on the site [14]. By the classification of semiregular polytopes in $\mathbb{R}^n$, $n\geqslant 4$ (see [13] and [11]), in addition to regular polytopes, there are three semiregular polytopes in $\mathbb{R}^4$ and there is one semiregular polytope in $\mathbb{R}^n$ for $n=5,6,7,8$. For $n=4$, the semiregular polytopes are full truncations, that is, the convex hulls of the midpoints of the edges of the four-dimensional regular simplex, of the $600$-cell, as well as of a snub $24$-cell (for details, see the next section). We denote a unique (up to similarity) semiregular Gosset polytope in $\mathbb{R}^n$ for $n\in\{5,6,7,8\}$ by the symbol $\operatorname{Goss}_n$. Now we can formulate the main result of the present paper (all necessary auxiliary definitions are given in the next section). Theorem 1. Let $P$ be one of the semiregular Gosset polytopes in $\mathbb{R}^n$, $n\geqslant 4$, and let $M$ be the set of its vertices endowed with the metric induced from $\mathbb{R}^n$. Then the following assertions hold: 1) $M$ is Clifford–Wolf homogeneous if $P=\operatorname{Goss}_5$ or $P=\operatorname{Goss}_8$; 2) $M$ is normally homogeneous, but not Clifford–Wolf homogeneous, if $P$ is the full truncation of a four-dimensional regular simplex, $P=\operatorname{Goss}_6$, or $P=\operatorname{Goss}_7$; 3) $M$ is not a normal homogeneous metric space if $P$ is the full truncation of the $600$-cell or the snub $24$-cell. This theorem completes the description of the metric properties of the vertex sets of regular and semiregular polytopes in Euclidean spaces from the point of view of normal homogeneity and Clifford–Wolf homogeneity (the exact statements of previously obtained results are given in the next section). All available information on this topic is summarised in Table 1, where $n$ is the dimension of the polytope, the degrees of regularity are $\mathrm{R}$ (a regular polytope) and $\mathrm{SR}$ (a semiregular (non-regular) polytope), and the metric properties of the set of vertices are $\mathrm{NH}$ (the normal homogeneity) and $\mathrm{CWH}$ (the Clifford–Wolf homogeneity). Table 1.Metric properties of regular and semiregular polytopes
Theorem 1 follows directly from Proposition 20, Corollaries 4 and 6, and Theorems 4, 5, 7, 8, and 9 proved below. It should be noted that, in this paper, we also obtain results for other classes of polytopes (see, for example, Propositions 12, 14, and 19) and study some special properties of Gosset polytopes (Theorem 6 and Propositions 21 and 22). In addition, a number of unsolved problems are formulated in different parts of the text. The structure of this paper is as follows. In § 2 we present necessary definitions and earlier results on the metric properties of sets of vertices of regular and semiregular polytopes in Euclidean spaces. In § 3, Gosset lattices and polytopes are discussed in the dimensions $6$, $7$, and $8$. Section 4 is devoted to the proof of some auxiliary assertions. Finally, in §§ 5–8, the Gosset polytopes are studied in the dimensions $4$, $5$, $6$, $7$, and $8$. § 2. PreliminariesWe assume that the space $\mathbb{R}^n$ is equipped with the standard inner product and the metric induced by it. Below, $\rho$ stands for the Euclidean distance between points in $\mathbb{R}^n$, $n\geqslant 1$. We need some definitions and results of [1] and [6]. Definition 1. A finite metric space $(M,d)$ is said to be homogeneous if for any points $x,y\in M$ there is an isometry $f$ of the space $(M,d)$ onto itself such that $f(x)=y$. Definition 2. A mapping $f\colon (M_1,d_1)\to (M_2,d_2)$ of metric spaces is said to be submetric if $f(B_{M_1}(x_1,r))=B_{M_2}(f(x_1),r)$ for every point $x_1\in M_1$ and every number $r\geqslant 0$, where $B_{M_i}(x_i,r)$ stands for the closed ball in $M_i$ with the centre $x_i$ and of radius $r$ for $i=1,2$ [15]. Definition 3. A finite homogeneous metric space $(M,d)$ is called normal homogeneous if for the group $G$ of isometries of $(M,d)$ and the stabiliser $H$ in $G$ of some point $x\in M$ there is a subgroup $\Gamma$ of $G$ transitive on $M$ and a bi-invariant metric $\sigma$ on $\Gamma$ such that the canonical projection $\pi\colon (\Gamma,\sigma)\to (\Gamma/(\Gamma\cap H),d)=(M,d)$ is a submetry. Definition 4. A finite metric space $(M,d)$ is said to be generalised normal homogeneous if for any points $x,y\in M$ there is an isometry $f$ (which is called a $\delta$-shift at a point $x$) of the space $(M,d)$ onto itself such that $f(x)=y$ and $d(x,f(x))\geqslant d(z,f(z))$ for any $z\in M$. Definition 5. By the Clifford–Wolf translation of a metric space $(M,d)$ we mean an isometry $f$ of the space $(M,d)$ onto itself that shifts all points of $M$ by the same distance, that is, $d(x,f(x))=d(y,f(y))$ for any $x,y\in M$. Definition 6. A finite metric space $(M,d)$ is called Clifford–Wolf homogeneous (briefly, $\mathrm{CW}$-homogeneous) if for any points $x,y\in M$ there is a Clifford–Wolf translation $f$ of the space $(M,d)$ such that $f(x)=y$. Let $\mathrm{FCWHS}$, $\mathrm{FGNHS}$, $\mathrm{FNHS}$, $\mathrm{FHS}$ denote the classes of finite $\mathrm{CW}$-homogeneous spaces, finite generalised normal homogeneous spaces, finite normal homogeneous spaces, and finite homogeneous spaces, respectively. In Theorem 1 of [1] it is proved that
$$
\begin{equation*}
\mathrm{FCWHS}\subset \mathrm{FGNHS} = \mathrm{FNHS}\subset \mathrm{FHS};
\end{equation*}
\notag
$$
here the indicated inclusions are strict. We stress that the properties of normal homogeneity and generalised normal homogeneity for finite metric spaces are equivalent (which is not the case, say, in the case of Riemannian manifolds [2]). We note that the verification of the condition of generalised normal homogeneity is simpler than the verification of the condition of normal homogeneity, and we shall use this circumstance below without additional comments. We now recall the definitions of homogeneous, regular, and semiregular polytopes in Euclidean spaces. We say that an $n$-dimensional polytope $P$ in $\mathbb{R}^n$ is homogeneous (or vertex-transitive) if its isometry group acts transitively on the set of its vertices. Further, $P$ is called a polytope with regular facets (a polytope with congruent facets, respectively) if all its facets are regular polytopes (congruent polytopes, respectively). A one-dimensional polytope is a closed interval bounded by two endpoints. It is regular by definition. The 2D regular polytopes are regular polygons on a Euclidean plane. a convex $n$-dimensional polytope for $n\geqslant 3$ is called regular if it is homogeneous and all its facets are regular polytopes congruent to each other. As is well known, there are only five regular three-dimensional polytopes: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. These polytopes are traditionally called Platonic solids. In [1] and [6], metric properties of the set of vertices are studied for every regular Euclidean polytope. Obviously, the vertices of a one-dimensional polytope (closed interval) and the vertices of every regular two-dimensional polytope (regular polygon) form Clifford–Wolf homogeneous metric spaces. In the three-dimensional case, we have a more curious situation. The vertex sets of a tetrahedron, cube, or octahedron in 3D Euclidean space form Clifford–Wolf homogeneous metric spaces [1]. Here the set of vertices of an icosahedron forms a normal homogeneous and not Clifford–Wolf homogeneous metric space, and the set of vertices of the dodecahedron forms a metric space that is not even normal homogeneous (see [1; Theorems 7 and 8]). For the multidimensional case, the following theorem holds. Theorem 2 (see [6; Theorem 1]). The set of vertices $M$ of every regular polytope of dimension $n \geqslant 4$ with the metric induced from $\mathbb{R}^n$ is Clifford–Wolf homogeneous, except for the $120$-cell in $\mathbb{R}^4$, whose set of vertices is even not normal homogeneous. We now recall the definition of a wider class of semiregular convex polytopes. For $n=1$ and $n=2$ the semiregular polytopes are defined as regular ones. A convex $n$-dimensional polytope for $n\geqslant 3$ is said to be semiregular if it is homogeneous and all its facets are regular polytopes. In three-dimensional space (in addition to the Platonic solids) there are the following semiregular polytopes: $13$ Archimedean solids and two infinite series of right prisms and antiprisms. Theorem 3 (see Theorem 9 of [1]). The set of vertices of every Archimedean solid is a homogeneous and not normal homogeneous metric space. It should also be noted that the set of vertices of any right prism and the set of vertices of any right antiprism are Clifford–Wolf homogeneous, which essentially distinguishes prisms and antiprisms from Archimedean solids. We now describe the multidimensional Gosset polytopes in more detail. As already noted, in $\mathbb{R}^n$, $n\geqslant 4$, in addition to the correct polytopes, there are three semiregular polytopes in $\mathbb{R}^4$ and by one semiregular polytope in $\mathbb{R}^n$ for $n=5,6,7,8$ (see [13] and [11]). Let us begin with the case $n=4$. The classification of regular polytopes in $\mathbb{R}^4$ is given in [6; Table 2]. There are six regular polytopes of this kind: the four-dimensional regular simplex ($5$-cell), hypercube ($8$-cell), hyperoctahedron ($16$-cell), $24$-cell, $120$-cell, and $600$-cell, whose facets are, respectively, regular tetrahedra, cubes, tetrahedra, octahedra, dodecahedrons, and tetrahedra according to their Schläfli symbols $\{3,3,3\}$, $\{4,3,3\}$, $\{3,3,4\}$, $\{3,4,3\}$, $\{5,3,3\}$, $\{3,3,5\}$. The full truncations of exactly three regular 4-polytopes (the regular simplex, hyperoctahedron, and 600-cell) are semiregular polytopes. In this case, the full truncation of the hyperoctahedron has $16+8 = 24$ octahedra as facets (according to its Schläfli symbol and by the number of its facets and vertices) and $24$ vertices by the number of its edges, as is the case for a regular $24$-cell, and it is a regular $24$-cell indeed [6]. The full truncations of the regular simplex and $600$-cell are semiregular Gosset polytopes. The full truncation of the regular four-dimensional simplex has five tetrahedra (three-dimensional regular simplices) and five octahedra as facets and $C_5^2=10$ vertices according to his Schläfli symbol and the number of edges. The full truncation of the $600$-cell has $120$ icosahedrons and $600$ octahedrons as facets and $720$ vertices according to its Schläfli symbol and the number of vertices, facets, and edges of the $600$-cell according to Table 2 in [6]. This agrees with the descriptions of the last two resulting polytopes in [11]. The third Gosset polytope is a snub $24$-cell (see [16]). According to its description in [11], it has $24$ icosahedrons, $120$ tetrahedra (totally $24+120=144=12^2$ facets), $480$ triangles, $432$ edges, and $96$ vertices. In addition, in [11] it is said about the intersections of faces of different dimension and about the fact that the radius of the circumscribed sphere is equal to the length of its edge multiplied by the golden ratio $\varphi=(1+\sqrt{5}\,)/2$. The snub cube (the dodecahedron) is the convex hull of all regular $4$-gons (5-gons) each of which is located inside $6$ ($12$) $4$-angle ($5$-angle) faces of the cube (dodecahedron). At the same time, it follows from the description of the Gosset polytope under consideration that it was obtained in a somewhat different way than the snub cube and snub dodecahedron. Since the original $24$-cell $\{3,4,3\}$ has $96$ edges, a conjecture arises that $96$ vertices of the snub $24$-cell $s\{3,4,3\}$ are located on the edges of $\{3,4,3\}$; however, not at the middle of them, but in some other way. It turns out that this is indeed the case. Here is a quote from H. Coxeter’s paper [9] about a result proved in his book [4]: Every vertex of $s\{3,4,3\}$ lies on an edge of $\{3,4,3\}$, but not at the middle of the edge. In fact, $96$ vertices of $s\{3,4,3\}$ divide $96$ edges of $24$-cell $\{3,4,3\}$ according to the golden ratio $\varphi:1$ or $\varphi^{-1}:\varphi^{-2}$ ($\varphi^{-1}+\varphi^{-2}\,{=}\,1$). In $\{3,4,3\}$, given by permutations of coordinates of the points $(\pm 1,\pm 1,0,0)$, a typical edge $[(1,1,0,0),(1,0,1,0)]$ is divided in the specified ratio by the point $(1,\varphi^{-1}, \varphi^{-2},0)$. Thus, the 96 vertices of $s\{3,4,3\}$ are given by even permutations of the coordinates of all possible points of the form $(\pm 1,\pm \varphi^{-1},\pm \varphi^ {-2},0)$, or, after a suitable stretch, of the form There is a universal two-sheeted covering epimorphism of the Lie groups
$$
\begin{equation*}
\operatorname{pr}\colon \operatorname{SU}(2)=\operatorname{Sp}(1)\to \operatorname{SO}(3)
\end{equation*}
\notag
$$
with central kernel $\{\pm 1\}$. By the binary tetrahedral and binary icosahedral groups one means the groups $\operatorname{BT}=\operatorname{pr}^{-1}(\operatorname{T})$ and $\operatorname{BI}=\operatorname{pr}^{-1}(\operatorname{I})$, respectively, where $\operatorname{T}$ and $\operatorname{I}$ are the groups of proper isometries of the regular tetrahedron and regular icosahedron, respectively. On the other hand, as is known (see [6]), the convex hulls of the binary tetrahedral and binary icosahedral subgroups
$$
\begin{equation*}
\operatorname{BT}\subset \operatorname{BI}\subset \operatorname{SU}(2)=\operatorname{Sp}(1)=S^3(0,1)\subset \mathbb{R}^4
\end{equation*}
\notag
$$
are correspondingly a regular $24$-cell and a regular $600$-cell with the Schläfli symbols $\{3,4,3\}$ and $\{3,3,5\}$. Here the difference $\operatorname{BI}-\operatorname{BT}$ between the binary icosahedral subgroup $\operatorname{BI}$ of order $120$ and the binary tetrahedral subgroup $\operatorname{BT}$ of order $24$ in the group $\operatorname{Q}$ of unit quaternions is obtained by even permutations of coordinates of all possible points of the form The formulae (1) and (2) show that, up to similarity, there are two different enantiomorphic forms of the Gosset semiregular polytope: $s\{3,4,3\}$ and $\operatorname{conv}(\operatorname{BI}-\operatorname{BT})$. According to [11], the Gosset semiregular polytopes in $\mathbb{R}^n$ for $n=8,7,6,5$ have $240$, $56$, $27$, and $16$ vertices, respectively. It is remarkable that they are all convex hulls of the corresponding effective homogeneous spaces of the Weyl groups $\operatorname{W}(\operatorname{E}_8)/\operatorname{W}(\operatorname{E}_7)$, $\operatorname{W}(\operatorname{E}_7)/\operatorname{W}(\operatorname{E}_6)$, $\operatorname{W}(\operatorname{E}_6)/\operatorname{W}(\operatorname{D}_5)$ [10], and $\operatorname{W}(\operatorname{D}_5)/\operatorname{W}(\operatorname{A}_4)$, where the reduced root systems of the corresponding simple compact Lie algebras are indicated in parentheses. § 3. Gosset polytopes, root systems, and latticesThe systems of the types $\operatorname{A}$, $\operatorname{D}$, and $\operatorname{E}$ are exactly the reduced root systems for simple compact Lie algebras all of whose roots have the same length. Each of these systems has a simple geometric description as the set of non-zero vectors of the lattice of the same title whose elements are the centres of balls of some packing in Euclidean space of the corresponding dimension that are the least distant from zero. As a consequence, the number of roots is equal to the contact number $\tau$ of this spherical lattice packing, and the corresponding Weyl group is a linear (sub)group $G_0$ of automorphisms of the lattice [17]. Here $\tau$ is equal to the number of all balls in the packing tangent to a given ball of the packing. We give a description of these lattices according to [17]. The standard cubic lattice in $\mathbb{R}^n$ is $\mathbb{Z}^n$. For $l\geqslant 3$ we write
$$
\begin{equation}
\operatorname{D}_l=\{(x_1,\dots,x_l)\in\mathbb{Z}^l\colon x_1+\dots + x_l\text{ is even}\}.
\end{equation}
\tag{3}
$$
For $l\geqslant 1$ we write
$$
\begin{equation}
\operatorname{A}_l=\{(x_1,\dots,x_{l+1})\in\mathbb{Z}^{l+1}\colon x_1+\dots + x_{l+1}=0\}.
\end{equation}
\tag{4}
$$
We write
$$
\begin{equation}
\operatorname{E}_8 =\Bigl\{(x_1,\dots,x_8)\in\mathbb{Z}^8\colon \text{all } x_i\text{ are even or odd},\ \sum x_i=0\ (\operatorname{mod} 4)\Bigr\},
\end{equation}
\tag{5}
$$
$$
\begin{equation}
\operatorname{E}_7 = \{(x_1,\dots,x_8)\in\operatorname{E}_8\colon x_1+\dots +x_8=0\},
\end{equation}
\tag{6}
$$
$$
\begin{equation}
\operatorname{E}_6 = \{(x_1,\dots,x_8)\in \operatorname{E}^8\colon x_1+x_8=x_2+\dots+ x_7=0\}.
\end{equation}
\tag{7}
$$
The contact numbers and the orders of Weyl groups of these lattices are
$$
\begin{equation*}
\begin{gathered} \, \tau(\operatorname{A}_l)=l(l+1),\quad |\operatorname{W}(\operatorname{A}_l)|=(l+1)!, \qquad \tau(\operatorname{D}_l)=2l(l-1),\quad |\operatorname{W}(\operatorname{D}_l)| =2^{l-1}\cdot l!, \\ \tau(\operatorname{E}_6)=72,\quad |\operatorname{W}(\operatorname{E}_6)| =2^7\cdot3^4\cdot 5, \qquad\tau(\operatorname{E}_7)=126, \quad |\operatorname{W}(\operatorname{E}_7)|=2^{10}\cdot3^4\cdot 5\cdot 7, \\ \tau(\operatorname{E}_8)=240,\quad |\operatorname{W}(\operatorname{E}_8)| =2^{14}\cdot 3^5\cdot 5^2\cdot 7. \end{gathered}
\end{equation*}
\notag
$$
We recall that $\operatorname{A}_l$ is the reduced root system of the simple compact Lie algebra $\mathfrak{su}(l+1)$, $\operatorname{D}_l$ is the reduced root system of the simple compact Lie algebra $\mathfrak{so}(2l)$, and the other systems are the reduced root systems of the similarly named simple compact Lie algebras (with small Gothic letters). The lattices $\operatorname{E}_6$, $\operatorname{E}_7$, and $\operatorname{E}_8$ are called Gosset lattices in [17]. A lattice or a quadratic form is said to be integer in [17] if the inner product of any two lattice vectors is an integer. Equivalently, a lattice $\Lambda$ is integer if and only if $\Lambda\subset \Lambda^{\ast}$, where
$$
\begin{equation*}
\Lambda^{\ast}_n=\{x\in \mathbb{R}^n\colon x\cdot u\in\mathbb{Z}\text{ for all } u\in\Lambda_n\}.
\end{equation*}
\notag
$$
An integer lattice $\Lambda$ is said to be unimodular or self-dual if $\Lambda=\Lambda^{\ast}$. All lattices $A$, $D$, $E$ are integer. We note that the lattices $E$ defined here are obtained from those defined in [17] by multiplying by $2$. When following the definitions of [17], the lattice $E_8$ is unimodular, $\tau(E^{\ast}_7)=56$, and $\tau(E^{\ast}_6)=54$. We can say that the Gosset polytopes in the dimensions $8$ and $7$ are the convex hulls of the shortest non-zero lattice vectors of $E^{\ast}_8$ and $E^{\ast}_7$, respectively, and the Gosset polytope in the dimension $6$ is the convex hull of an arbitrary maximal subset of pairwise non-adjacent vertices of the convex hull of the shortest non-zero vectors of the lattice $E^{\ast}_6$. Thus, we obtain a geometric description of these Gosset polytopes. In Chap. 21, § 3, Sect. 3.D of [17] it is stated that the Voronoi polytopes for the lattices $E_8$, $E_7$, and $E_6$ are dual, respectively, to the Gosset polytopes $4_{21}$, $2_{31}$, $1_{22}$ described in § 11.8 of [4]. In other words, these Gosset polytopes are Delaunay polytopes [18] for the lattices $E_8$, $E_7$, and $E_6$. The Voronoi polytopes of an arbitrary lattice in $\mathbb{R}^n$ have a special property, namely, they are space-filling polytopes: the entire space $\mathbb{R}^n$ can be filled with disjoint shifted copies of the same polytope. It is remarkable that, for a Gosset polytope in $\mathbb{R}^8$, one can take the root system of type $\operatorname{E}_8$ ([10], [19]). The Gosset polytope in $\mathbb{R}^8$ is also realised in the form of some multiplication-closed system of unit octaves, the so-called integer Cayley numbers ([10], [17]). Moreover, in [10], the famous 2D McMullen orthogonal projections of Gosset polytopes in $\mathbb{R}^8$ and $\mathbb{R}^7$ are presented, where the vertex projections are located correspondingly on eight concentric circles (Gosset circles) with $30$ vertices on each of them [19] and on three concentric circles with $18$ vertices on each of them. The first projection is obtained from the drawing on the frontispiece of the book [5] by adding certain $240$ lines [10]. The McMullen projection can be generalised to any complex simple Lie algebras; the Gosset circles are generalised by means of the root orbits with respect to the Coxeter element. B. Kostant found a readily defined operator on the Cartan algebra whose eigenvalue ratios are the same as those for squared radii of the corresponding generalised Gosset circles (this result is given in [19] along with a useful discussion of it). The mass ratios of $8$, $7$, and $6$ particles in the Fateev–Zamolodchikov models [20] of the conformal field theory for $E_8$, $E_7$, and $E_6$ are exactly the ratios of the radii of the corresponding (generalised) Gosset circles. These models have received an experimental confirmation [19]. According to Chap. 14, § 1 of [17], in the lattice packing in $\mathbb{R}^8$ with centres at the points of the lattice $\operatorname{E}_8$ there are $240$ balls tangent to a single ball, $56$ balls tangent to either of balls in a pair of tangent balls, $27$ balls tangent to every ball in a triple of pairwise tangent balls, $16$ balls tangent to every ball of a quadruple of pairwise tangent balls, $10$ balls tangent to every ball from a quintuple of pairwise tangent balls, and $6$ balls tangent to every ball from a hextuple of pairwise touching balls. In fact, this gives a uniform way of constructing semiregular Gosset polytopes: eight-dimensional, seven-dimensional, six-dimensional, five-dimensional, and one of the three four-dimensional ones, namely, the full truncation of a four-dimensional regular simplex; $6$ is the number of vertices of the regular octahedron. For $\operatorname{A}_4$ we obtain $4\cdot 5=20$ roots and the Weyl group of order $(4+1)!=120$. Question 1. Is it true that for the vertices of a full truncation of the four-dimensional regular simplex ($5$-cell) we can take a half of the system $\operatorname{A}_4$? Proposition 1. The answer to Question 1 is negative. Proof. The vertices of a truncated four-dimensional regular simplex can be represented as points obtained from the point $(1,1,0,0,0)$ by permuting the coordinates. The angle between two radius vectors of these points is either $\pi/2$ or $\pi/3$. Here it is easy to choose three radius vectors of this kind in such a way that the angle between any two of them is equal to $\pi/3$ (for example, we can take $(1,1,0,0,0)$, $(1,0,1,0,0)$, and $(0,1,1,0,0)$).
The roots of the root system $A_4$ are obtained from the vector $(1,-1,0,0,0)$ by the permutations of the coordinates. The angle between any vectors of this kind is equal to one of the following three values: $\pi/2$, $\pi/3$, or $2\pi/3$. On the other hand, it is impossible to choose three roots of this kind with the angle $\pi/3$ between any two of them. Really, suppose that there are three root vectors of this kind and denote them by $a$, $b$, and $c$. Thus, $(a,b)=(a,c)=(b,c)=1$. Let the vector $a$ have non-zero components $a_i$ and $a_j$, that is, $\{a_i,a_j\}=\{-1,1\}$. Then the vector $b$ has non-zero components $b_i$ and $b_k$ for some $k\notin \{i,j\}$, and for the vector $c$ the non-zero components are $c_j$ and $c_k$. Here $1=(a,b)=(a,c)=(b,c)=a_i\cdot b_i=a_j \cdot c_j=b_k\cdot c_k$. Then $1=(a_i\cdot b_i)\cdot (a_j \cdot c_j)\cdot(b_k\cdot c_k)=(a_i\cdot a_j)\cdot (b_j \cdot b_k)\cdot(c_j\cdot c_k)=-1$, which is impossible. This completes the proof of the proposition. Remark 2. It can be proved similarly that we cannot take a half of the system $\operatorname{A}_n$, $n\geqslant 2$, for the vertices of the total truncation of an $n$-dimensional regular simplex. Proposition 2. The group $\operatorname{BT}$ is isomorphic to a simply transitive subgroup of automorphisms of the root system $\operatorname{D}_4$, which is a subgroup of index $8$ in $\operatorname{W}(D_4)$. Proof. Figure 8.1 in Ch. 8, § 4 of [17] shows a two-dimensional projection of a self-dual regular polytope $\{3,4,3\}$ ($24$-cell) whose vertices are $24$ minimal vectors of the lattice $\operatorname{D}_4$ represented as elements of the lattice $\mathbb{Z}[\zeta]$, where $\zeta=e^{\pi i/4}$. According to [6], this implies that the root system $\operatorname{D}_4$ is isomorphic to the binary tetrahedral group $\operatorname{BT}$. Therefore, we can assume that $\operatorname{BT}$ is a simply transitive subgroup of automorphisms of the root system $\operatorname{D}_4$, which is a subgroup of the index $|\operatorname{W}(\operatorname{D}_4)|/|\operatorname{BT}|=192/24=8$ in the Weyl group $\operatorname{W}(D_4)$. We note that the projections of the $24$ vertices of the root system $\operatorname{D}_4$ (vertices of the projection) in Fig. 8.1 of [17] are located on three concentric circles with eight vertices on each of them. The assertion has been proved. Proposition 3. The groups $\operatorname{W}(\operatorname{A}_4)$ and $\operatorname{BI}$ (of order 120) are not isomorphic. Proof. For the proof, it suffices to establish the possible orders of elements of these groups. It readily follows from the formula (4) that the group $\operatorname{W}(\operatorname{A}_4)$ is the symmetric group $\mathbf{S}_5$ of permutations of five coordinates [17]. The well-known assertion that every permutation $p\in \mathbf{S}_5$ is a product of pairwise commuting independent cycles of length at least $1$ with the sum of their lengths equal to $5$, and the order of the element $p$ is the greatest common divisor of the lengths of these cycles, is almost obvious. This implies that the orders of elements of $\mathbf{S}_5$ can be be only the numbers $1$, $2$, $3$, $4$, $5$, $6$.
We recall that $\operatorname{BI}=\operatorname{pr}^{-1}(\operatorname{I})$, where $\operatorname{pr}\colon \operatorname{SU}(2)=\operatorname{Sp}(1)\to \operatorname{SO}(3)$ is the universal double-sheeted covering epimorphism of Lie groups with central kernel $\{\pm 1\}$, and $\operatorname{I}$ is the group of proper isometries of the regular icosahedron. Then $\operatorname{pr}\colon \operatorname{BI}\to \operatorname{I}$ is an epimorphism with the same kernel. The possible orders of elements of $\operatorname{I}$ are $1$, $2$, $3$, $5$. We claim that there is an element of order $10$ in $\operatorname{BI}$. Really, let $x\in \operatorname{I}$ have the order $5$ and $y\in \operatorname{pr}^{-1}(x)$. Then $y^5=1$ or $y^5=-1$. In the latter case, $y$ has order $10$, and everything is proved. In the first case, we set $z=(-1)y$. Then $z^5=-1$, and $z$ has the order $10$. The proposition is proved. Below in the text we use the notation $C_n$ for the cyclic group of order $n$. Proposition 4. Let $C_{10}$ be the cyclic subgroup of order $10$ generated by the element $y$ or $z$ from the proof of 3. Then: 1) $C_{10}$ is a subgroup of some $1$-parameter subgroup of $\operatorname{SU}(2)$, which is geodesic in $S^3(0,1)\subset \mathbb{R}^4$, and $C_{10}$ divides the image of the $1$-parameter subgroup into arcs of the same length; 2) $\operatorname{BT}{\cap}\, C_{10}=\{\pm 1\}$, and $\operatorname{BI}=\operatorname{BT}{\cdot}\, C_{10}$ and $\operatorname{BI}=C_{10}\,{\cdot} \operatorname{BT}$ are decompositions of $\operatorname{BI}$ with respect to left and right cosets of the groups $C_{10}$ and $\operatorname{BT}$ with every coset doubled; 3) the decompositions in the part 2) give two different partitions of the group $\operatorname{BI}$ (in other words, the sets of vertices of a regular $600$-cell in $\mathbb{R}^4$) into $12$ ten-point regular subsets of the great circles in $S^3(0,1)$ having two-point intersections with $\operatorname{BT}$ and translatable into each other only by left or right shifts by elements of the subgroup $\operatorname{BT}$ that are Clifford–Wolf translations in $S^3(0,1)$; 4) the decompositions in the part 2) give two different partitions of the group $\operatorname{BI}$ (in other words, the sets of vertices of a regular $600$-cell in $\mathbb{R}^4$) into five subsets consisting of vertices of regular $24$-cells and having two-point intersections with $C_{10}$, as well as translatable into each other only by right or left shifts by elements of the subgroup $C_{10}$ that are Clifford–Wolff translations in $S^3(0,1)$; 5) there are $12$ partitions for $\operatorname{BI}$ similar to those given in each of the parts 3) and 4). Proof. 1) The subgroup of order $5$ generated by the element $x$ from the proof of Proposition 3 is a subgroup of a unique, up to reparametrization, $1$-parameter subgroup $\gamma=\gamma(t)$, $t\in\mathbb{R}$, of rotations around the axis connecting some pair of diametrically opposite vertices of the icosahedron. Then there is only one periodic $1$-parameter subgroup of $\widetilde{\gamma}(t)$ in $\operatorname{SU}(2)=\operatorname{Sp}(1)=S^3(0,1)\subset \mathbb{R}^4$ such that $\operatorname{pr}(\widetilde{\gamma}(t))=\gamma(t)$, $t\in\mathbb{R}$. It is clear that $C_{10}$ is a subgroup of the $1$-parameter group $\widetilde{\gamma}(t)$, $t\in\mathbb{R}$. On $\operatorname{Sp}(1)=S^3(0,1)$, there is a unique bi-invariant Riemannian metric coinciding with the intrinsic metric $d$ of the unit sphere $S^3(0,1)$. Consequently (see [2]), $\widetilde{\gamma}(t)$, $t\in \mathbb{R}$, is a geodesic in $S^3(0,1)$, and $C_{10}$ is a subset of some great circle in $S^3(0,1)$ splitting the circle into arcs of the same length.
2) It can readily be seen that every element of the group $C_{10}$, except for $1$ and $-1$, has the order $5$ or $10$. Then $\operatorname{BT}{\cap}\, C_{10}=\{\pm 1\}$, since the group $\operatorname{BT}$ of order $24$ cannot contain elements of order $5$ or $10$. This implies all assertions of the part 2). Assertions 3) and 4) follow from 1) and 2). Assertion 5) follows from what has been said and from the fact that a regular icosahedron has six different pairs of diametrically opposite vertices. The proposition has been proved. Corollary 1. There are expansions $\operatorname{BI}-\operatorname{BT}=\operatorname{BT}{\cdot}\, (C_{10}-\{\pm 1\})$ and $\operatorname{BI}-\operatorname{BT}=(C_{10}-\{\pm 1\})\cdot \operatorname{BT}$ of the vertices of a semiregular Gosset polytope with the duplication of left and right shifts of the set $C_{10}-\{\pm 1\}$. There are 12 different ways of partitioning the set $\operatorname{BI}-\operatorname{BT}$ into 12 eight-point subsets of great circles in $S^3(0,1)$, which can be translated into each other only by right or left shifts by elements of the group $\operatorname{BT}$ that are Clifford–Wolf translations in $S^3(0,1)$. Corollary 2. The projection $\operatorname{pr}$ of the decompositions specified in part 2) of Proposition 4 gives direct decompositions into products of subgroups $\operatorname{I}=\operatorname{T}{\cdot}\, C_5 $ and $\operatorname{I}=C_5\,{\cdot}\operatorname{T}$, where $\operatorname{T}$ is the group of proper isometries of a regular simplex. Proposition 5. The root system $\operatorname{E}_8$ contains the point $v=(1,1,\dots,1)$ whose stationary group is the subgroup $\operatorname{W}(\operatorname{E}_7)\subset \operatorname {W}(\operatorname{E}_8)$. Consequently, $\operatorname{E}_8$ is realised as the homogeneous space $\operatorname{W}(\operatorname{E}_8)/\operatorname{W}(\operatorname{E}_7)$. Proof. By [17], the radius of the lattice packing of balls for the lattice $\operatorname{E}_8$ is equal to $\sqrt{2}$. This, together with the formula (5), implies that $v$ is an element not only of the lattice $\operatorname{E}_8$ but also of the root system $\operatorname{E}_8$. As is known, the Weyl group of the root system of a simple compact Lie algebra acts transitively on every subset of its roots of the same length. Therefore, the root system $\operatorname{E}_8$ is the orbit of the point $v$ with respect to the Weyl group $\operatorname{W}(\operatorname{E}_8)$. It follows from the formula (6) that the vector $v$ is orthogonal to the lattice $\operatorname{E}_7$. The lattice $\operatorname{E}_8$ is layered [17]; its layers are some parallel translations of the lattice $\operatorname{E}_7$. Therefore, the action of every element of $\operatorname{W}(\operatorname{E}_7)$ on $\operatorname{E}_7$ can uniquely be extended to a linear automorphism of the lattice $\operatorname{E}_8$ keeping the vector $v$, that is, to the action of some element of the group $\operatorname{W}(\operatorname{E}_8)$ on $\operatorname{E}_8$ with the fixed vector $v$. Thus, we can assume that $\operatorname{W}(\operatorname{E}_7)\subset\operatorname{W}(\operatorname{E}_8)$. Moreover, $\operatorname{W}(\operatorname{E}_7)$ is the stabiliser of the point $v$ in the group $\operatorname{W}(\operatorname{E}_8)$, since $\operatorname {W}(\operatorname{E}_7)$ has no fixed vectors in the linear span of the lattice $\operatorname{E}_7$. The second assertion of Proposition 5 follows from what has been said. The proposition has been proved. Proposition 5 and the description of the Weyl group $W(E_8)$ in Ch. 4, Sect. 8.1 of [17] imply the following assertion. Corollary 3. All elements of the root system $\operatorname{E}_8$ are obtained from $v$ by applying the composition of the following actions in any order: all permutations of the eight coordinates, all even sign changes, and the (left) multiplication by the matrix $\operatorname{diag}(H,H)$, where $H$ is the $(4\times 4)$ Hadamard matrix from Ch. 4, Subsec. 7.1 of [17]. Proposition 6. The lattice $\operatorname{E}_7$ contains the point $w=(-3,1,1,1,1,1,1,-3)$ whose stationary group is the subgroup $\operatorname{W}(\operatorname{ E}_6)\subset \operatorname{W}(\operatorname{E}_7)$. Therefore, the homogeneous space $\operatorname{W}(\operatorname{E}_7)(w)= \operatorname{W}(\operatorname{E}_7)/\operatorname{W}(\operatorname{E}_6)$ is a Gosset semiregular polytope in $\mathbb{R}^7$. Proof. By the formulae (6) and (7), the lattice $\operatorname{E}_6$ is a sublattice of the lattice $\operatorname{E}_7$ and $w$ is an element of the lattice $\operatorname{E}_7$ orthogonal to the lattice $\operatorname{E}_6$. The lattice $\operatorname{E}_7$ is layered [17]; its layers are some parallel shifts of the lattice $\operatorname{E}_6$. The subsequent proof is the same as that for Proposition 5. The proposition has been proved. Proposition 7. The formula defines a linear non-isometric isomorphism of the lattice $\operatorname{D}_5$ onto the lattice Here $\operatorname{W}(\operatorname{D}_5')\cong \operatorname{W}(D_5)$. Due to the last statement of this proposition, $\operatorname{D}_5$ in the following proposition is identified with $\operatorname{D}_5'$. Proposition 8. The lattice $\operatorname{E}_6$ contains the point $u=(2,0,0,0,0,0,0,-2)$ whose stationary subgroup is the group $\operatorname{W}(\operatorname{D}_5)\subset \operatorname{W}(\operatorname{E}_6)$. Therefore, the homogeneous space $\operatorname{W}(\operatorname{E}_6)(u)= \operatorname{W}(\operatorname{E}_6)/\operatorname{W}(\operatorname{D}_5)$ is a Gosset semiregular polytope in $\mathbb{R}^6$. Proof. Due to Proposition 7, the lattice $\operatorname{D}_5$ is a sublattice of the lattice $\operatorname{E}_6$ and $u$ is an element of the lattice $\operatorname{E}_6$ orthogonal to the lattice $\operatorname{D}_5$. The lattice $\operatorname{E}_6$ is layered [17]; its layers are some parallel shifts of the lattice $\operatorname{D}_5$. The subsequent proof is the same as that for Proposition 5. Proposition 9. The lattice $\operatorname{D}_5$ contains the point $q=(2,2,2,2,2)$ whose stationary subgroup is the group $\operatorname{W}(\operatorname{A}_4)\subset \operatorname{W}(\operatorname{D}_5)$. Here the homogeneous space $\operatorname{W}(\operatorname{D}_5)(q)= \operatorname{W}(\operatorname{D}_5)/\operatorname{W}(\operatorname{A}_4)$ is the largest subset $C$ of pairwise non-adjacent hypercube vertices in $\mathbb{R}^5$ centred at zero and with the edge of length $4$ whose facets are parallel to the coordinate hyperplanes containing the vertex $q$. Therefore, the convex hull of the set $C$ is the well-known Gosset semiregular polytope in $\mathbb{R}^5$ containing $16$ hypercube vertices, the so-called semihypercube or semipenteract. Proof. By the formulae (3) and (4), the lattice $\operatorname{A}_4$ is a sublattice of the lattice $\operatorname{D}_5$ and $q$ is an element of the lattice $\operatorname{D}_5$ orthogonal to $\operatorname{A}_4$. The lattice $\operatorname{D}_5$ is layered [17]; its layers are some parallel shifts of the lattice $\operatorname{A}_5$. As in the case of Proposition 5, we obtain $C:=\operatorname{W}(\operatorname{D}_5)(q)= \operatorname{W}(\operatorname{D}_5)/\operatorname{W}(\operatorname{A}_4)$. It can readily be seen that $|\operatorname{W}(\operatorname{D}_5)|/|\operatorname{W}(\operatorname{A}_4)|=16$. Moreover, by Ch. 4, Sec. 7.1 of [17], the group $\operatorname{W}(\operatorname{D}_l)$ is generated for all $l\geqslant 3$ by all permutations and sign changes for an even number of coordinates. All what has been said implies the description of the convex hull of the set $C$ given in Proposition 9, which is a well-known description of the Gosset semiregular polytope in $\mathbb{R}^5$, the so-called semihypercube or semipenteract. The proposition has been proved. Remark 3. The convex hulls of the spaces $\operatorname{W}(\operatorname{D}_{l+1})/\operatorname{W}(\operatorname{A}_l)$, $\operatorname{W}(\operatorname{A}_{l+1})/\operatorname{W}(\operatorname{A}_l)$ and $\operatorname{W}(\operatorname{D}_{l+1})/\operatorname{W}( \operatorname{D}_l)$ obtained in a similar way for all $l\,{\geqslant}\, 3$ are semihypercubes, proper simplices, and regular hyperoctahedra, respectively. Question 2. Every lattice in the sequence below, except for $\operatorname{A}_2$, contains the following one: Is it true that for every lattice $L$ in this sequence, except for its extreme terms, the composition of the orthogonal projections of every linear span of the lattice along its non-zero vector (starting from $L$ and ending with the lattice $\operatorname{A}_3$) orthogonal to the linear span of the next lattice onto the last span takes the set of non-zero vectors in $L$ of minimum length in a one-to-one way (that is, the root system of the same name as $L$) into the linear span of the lattice $\operatorname{A}_2$? Remark 4. It follows from the formulae (5) and (6) and Proposition 5 that for $\operatorname{E}_8$ the answer to an analogue of the question 2 is negative. The densest six-dimensional (eight-dimensional) lattice $\operatorname{E}_6$ ($\operatorname{E}_8$, respectively) is readily described as a realification of the lattice in $\mathbb{C}^3$ (in $\mathbb{H }^2$) over the ring $\mathcal{E}$ ($\mathcal{H}$) of integer Eisenstein (Hurwitz, respectively) numbers (see Ch. 2, the end of § 2.6 in [17]). Let us present a non-trivial and interesting method proposed in [17] for constructing the lattice $\operatorname{E}_8$. The binary icosahedral group is called in [17] the group of icosians. By the way, we note that the icosahedral group which is two-sheeted covered by $\operatorname{BI}$ is isomorphic to the alternating group $\mathbf{A}$$_5$ of even permutations on five characters. The ring of icosians $\mathcal{J}$ is the set of all finite sums of elements of the group of icosians. The elements of the ring $\mathcal{J}$ are called simply icosians. a typical icosian has the form $q=\alpha + \beta i +\gamma j + \delta k$, where the coordinates $\alpha$, $\beta$, $\gamma$, and $\delta$ belong to the golden section field $\mathbb{Q}(\varphi)$ and, therefore, have the form $a+b\sqrt{5}$, where $a,b\in\mathbb{Q}$. The conjugate icosian is $\overline{q}=\alpha -\beta i-\gamma j - \delta k$, and $q\overline{q}=\alpha^2+\beta^2+\gamma^2+\delta^2$. For $v\in\mathcal{J}$ two different norms are used: the quaternion norm $\operatorname{QN}(v,v)=(v,v):=v\overline{v}$, which is a number of the form $a+b\sqrt{5}$ with $a,b\in\mathbb{Q}$, and the Euclidean norm $\operatorname{EN}(v)=a+b\geqslant 0$. The icosians with quaternionic norm $1$ form the group of icosians. With respect to the quaternionic norm, the icosians belong to the four-dimensional space over $\mathbb{Q}(\sqrt{5})$; with respect to the Euclidean norm, they lie in the eight-dimensional space. In fact, in the Euclidean norm, the ring $\mathcal{J}$ is isomorphic to the lattice $\operatorname{E}_8$ in this space. Table 8.1 in Ch. 8, § 2 [17] (which takes up a whole page) shows this isomorphism. § 4. Auxiliary assertionsLet us prove a technical assertion, which is useful when proving the normal homogeneity of the set of vertices of some polytopes. Proposition 10. Let a convex polytope $P \subset \mathbb{R}^n$ have the following properties: 1) $P$ is homogeneous (vertex-transitive); 2) the distance between pairs of different vertices $P$ can take only the values $d_i$, $1\leqslant i \leqslant s$, $d_1<d_2<\cdots <d_{s-1}<d_s$; 3) the isotropy group of some vertex $v \in P$ acts transitively on the set of vertices $P$ at the distance $d_i$ from $v$ for any $1\leqslant i \leqslant s-1$; 4) for all $1\leqslant i \leqslant s-1$ there is an isometry $\psi$ of the polytope $P$ with the maximal displacement of the vertices $P$ by the distance $d_i$. Then the set of vertices of $P$ is a normal homogeneous metric space. Proof. We are to prove that for any vertices $v_1,v_2 \in P$, $v_1\neq v_2$, there is an isometry $\eta$ of the polytope $P$ taking $v_1$ to $v_2$ and having the maximal displacement at the point $v_1$.
If $\rho(v_1,v_2)=d_s$, then for $\eta$ one can take an arbitrary isometry of the polytope $P$ taking $v_1$ to $v_2$ (such an isometry exists due to the homogeneity of the polytope). By condition 2), we need to consider only the case $\rho(v_1,v_2)=d_i$ for some $1\leqslant i \leqslant s-1$. According to condition 4), there is an isometry $\psi$ of the polytope $P$ with the maximal displacement of vertices $P$ by the distance $d_i$. By conditions 1) and 3), we can assume without loss of generality that $\psi(v_1)=v_2$. Thus, $\psi$ is the desired isometry. This completes the proof of the proposition. As is known, the vertices of any homogeneous convex polytope in $\mathbb{R}^n$ lie on some hypersphere with centre at the centre of gravity of the polytope, which is called below the centre of the polytope (on the circumscribed hypersphere). Let us now introduce the concept of almost perfect polytope, which plays an important role in the study of the six-dimensional Gosset polytope. Definition 7. a non-degenerate homogeneous polytope $P\subset \mathbb{R}^n$ is said to be almost perfect if every second-order hypersurface in $\mathbb{R}^n$ containing all vertices of $P$ and symmetric with respect to the centre of $P$ coincides with the circumscribed hypersphere of $P$. Lemma 1. a homogeneous polytope $P\subset \mathbb{R}^n$ is almost perfect if and only if for any ellipsoid $E$ in $\mathbb{R}^n$ with the centre coinciding with the centre of $P$ and different from the hypersphere circumscribed around $P$ there is a vertex of $P$ that does not lie on $E$. Proof. Without loss of generality, we can assume that the centre of $P$ coincides with the origin $O$ of some Cartesian coordinate system. If there is an ellipsoid centred at $O$ that contains all the vertices $P$ and differs from the hypersphere circumscribed around $P$, then $P$ is not nearly perfect, because the ellipsoids are second-order surfaces. Let us now prove the assertion in the other direction. Assume that a polytope $P$ is not nearly perfect. Let $\sum_{i=1}^n x_i^2-r^2=0$ be the equation of the hypersphere circumscribed around $P$, and let $F\colon \mathbb{R}^n \to \mathbb{R}$ be a quadratic form such that $F(A)-1=0$ for any vertex $A$ of $P$ and $r^2F(X) \neq \sum_{i=1}^n x_i^2$. It is clear that for any $\mu \in \mathbb{R}$ the second-order hypersurface given by the equation $\sum_{i=1}^n x_i^2-r^2+\mu(F(x)-1)=0$ contains all vertices of the polytope $P$, differs from a hypersphere, and is symmetric with respect to the centre $O$. Moreover, if the absolute value of $\mu$ is sufficiently small, then this the hypersurface is an ellipsoid. This completes the proof of the lemma. It can readily be proved that every regular $n$-gon in a plane is almost perfect for $n\geqslant 5$ (different second-order curves intersect at most at four points). However, this is not true for $n=4$. Really, let $L\in \mathbb{R}$, then the second-order curve $Lx^2+(2-L)y^2=1$ contains four points of the form $(\pm 1/\sqrt{2},\pm 1/\sqrt{2})\in \mathbb{R}^2$, that is, the vertices of a square with the side length $\sqrt{2}$. The circle circumscribed around this square is obtained for $L=1$. We note that a regular triangle is also almost perfect. Really, any second-order curve containing the vertices of a regular triangle and symmetric about its centre contains also points symmetric to its vertices with respect to the centre of the triangle, that is, it contains the vertices of a regular hexagon, and hence coincides with the circumscribed circle. Lemma 2. Let $a>0$ and let $F\colon \mathbb{R}^n \to \mathbb{R}$, $n\geqslant 2$, be a quadratic form such that $F(\pm a, \pm a,\dots, \pm a)=1$ for any sign $\pm$, that is, the hypersurface $F(x)=1$ contains all vertices $(\pm a, \pm a,\dots, \pm a)$ of the $n$-dimensional hypercube. Then $F(x)=\sum_{i=1}^n l_i x_i^2$, where $\sum_{i=1}^nl_i=1/a^2$; in particular, the principal directions of the hypersurfaces are parallel to the edges of the hypercube. Conversely, any quadratic form of the specified form contains all vertices of the $n$-dimensional hypercube under consideration. Proof. Let $F(x)=\sum_{i,j=1}^n a_{ij} x_ix_j$, where $a_{ij}=a_{ji}$. Let us fix arbitrary $i\neq j$ and prove that $a_{ij}=0$. Consider four vertices of the hypercube, $A_0$, $A_i$, $A_j$, and $A_{i,j}$, such that all coordinates of $A_0$ are equal to $a$, $A_i$ (and $A_j$) differs from $A_0$ by the sign of the $i$-th ($j$-th, respectively) coordinate, and $A_{i,j}$ differs from $A_0$ by the signs of $i$-th and $j$-th coordinates. It is clear that these points are the vertices of some square face. The equations are equivalent, correspondingly, to the equations Subtracting the third equation from the sum of the first two of these equations, we obtain $2a_{ij}=0$. This argument proves the first assertion of the lemma; the other one is obvious. This completes the proof of the lemma. Proposition 11. Every $\mathrm{CW}$-translation $f$ of the vertex set $M$ of an almost perfect homogeneous polytope $P^n$ in $\mathbb{R}^n$, $n\geqslant 2$, with the circumscribed sphere $S^{n-1}=S^{n-1}(0,1)\subset \mathbb{R}^n$ at a distance $r$, where $0<r \leqslant 2$, is also a $\mathrm{CW}$-translation of the sphere $S^{n-1}$ to $r$. If $0<r<2$, then $n=2m$, $\mathbb{R}^n$ splits into a direct orthogonal sum $\mathbb{R}^n=\bigoplus_{l=1}^m\mathbb{R}^2_l$, and the restriction $f$ to every Euclidean $2$-plane $\mathbb{R}^2_l$, $l=1,\dots, m$, is a rotation by an angle $2\pi t/k$, where $k\geqslant 3$ is a divisor of the number of the vertices $|M|$, $t\in [1,k/2)$ is a positive integer such that $t/k$ is an irreducible fraction, and $r=2\sin(\pi t/k)$. Proof. If $r=2$, then $f$ is a central symmetry on $M$, which is a restriction of a central symmetry on $S^{n-1}$.
In the other case, $\mathbb{R}^n$ is decomposed into a direct orthogonal sum
$$
\begin{equation}
\mathbb{R}^n=\mathbb{R}^q\oplus\mathbb{R}^{s}\oplus \biggl( \bigoplus_{i=1}^{j}\mathbb{R}^2_{i} \biggr),
\end{equation}
\tag{8}
$$
where $q,s\geqslant 0$, and $q+s\geqslant 1$ if $n$ is odd; $n=q+s+2j$, the restriction of $f$ to $\mathbb{R}^q$ (to $\mathbb{R}^s$, respectively) has the eigenvalue $-1$ ($1$, respectively), and $\mathbb{R}^2_i$, $i=1,\dots,j$, corresponds to a pair of complex conjugate values that differ from $1$ for $f$, and thus the restriction of $f$ to $\mathbb{R}^2_i$ is a rotation by the angle $2\pi t_i/k_i$, $k_i\geqslant 3$, $t_i\in [1,k_i/2)$. Let $(x_1,\dots,x_n)$ be the system of coordinates with respect to an orthonormal basis in $\mathbb{R}^n$ adapted to the decomposition (8). Then for $a_i:=2\sin(\pi t_i/k_i)$, $i=1,\dots, j$, we have for an arbitrary vertex o $M$ with the coordinates $(x_1,\dots,x_n)$. Since the polytope $P^n$ is almost perfect, it follows that $q+s=0$, $j=m$, $n=2m$, $a_i^2=r^2$, $k_i= k$, $t_i= t$, $i=1,\dots,m$, and $r=2\sin(\pi t/k)$. This completes the proof of the proposition. § 5. Four-dimensional Gosset polytopesWe recall that the four-dimensional Gosset polytopes are the full truncations of the four-dimensional regular simplex and of the $600$-cell, and also the snub $24$-cell. Let us first study the full truncations of regular simplices (in spaces of arbitrary dimension). The regular $n$-dimensional simplex $T_n=T_n(\sqrt{2}\cdot l)$ with the edge length $\sqrt{2}\cdot l$ can be represented as the convex hull of the points $A_i \in \mathbb{R}^{n+1}$, $i=1,\dots,n+1$, such that all coordinates of the point $A_i$ are equal to zero except for the $i$-th coordinate, which is equal to $l$. It is clear that all the coordinates of the centre $C$ of this simplex are equal to $l/(n+1)$. The distance from $C$ to every point $A_i$ is equal to $\sqrt{n/(n+1)}\,{\cdot}\, l$ and coincides with the radius of the circumscribed hypersphere for $T_n(\sqrt{2}\cdot l)$. The full truncation of the regular simplex $T_n(2\sqrt{2})$ can be represented as the convex hull of the points $B_{i,j} \in \mathbb{R}^{n+1}$, $i,j=1,\dots,n+1$, $i<j$, such that all coordinates of the point $B_{i,j}$ are zero except for the $i$-th and $j$-th coordinates which are equal to $1$. It is clear that there are exactly $C_n^2=n(n-1)/2$ such points. The distance between (different) points $B_{i,j}$ and $B_{s,t}$ is equal to $d_1:=\sqrt{2}$ for $i=s$ or $j=t$ and to $d_2:=2$ for $\{i,j\}\cap \{s,t\}=\varnothing$. Proposition 12. The set $M$ of vertices of the full truncation of the regular simplex $T_n$ of arbitrary dimension is normally homogeneous. Proof. It is sufficient for us to deal with the implementation of the full truncation of the regular simplex $T_n(2\sqrt{2})$ given above. It is obvious that the set $M$ under consideration is homogeneous. Let us fix two points $B_{i,j}, B_{s,t} \in M$ and prove the existence of a $\delta$-shift $\psi$ at the point $B_{i,j}$ translating it to the point $B_{s,t}$.
If $d(B_{i,j}, B_{s,t})=d_2$, then we can take for $\psi$ any isometry of $M$ taking $B_{i,j}$ to $B_{s,t}$. Let $d(B_{i,j}, B_{s,t})=d_1$. This means that the points $B_{i,j}$ and $B_{s,t}$ are different in exactly two coordinates (that is, $i=s$ and $j \neq t$ or $i\neq s$ and $j=t$). Consider the isometry $\psi$ of the set $M$ generated by the permutation of these two coordinates. It is clear that $\psi(B_{p,q})=B_{p,q}$ if these coordinates of the point $B_{p,q}$ are equal, while $d(B_{p,q}, \psi(B_{p,q}))=d_1$ if these coordinates of the point $B_{p,q}$ are different. Thus, all points either are kept or are shifted by the distance $d_1$. Therefore, $\psi$ is a $\delta$-shift at the point $B_{i,j}$ taking it to $B_{s,t}$. This completes the proof of the proposition. It can readily be seen that the convex polytope $K= \operatorname{conv}(M)$, where has $n+1$ facets $G_i=\{x=(x_1,x_2, \dots, x_{n+1}) \in K \mid x_i=1\}$ each of which is an $(n-1)$-dimensional regular simplex with edge length $d_1=\sqrt{2}$ and $n+1$ facets $O_i=\{x=(x_1,x_2, \dots, x_{n+1}) \in K \mid x_i=0\}$ each of which is an $(n-1)$-dimensional total truncation of an $(n-1)$-dimensional regular simplex with edge length $d_1=\sqrt{2}$. We note that the total truncation of $T_n$ for $n=2$ is a triangle, and for $n=3$ this is an octahedron.For $n\geqslant 4$ the $(n-1)$-dimensional full truncation of an $(n-1)$-dimensional regular simplex is not an $(n-1)$-dimensional regular simplex (in the full truncation, the distance between two different vertices is not constant). This simple observation implies the following assertion. Lemma 3. For every isometry of the polytope $K$, $n\geqslant 4$, every facet $G_i$, $1\leqslant i \leqslant n+1$, is mapped onto a facet $G_j$ for some $1\leqslant j \leqslant n+1$. Remark 5. For $n=3$ the face $G_i$ can also be mapped onto some face $O_j$, since all faces of $K$ in this dimension are regular triangles. Moreover, we know that the set of vertices $M$ of the octahedron is a Clifford–Wolf homogeneous metric space. Proposition 13. For every isometry of the set $M$ of vertices of the full truncation of the regular simplex $T_n$, $n \geqslant 4$, there is a point in $M$ which is either stationary or shifted by $d_1$. Proof. Let us assume that there is an isometry $\psi$ of $M$ shifting every point by the distance $d_2$. It is clear that this isometry has no fixed points. Moreover, it extends naturally to an isometry of the polytope $K=\operatorname{conv}(M)$, which we also denote by the symbol $\psi$. Let $\eta$ be a permutation on the set $\{1,2,\dots, n, n+1\}$ such that $\psi(G_i)=G_{\eta(i)}$, $1\leqslant i \leqslant n+1$ (see Lemma 3).
Suppose that $\eta(i)=i$ for some index. Then $\psi(G_i)=G_i$ and, since all vertices in the facet $G_i$ are at the distance $d_1$ from each other, we have found a vertex that does not move at the distance $d_2$ under the action of the isometry $\psi$, which contradicts our assumption. Thus, $\eta(i)\neq i$ for all indices. Let $j=\eta(1)\neq 1$ and $k=\eta(j)\neq j$. Then $\psi(G_1)=G_j$ and $\psi(G_j)=G_k$. It is clear that $B_{1,j}\in G_1 \cap G_j$, and hence $\psi(B_{1,j})\,{\in}\,\psi(G_1)\,{\cap}\, \psi(G_j)=G_j\,{\cap}\,G_k$. Thus, $\psi(B_{1,j})=B_{j,k}$ for $j<k$ or $\psi(B_{1,j})=B_{k,j}$ for $j>k$. Therefore, the points $B_{1,j}$ and $\psi(B_{1,j})$ coincide for $k=1$ and differ in two coordinates (with numbers $1$ and $k$) for $k \neq 1$. In both the cases, the distance between $B_{1,j}$ and $\psi(B_{1,j})$ does not exceed $d_1$. The resulting contradiction proves the proposition. Proposition 14. The set $M$ of the vertices of the full truncation of the regular simplex $T_n$, $n \geqslant 4$, is not a Clifford–Wolf homogeneous metric space. Proof. Since there are vertices in $M$ that are at the distance $d_2$ from each other (for example, $B_{1,2}$ and $B_{3,4}$), it follows that, for the Clifford–Wolf homogeneity, one needs the existence of an isometry $\eta$ that displaces all points of $M$ at the distance $d_2$, which is impossible by Proposition 13. This completes the proof of the proposition. Propositions 12 and 14 for $n=4$ imply the following assertion. Corollary 4. The set of vertices of the full truncation of the regular four-dimensional simplex is normally homogeneous, but not Clifford–Wolf homogeneous. Now we begin to study the snub 24-cell (we assume that the set of its vertices coincides with the set $\operatorname{BI}-\operatorname{BT}$). Lemma 4. Every non-trivial proper movement of the tetrahedron is a rotation of order two or three about some axis in $\mathbb{R}^3$. Proof. Let $O$ be the centre of the tetrahedron. Then every movement of the tetrahedron keeps the point $O$. As is well known, every non-trivial proper movement in $\mathbb{R}^3$ with the fixed point $O$ is a rotation about some axis passing through $O$. If it is also an isometry of the tetrahedron onto itself, then this axis must pass either through the midpoints of some pair of non-intersecting edges of the tetrahedron or through some of its vertex and the centre of the opposite face. In the first case, we obtain a rotation of order two; in the other, of order three. This completes the proof of the lemma. Theorem 4. The set $M=\operatorname{BI}-\operatorname{BT}$ of vertices of the snub 24-cell is not normal homogeneous. Proof. As in the proofs of Propositions 3 and 4, it follows from Lemma 4 that every element of $\operatorname{BT}-\{\pm 1\}$ generates one of cyclic subgroups of the form $C_{4}$, $C_3$, or $C_{6}$ consisting of elements lying on some great circle in $S^3(0,1)\subset \mathbb{R}^4$. The subgroups $C_{4}$ and $C_{6}$ include the elements $\pm 1$, and $C_3$ includes $1$ but not $-1$. The minimum distances between different elements of the last three groups are respectively $2\sin(2\pi/8)=2\sin(\pi/4)$, $2\sin(\pi/6)$, $2\sin(\pi/3)$; the minimum of them is equal to $2\sin(\pi/6)$. At the same time, the minimum distance between different elements of the set $C_{10}-\{\pm 1\}$ is equal to $2\sin(\pi/10)$, which is less than $2\sin(\pi/6)$. By Corollary 1, $\operatorname{BI}-\operatorname{BT}$ is decomposed into left shifts of the set $C_{10}-\{\pm 1\}$ by elements of $\operatorname{BT}$, which are Clifford–Wolf translations. However, it follows from what was said above that the distance from any element in $C_{10}-\{\pm 1\}$ to any element of $(\operatorname{BI}-\operatorname{BT})-(C_{10}-\{\pm 1\})$ is at least $2\sin(\pi/6)$. Therefore, if $\operatorname{BI}-\operatorname{BT}$ is normal homogeneous, then for any pair $x$, $y$ of points in $C_{10}-\{\pm 1\}$ with the distance $d(x,y)=2\sin(\pi/10)$ there must be an isometry $f$ of the polytope $\operatorname{BI}-\operatorname{BT}$ onto itself such that $f(x)=y$ and $d(z,f(z))\leqslant 2\sin(\pi/10)$ for any point $z\in (\operatorname{BI}-\operatorname{BT})$, and $f(C_{10}-\{\pm 1\})=C_{10}-\{\pm 1\}$. However, it is clear that such an isometry must be a rotation of the great circle in $S^3(0,1)$ containing $C_{10}-\{\pm 1\}$ by the angle $\pm 2\pi/10=\pm\pi/5$. Then $f(z)=1$ for one of the two points in $C_{10}-\{\pm 1\}$ located at the distance $2\sin(\pi/10)$ from the point $1$, which should not be. This completes the proof of the theorem. Next, let us consider the full truncation of the regular $600$-cell in $\mathbb{R}^4$. The homogeneity of the set $M$ of its vertices follows from the fact that every isometry of the regular $600$-cell in $\mathbb{R}^4$ induces an isometry of its full truncation. The converse to the highlighted statement is also valid (see Sec. 5 of [21]). Proposition 15. The full truncation of the regular $600$-cell has the same group of symmetries of the order of $14400$ as the $600$-cell itself. Below, in Propositions 16–18, we successively prove the following theorem. Theorem 5. The set $M$ of vertices of the full truncation of a $600$-cell is not normal homogeneous. Proposition 16. The edge length of a regular $600$-cell inscribed in the sphere $S^3(0,1)$ is equal to $2\sin(\pi/10)$, and that of its total truncation is $\sin(\pi/10)$. Proof. Similarly to the proof of Lemma 4, it is established that the non-trivial proper motions of the icosahedron are rotations of orders $2$, $3$, or $5$. Then the first assertion follows from the fact that a regular $600$-cell of this kind can be identified with $\operatorname{BI}$ and from the proofs of Propositions 3 and 4 and Theorems 4. The other assertion follows from the first one and from the fact that every edge of the full truncation of this $600$-cell is one of the three midlines of some two-dimensional face of the $600$-cell which is a regular triangle. This completes the proof of the proposition. Corollary 5. The length of an edge of the full truncation of the $600$-cell inscribed into the unit sphere $S^3(0,1)\subset \mathbb{R}^4$, is Proof. The first equation follows from Proposition 16 and the from the fact that the distance from the midpoint of the edge of a $600$-cell to its centre is equal to $\cos(\pi/10)$. The other equation follows from the fact that $\cos(\pi/5)=\varphi/2$ (see [6]) and from the well-known expression of sines and cosines in terms of the cosine of the double angle. This completes the proof of the proposition. Proposition 17. If there is a non-trivial isometry $f$ of the full truncation of the $600$-cell such that $\rho(v,f(v))\leqslant l$ (the number $l$ is given in Corollary 5) for the Euclidean metric $\rho$ and all $v\in M$, then $\rho(y,f(y))\leqslant 2/(\sqrt{5}\,\varphi^2)$ for all $y\in S^3(0,1)$. Proof. The facets of the truncated $600$-cell are regular icosahedrons and octahedra with the edge length $l$. Then the radii of the circumscribed sphere of the icosahedron and octahedron are respectively $(l/2)\bigl((\sqrt{5+\sqrt{5}})/\sqrt{2}\bigr)$ and $(l/2)\sqrt{2}$, and the first number exceeds the other one. For every point $x\in \mathbb{R}^4$ we have $\rho(x,f(x))=|f(x)-x|$, and the function $x\to f(x)-x$ is linear on $\mathbb{R}^4$. Let $x$ be an arbitrary point of some facet the polytope under consideration (an octahedron or an icosahedron). Then $x=\sum_{k=1}^6x_kv_k$ or $x=\sum_{k=1}^{12}x_kv_k$, where $x_k\geqslant 0$, $\sum_kx_k=1$, and $v_k$ are the vertices of the octahedron or icosahedron, and
$$
\begin{equation*}
\rho(x,f(x))=|f(x)-x|=\biggl|\sum_kx_k(f(v_k)-v_k)\biggr|\leqslant \sum_kx_k|f(v_k)-v_k|\leqslant \sum_kx_kl=l.
\end{equation*}
\notag
$$
Here the value of the displacement of the central projection $y$ of the point $x$ to the sphere $S^3(0,1)$ does not exceed the quotient of $l$ divided by the distance from the centre of the facet to $0$. The last distance of the icosahedron is less than that of the octahedron; therefore, when obtaining the upper bound for $\rho(y,f(y))$, we consider the icosahedron. Then, as a consequence of what was said above, and also of the equations $l^{-2}=5+2\sqrt{5}$ and $2\varphi^4=7+3\sqrt{5}$, we have
$$
\begin{equation*}
\begin{aligned} \, &\cos\biggl(\arcsin \biggl(\frac{l}2\frac{\sqrt{5+\sqrt{5}}}{\sqrt{2}}\biggr)\biggr) =\sqrt{1-\frac{(5+\sqrt{5})l^2}{8}} \\ &\qquad=\frac{1}{2\sqrt{2}}\,\sqrt{8-\frac{5+\sqrt{5}}{5+2\sqrt{5}}}= \frac{\sqrt{5}}{2\sqrt{2}}\, \sqrt{\frac{7+3\sqrt{5}}{5+2\sqrt{5}}} =\frac{\sqrt{5}}{2}\, \sqrt{\varphi^4l^2}=\frac{\sqrt{5}\,l\varphi^2}{2}. \end{aligned}
\end{equation*}
\notag
$$
This implies that This completes the proof of the proposition. Proposition 18. There is no isometry $f$ satisfying Proposition 17. Proof. Suppose that such an isometry $f$ exists. Then $f(v_1)=v_2$ for some $v_1,v_2\in M$ such that $\rho(v_1,v_2)=l$. Hence, $v_1$ and $v_2$ are the midpoints of some two adjacent edges of a $600$-cell connecting vertices $w_0$, $w_1$ and $w_0$, $w_2$, respectively. It is clear that, by Proposition 15, two cases are possible:
1) $f(w_0)=w_0$, $f(w_1)=w_2$; 2) $f(w_0)=w_2$, $f(w_1)=w_0$. In the case 1), $w_0$, $w_1$, and $w_2$ are the vertices of an equilateral spherical triangle on some two-dimensional sphere $S^2(0,R)\subset S^3(0,R)\subset\mathbb{R}^4$, where $S^3(0,R)$ is the circumscribed sphere for the $600$-cell. The extensions of the sides $[w_0,w_1]$, $[w_0,w_2]$ of this spherical triangle cut points $Z_1$ and $Z_2$ such that $f(Z_1)=Z_2$ on the equator of the sphere $S^2(0,R)$ with a pole $w_0$. Here $\rho(Z_1,Z_2)>R$, since the angle $C$ of this equilateral spherical triangle is greater than $\pi/3$. Then the intersections of the segments $[0,Z_1]$ and $[0,Z_2]$ with the sphere $S^3(0,1)$ give points $z_1$ and $z_2$ such that $f(z_1)=z_2$ and $\rho(z_1,z_2)>1$. However, by Proposition 17, we should have $\rho(z_1,f(z_1))\leqslant 2/(\sqrt{5}\varphi^2)< 1$, a contradiction. In the case 2), more precise estimates are to be made, including the calculation of the angle $C$. Let us make a stretch with the coefficient $1/R$ in such a way that $S^3(0,1)$ is the circumscribed sphere for the $600$-cell. We keep the same notation for $Z_1$, $Z_2$ (instead of $z_1$, $z_2$) and the vertices $w_0$, $w_1$, $w_2$ of the triangle under consideration. Then the length of the side of the triangle is $a=2\arcsin(l/R)$, where
$$
\begin{equation*}
\frac{1}{R}=\cos \frac{\pi}{10} =\sqrt{\frac{1+\cos(\pi/5)}{2}} =\sqrt{\frac{1+\varphi/2}{2}} =\frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}.
\end{equation*}
\notag
$$
Due to the cosine theorem of spherical geometry, for the angle $C$ of the triangle under consideration we have
$$
\begin{equation*}
\begin{gathered} \, \cos C=\frac{\cos a - \cos^2a}{\sin^2a} =\frac{\cos a(1-\cos a)}{1-\cos^2a}=\frac{\cos a}{1+\cos a}, \\ \cos a = 1 - 2\sin^2\biggl(\arcsin\biggl(\frac{l}{R}\biggr)\biggr) =1-\frac{2l^2}{R^2} =1-\frac{\sqrt{5}+1}{4(2+\sqrt{5})} =\frac{7+3\sqrt{5}}{4(2+\sqrt{5})}, \\ \cos C=\biggl(\frac{7+3\sqrt{5}}{4(2+\sqrt{5})}\biggr)\cdot \biggl(1+\frac{7+3\sqrt{5}}{4(2+\sqrt{5})}\biggr)^{-1}= \frac{7+3\sqrt{5}}{15+7\sqrt{5}}=\frac{1}{\sqrt{5}}. \end{gathered}
\end{equation*}
\notag
$$
This time, $f(Z_1)$ is equal to the shift of $-Z_2$ by the angle $a$ in the direction towards $Z_2$ along the semicircle $[(-Z_2)Z_2]$ with midpoint $w_0$ rather than to $Z_2$. Then the spherical distance is
$$
\begin{equation*}
d(Z_1,f(Z_1))\geqslant \pi -C -a =\pi-(a+C)= \pi -\biggl(\arccos\biggl(\frac{7+3\sqrt{5}}{4(2+\sqrt{5})}\biggr) + \arccos\biggl(\frac{1}{\sqrt{5}}\biggr)\biggr),
\end{equation*}
\notag
$$
while, as a consequence of Proposition 17, it should be
$$
\begin{equation*}
d(Z_1,f(Z_1))\leqslant \arccos\biggl(\frac{31+15\sqrt{5}}{35+15\sqrt{5}}\biggr).
\end{equation*}
\notag
$$
Computer calculations show that the right-hand side of the former inequality is greater than the right-hand side of the latter by $1.062800925\dots$ . The contradictions obtained in both the cases prove Proposition 18. § 6. Five-dimensional Gosset polytope $\operatorname{Goss}_5$Let us prove a general assertion for spaces of arbitrary dimension. Proposition 19. The set of vertices of every semihypercube in $\mathbb{R}^n$ with the metric induced from $\mathbb{R}^n$ is Clifford–Wolf homogeneous. Proof. We recall that an $n$-dimensional hypercube with edge length $2$ can be represented as the convex hull of points of the form $(\pm1,\pm1,\dots,\pm 1)\in \mathbb{R}^n$ with arbitrarily chosen signs of every coordinate. It is clear that the group $(\mathbb{Z}_2)^n$ acts simply transitively on the set of the vertices of this hypercube, where $\mathbb{Z}_2=\{-1,1\}$ (the vertices of the hypercube are naturally identified with elements of $(\mathbb{Z}_2)^n$), which consists of Clifford–Wolff translations of the set of vertices of the hypercube.
Consider the subset of vertices $G$ of the hypercube that consists of all vertices with even number of characters “$-$” in the coordinates and identify this set with the subset of the group $(\mathbb{Z}_2)^n$, which is also denoted by $G$. It is not difficult to see that the product of elements in $G$ remains in $G$. Thus, $G$ is a subgroup of index $2$ in the group $(\mathbb{Z}_2)^n$. The convex hull of all vertices in $G$ is a semihypercube by definition. The group $G$ is commutative and acts on the set of hypercube vertices by Clifford–Wolf translations. Moreover, $G$ is simply transitive on $G$ and, by the commutativity of $G$, consists of the Clifford–Wolf translations of the set of vertices of the semihypercube. Really, choose some element $b\in G$. For arbitrary $x,y \in G$, there is an element $a\in G$ such that $y=a\cdot x$ (the composition in the group $G$). Since $a\cdot b = b \cdot a$ and the metric $\rho$ is invariant under the left shifts of $G$, we have $\rho(y,b\cdot y)=\rho(a\cdot x,b\cdot a\cdot x)=\rho(a\cdot x,a\cdot b \cdot x)=\rho(x,b \cdot x)$, that is, the multiplication by $b$ is a Clifford–Wolf translation. This completes the proof of the proposition. Since the five-dimensional Gosset polytope is a semihypercube, we immediately obtain the following assertion. Corollary 6. The set of vertices of the Gosset polytope $\operatorname{Goss}_5$ in $\mathbb{R}^5$ is Clifford–Wolf homogeneous. § 7. Six-dimensional Gosset polytope $\operatorname{Goss}_6$The six-dimensional Gosset polytope $\operatorname{Goss}_6$ can be implemented in different ways. Let us define it with by the coordinates of the vertices in $\mathbb{R}^6$, as is done in [12]. Let $a=\sqrt{2}/4$ and $b=\sqrt{6}/12$. Define the points $A_i \in \mathbb{R}^6$, $i=1,\dots,27$, as follows:
$$
\begin{equation*}
\begin{alignedat}{2} A_1 &=(0,0,0,0,0,4b), &\qquad A_2 &=(a,a,a,a,a,b), \\ A_3 &=(-a,-a,a,a,a,b), &\qquad A_4 &=(-a,a,-a,a,a,b), \\ A_5 &=(-a,a,a,-a,a,b), &\qquad A_6 &=(-a,a,a,a,-a,b), \\ A_7 &=(a,-a,-a,a,a,b), &\qquad A_8 &=(a,-a,a,-a,a,b), \\ A_9 &=(a,-a,a,a,-a,b), &\qquad A_{10} &=(a,a,-a,-a,a,b), \\ A_{11} &=(a,a,-a,a,-a,b), &\qquad A_{12} &=(a,a,a,-a,-a,b), \\ A_{13} &=(-a,-a,-a,-a,a,b), &\qquad A_{14} &=(-a,-a,-a,a,-a,b), \\ A_{15} &=(-a,-a,a,-a,-a,b), &\qquad A_{16} &=(-a,a,-a,-a,-a,b), \\ A_{17} &=(a,-a,-a,-a,-a,b), &\qquad A_{18} &=(2a,0,0,0,0,-2b), \\ A_{19} &=(0,2a,0,0,0,-2b), &\qquad A_{20} &=(0,0,2a,0,0,-2b), \\ A_{21} &=(0,0,0,2a,0,-2b), &\qquad A_{22} &=(0,0,0,0,2a,-2b), \\ A_{23} &=(-2a,0,0,0,0,-2b), &\qquad A_{24} &=(0,-2a,0,0,0,-2b), \\ A_{25} &=(0,0,-2a,0,0,-2b), &\qquad A_{26} &=(0,0,0,-2a,0,-2b), \\ A_{27} &=(0,0,0,0,-2a,-2b). &\qquad & \end{alignedat}
\end{equation*}
\notag
$$
The Gosset polytope $\operatorname{Goss}_6$ is the convex hull of these points. It can readily be seen that $\rho(A_1, A_i)=1$ for $2\leqslant i \leqslant 17$ and $\rho(A_1, A_i)=\sqrt{2}$ for $18\leqslant i \leqslant 27$. Remark 6. Using the notation $\overline{a_1,a_2,\dots,a_m}$ for the set of all permutations obtained from the elements $a_1,a_2,\dots,a_m$, one can write out the set of points $A_i$ for $3\leqslant i \leqslant 27$ in a more compact way: It is clear that the points $A_2$–$A_{17}$ are the vertices of a five-dimensional semihypercube (the corresponding hypercube has $32$ vertices of the form $(\pm a, \pm a,\pm a,\pm a, \pm a,b)$), and the points $A_{18}$–$A_{27}$ are the vertices of a five-dimensional hyperoctahedron (orthoplex), which is a facet of the polytope $\operatorname{Goss}_6$ (lying in the hyperplane $x_6=-2b$). The origin $O=(0,0,0,0,0,0)\in \mathbb{R}^6$ is the centre of the hypersphere circumscribed around $\operatorname{Goss}_6$ with the radius $4b=\sqrt{ 2/3}$. We note that, among the vertices of the above five-dimensional hyperoctahedron, there are five pairs of vertices such that the points of each of them are at the distance $d_2=\sqrt{2}$ from each other. Together with the vertex $A_1$, every such pair forms a regular triangle with the side length $d_2$. There are $27\cdot 5/3=45$ such triangles. We call them large triangles in $\operatorname{Goss}_6$. Every such triangle has the origin $O$ as its centre of mass. Moreover, only these $45$ triangles have this property. Really, if we have three vertices $A_i,A_j,A_k \in \operatorname{Goss}_6$ such that $A_i+A_j+A_k=O=(0,0,0,0,0,0)$, then, without loss of generality, we can assume that $i=1$ (due to the homogeneity of the polytope $\operatorname{Goss}_6$). Since the last component of $A_1$ is equal to $4b$, it follows that $18 \leqslant i,j \leqslant 27$ and, moreover, the first non-zero component of $A_j$ must have the same number as the first non-zero component of $A_k$. Thus, the points $A_i$, $A_j$, $A_k$ form a regular triangle with the side length $d_2=\sqrt{2}$. It can readily be seen that the set of isometries of the space $\mathbb{R}^6$ that take $\operatorname{Goss}_6$ into itself and leave the straight line $Ox_6$ fixed acts transitively on the set of all vertices $A_i$ for $2\leqslant i \leqslant 17$ and for $18\leqslant i \leqslant 27$. Proposition 20. The set of vertices of the six-dimensional Gosset polytope $\operatorname{Goss}_6$ is normally homogeneous. Proof. Let us use the assertion of Proposition 10. As is well known, the Gosset polytope $\operatorname{Goss}_6$ is homogeneous, and the distance between its different vertices takes two values ($d_1=1$ and $d_2=\sqrt{2}$, up to similarity), see the calculations above.
We claim that (in the above notation) the isotropy group $I(A_1)$ of the vertex $A_1$ acts transitively on the set of points $A_i$, $i=2,\dots, 17$. We note that $I(A_1)$ contains all permutations of the coordinates $x_i$ and $x_j$, where $i,j =1,\dots,5$, $i\neq j$. Moreover, $I(A_1)$ contains the transformation
$$
\begin{equation*}
(x_1,x_2,x_3,x_4,x_5,x_6) \mapsto (-x_1,-x_2,x_3,x_4,x_5,x_6).
\end{equation*}
\notag
$$
This implies that some element of the group $I(A_1)$ moves a given point $A_i$ to another given point $A_j$, $i,j =2,\dots,17$. Thus, condition 3) of Proposition 10 is satisfied.
We note that the mapping $\eta$ acting according to the rule $(x_1,x_2,x_3,x_4,x_5,x_6)\mapsto (x_2,x_1,x_3,x_4,x_5,x_6)$ is an isometry of the Gosset polytope $\operatorname{Goss}_6$. Here we have $\rho (A_i, \eta(A_i)) \leqslant d_1=1$ for all $i=1,\dots,27$. Moreover, $\rho (B, \eta(B)) = d_1=1$, where $B=(2a,0,0,0,0,-2b)$. Then condition 4) of Proposition 10 is also satisfied. Now the assertion of our proposition follows from Proposition 10. This completes the proof of the proposition. Proof. Due to the compactness of the set $U$, the solution of the minimum problem under consideration exists. Without loss of generality, we can assume that $x_1\leqslant x_2\leqslant x_3\leqslant x_4\leqslant x_5$. It is easy to understand that the solution is obtained for the case in which the number $2x_5=2\max\{x_1,\dots,x_5\} = \sum_{i=1}^5 x_i$ is minimal. If $x_5= \sum_{i=1}^4 x_i \leqslant 1/\sqrt{2}$, then $\sum_{i=1}^5x_i^2\leqslant \bigl( \sum_{i=1}^4 x_i \bigr)^2+x_5^2\leqslant 1$, and the equality between the extreme parts is possible only for $x_4=x_5=1/\sqrt{2}$ and $x_1=x_2=x_3=0$. It is clear that, for these values, the desired quantity is equal to $\sqrt{2}$. This completes the proof of the lemma. Proposition 21. For an arbitrary two-dimensional vector subspace $P^2$ in $\mathbb{R}^6$, there is a unit vector in $\mathbb{R}^6$ for which its inner product by one of the vertices of $\operatorname{Goss}_6$ is not less than $1/2$, or, if the circumscribed sphere of this polytope is normalised to the unit sphere, not less than $\sqrt{3/8}$. Proof. It is clear that the subspace $P^2$ contains a $1$-dimensional subspace $P^1$ which lies in the hyperplane $x_6=0$ as well. Let $e$, $-e$ be the unit vectors in $P^1$. For one of these vectors, let it be $e$, all products of its coordinates with the corresponding coordinates of one of the vertices $A_2$–$A_{17}$ are non-negative. Then the inner product of these two vectors is the product of the sum of moduli of the coordinates of the vector $e$ by $a$. The inner product of $e$ and some vector in $A_{18}$–$A_{27}$ is equal to the product of the modulus of an arbitrarily chosen coordinate of the vector by $2a$. It follows from the assertion of Lemma 5 that the inner product of the vector $e$ by one of the vertices is not less than $\sqrt{2}\, a=1/2$. When normalizing the circumscribed sphere to one, we obtain the number $\sqrt{3}/(2\sqrt{2})$. This completes the proof of the proposition. Let us now establish another remarkable property of the polytope $\operatorname{Goss}_6$. We recall that the definition of almost perfect polytopes and a description of their properties are given in § 4. Proof. Let $F\colon \mathbb{R}^6 \to \mathbb{R}$ be a quadratic form such that $F(A_i)=1$ for $i=1,2,\dots,27$. Let $F(x)=\sum_{i,j=1}^6 a_{ij} x_ix_j$, where $a_{ij}=a_{ji}$ for all indices.
Since $F(A_1)=16b^2a_{66}=(2/3) a_{66}=1$, it follows that $a_{66}=3/2$. Further, $F(A_{18})=4a^2a_{11}+4b^2a_{66}-8aba_{16}=1$ and $F(A_{23})=4a^2a_{11}+4b^2a_{66}+8aba_{16}=1$ and, therefore, $a_{16}=0$ and $(1/2)a_{11}+(1/6)\cdot (3/2)=1$, that is, $a_{11}=3/2$. Considering similarly pairs of points $A_i$ and $A_{i+5}$ for $i=19,20,21,22$, we obtain $a_{k6}=0$ and $a_{kk}=3/2$ for $1\leqslant k \leqslant 5$. We note now that $F(x)=F(-x)$ and, therefore, together with every vertex of the polytope $\operatorname{Goss}_6$, the second-order hypersurface $F(X)=1$ contains a point symmetric to the point with respect to $O$. Since $a_{k6}=0$ for $1\leqslant k \leqslant 5$, it follows that the change of the sign of the sixth coordinate of an arbitrary point on this hypersurface gives again a point on the hypersurface. Thus, the hypersurface under consideration contains all points of the form $(\pm a, \pm a, \pm a, \pm a, \pm a, b)$ for any choice of the signs $\pm$. If we consider $G\colon \mathbb{R}^5 \to \mathbb{R}$, where $G(x)=\sum_{i,j=1}^5 a_{ij} x_ix_j$, then, for any vertices $B=(\pm a, \pm a, \pm a, \pm a, \pm a)$ of the five-dimensional hypercube, we have
$$
\begin{equation*}
G(B)=F(B')-a_{66}b^2=1-\frac{3}{2}\cdot \frac{1}{24}=\frac{15}{16},
\end{equation*}
\notag
$$
where the point $B'\in \mathbb{R}^6$ is obtained from $B$ by adding the sixth coordinate $b$. Now Lemma 2 implies the equality $a_{ij}=0$ for $1\leqslant i,j \leqslant 5$, $i\neq j$. Thus, we have $F(x)=(3/2) (x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2)$. This completes the proof of the theorem. Remark. Since the Gosset polytope $\operatorname{Goss}_6$ is a Delaunay polytope of the lattice $\operatorname{E}_6$, it follows that the result of the above theorem readily follows from Lemma 1 and from the fact that this polytope is perfect, that is, the Delaunay sphere of the corresponding lattice is a unique ellipsoid circumscribed around the polytope; see [22] for details. Theorem 7. The set $M$ of vertices of the polytope $\operatorname{Goss}_6$ admits no $\mathrm{CW}$-translation by the distance $1$ (and is not Clifford–Wolf homogeneous). Proof. To apply Proposition 11, it is convenient to normalise $\operatorname{Goss}_6$ in such a way that the radius of its circumscribed sphere is $1$. Then we are talking about a $\mathrm{CW}$-translation by $r_0\,{=}\,\sqrt{3}/\sqrt{2}:=2\sin(\alpha/2)$. In this case,
Since the polytope $\operatorname{Goss}_6$ is almost perfect by Theorem 6, it follows from Proposition 11 that $\mathbb{R}^6$ splits into a direct orthogonal sum $\mathbb{R}^6=\bigoplus_{l=1}^3\mathbb{R}^2_l$ in such a way that the restriction of $f$ to every Euclidean $2$-plane $\mathbb{R}^2_l$, $l=1,2,3$, is a rotation by an angle $2\pi t/k$, where $k\geqslant 3$ is a divisor of the number $27$, $t\in [1,k/2)$, and $r=2\sin(\pi t/k)$. Then $\alpha=2\pi t/k$, and the possible values of $k$ are $k=3,9,27$. For $k=3$ it should be $t=1$, and we obtain the $\mathrm{CW}$-translation of the unit sphere by $r=2\sin(\pi/3)=\sqrt{3}$. This $\mathrm{CW}$-translation for $M$ exists indeed, but $\sqrt{3}>\sqrt{3}/\sqrt{2}$. For $k=9$ and $t=1$, we obtain the $\mathrm{CW}$-translation of the unit sphere by
$$
\begin{equation}
r=2\sin\frac{\pi}9 < 2\sin\frac{\pi}8= \sqrt{2-\sqrt{2}}<1 <\frac{\sqrt{3}}{\sqrt{2}}.
\end{equation}
\tag{9}
$$
For $k=9$ and $t=2$, we obtain the Clifford–Wolf translation of the unit sphere by
$$
\begin{equation}
r=2\sin\frac{2\pi}9, \qquad \cos\frac{4\pi}9= \sin\frac{\pi}{18}= 0.1736481777\ldots < \frac14,\quad r>r_0.
\end{equation}
\tag{10}
$$
For $k=27$, taking into account the inequalities (9), (10), we see that it remains to consider the cases $t\,{=}\,4,5$:
$$
\begin{equation*}
\cos\frac{8\pi}{27} = 0.5971585916\ldots > \cos\frac{10\pi}{27}= 0.39607976657\ldots > \frac14, \quad r < r_0.
\end{equation*}
\notag
$$
The Clifford–Wolf inhomogeneity of the set $M$ now follows from the fact that $\rho(A_1,A_2)=1$. This completes the proof of Theorem 7. We can give an explicit example of the Clifford–Wolf translation with the shift $\sqrt{2}$ for the polytope $\operatorname{Goss}_6$. In the standard basis in $\mathbb{R}^6$, such a translation, for example, is given by the matrix (we assume that the mapping corresponding to a matrix $A$ is given by the formula $x \mapsto x \cdot A$)
$$
\begin{equation*}
\widetilde{A}= \frac{1}{2}\begin{pmatrix} -1 & 1 & 1 & -1 & 0 & 0 \\ -1 & -1 & 1 & 1 & 0 & 0 \\ -1 & -1 & -1& -1 & 0 & 0 \\ 1 & -1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & \sqrt{3} \\ 0 & 0 & 0 & 0 & -\sqrt{3} & -1 \end{pmatrix}.
\end{equation*}
\notag
$$
This matrix has two eigenvalues $(-1 \pm \mathrm{i}\sqrt{3})/2$ of multiplicity three and defines the Clifford–Wolf translation $I$ on the sphere $S(0,1)$. We note that the set of vertices $A_i$, $1\leqslant i \leqslant 27$, is divided into nine cycles of three elements with respect to the corresponding Clifford–Wolf translation. These cycles are as follows (we indicate the numbers of the vertices): $(1,27,22)$, $(2,15,25)$, $(3,17,19)$, $(4,9,26)$, $(5,14,18)$, $(6,24,10)$, $(7,12,23)$, $(8,16,21)$, and $(11,20,13)$. It is clear that each of these cycles defines a large triangle in the polytope $\operatorname{Goss}_6$. We note that, for pairwise orthogonal and $I$-invariant two-dimensional planes, we can consider
$$
\begin{equation*}
\begin{aligned} \, \pi_1 &= \{(x_1,x_2,x_3,0,0,0)\mid x_1-x_2+x_3=0\}, \\ \pi_2 &= \{(x_1,x_2,x_3,x_4,0,0)\mid x_1+x_2=0,\, x_1=x_3\}, \\ \pi_3 &= \{(0,0,0,0,x_5,x_6)\}. \end{aligned}
\end{equation*}
\notag
$$
This can readily be verified by taking into account the following displacements of points under the action of $I$:
$$
\begin{equation*}
\begin{gathered} \, (1,1,0,0,0,0) \to (-1,0,1,0,0,0) \to (0,-1,-1,0,0,0) \to (1,1,0,0,0,0), \\ (0,0,0,2,0,0) \to (1,-1,1,-1,0,0) \to (-1,1,-1,-1,0,0) \to (0,0,0,2,0,0). \end{gathered}
\end{equation*}
\notag
$$
Since the polytope $\operatorname{Goss}_6$ is homogeneous, and the isotropy group of an arbitrary vertex of $\operatorname{Goss}_6$ acts transitively on the set of vertices placed at the distance of $\sqrt{2}$ from this vertex (the same holds for the set of vertices placed at the distance of $1$ from the chosen vertex), it follows that the above example of the Cliffoord–Wolf translation shows that any vertex $u$ of the polytope $\operatorname{Goss}_6$ can be translated to any other vertex $v$ at the distance $\sqrt{2}$ from $u$ by a suitable Clifford–Wolf translation for the vertex set of $\operatorname{Goss}_6$. Consider another isometry, of the polytope $\operatorname{Goss}_6$, generated by the matrix
$$
\begin{equation*}
\widetilde{B}=\frac{1}{4}\begin{pmatrix} 1& 1 & -1 & -3 & -1& -\sqrt{3}\, \\ -3& 1 & -1 & 1 & -1& -\sqrt{3} \\ 1& 1 & -1 & 1 &3& -\sqrt{3} \\ 1& -3 & -1 & 1 &-1& -\sqrt{3} \\ 1& -1 & 3 & -1 &1& \sqrt{3} \\ \sqrt{3} &\sqrt{3} & -\sqrt{3} &\sqrt{3} & -\sqrt{3} & 1 \end{pmatrix}.
\end{equation*}
\notag
$$
This matrix has two eigenvalues $(1 \pm \mathrm{i}\sqrt{3})/2$ of multiplicity two and two eigenvalues $(-1 \pm \mathrm{i}\sqrt{3})/2$ of multiplicity one. In this case, the set of vertices $A_i$, $1\leqslant i \leqslant 27$, is split into five cycles under the action of this isometry. One of these cycles (we indicate the vertex numbers), $(2,15,25)$, has three elements that move by the distance $\sqrt{2}$ (the cyclic movement of the vertices of the large triangle). The remaining vertices move by the distance $1$ and form the cycles $(1,11,27,20,22,13)$, $(3,7,17,12,19,23)$, $(4,14,9,18,26,5)$, $(6,21,24,8,10,16)$. Each of these four cycles defines two large triangles, which are translated into each other by the isometry under consideration. Definition 8. A metric space $(M,d)$ is called a space with $\mathrm{MD}$-property (from “Minimum Displacement”) if, for any two points $u,v \in M$, there is an isometry $I$ of the space under consideration such that $I(u)=v$ and $d(u,v)=\min_{x\in M} d(x,I(x))$. Isometries of this kind are opposite in a certain sense to $\delta$-shifts (see Definition 4). Naturally, the question about the description of metric spaces with $\mathrm{MD}$-property arises. It is clear that all spaces of this kind are homogeneous and that every Clifford–Wolf homogeneous metric space has the $\mathrm{MD}$-property. However, not only $\mathrm{CW}$-homogeneous spaces have the $\mathrm{MD}$-property. Proposition 22. The set of vertices $M$ of the polytope $\operatorname{Goss}_6$ has the $\mathrm{MD}$-property. Proof. For arbitrary distinct points $u,v \in M$ we have $\rho(u,v)=1$ or $\rho(u,v)=\sqrt{2}$. If $\rho(u,v)=\sqrt{2}$, then, as the arguments above show, there is a $\mathrm{CW}$-translation $I$ of the set $M$ such that $I(u)=v$. It is clear that $\rho(u,v)=\min_{x\in M} d(x,I(x))$, since $d(x,I(x))\equiv \sqrt{2}$.
If $\rho(u,v)=1$, then we conclude, keeping in mind the above example of an isometry generated by the matrix $\widetilde{B}$, the homogeneity of the polytope $\operatorname{Goss}_6$, and the fact that the isotropy group of any of its vertices acts transitively on the set of vertices at the distance of $1$ from the vertex, that there is an isometry $J$ of the set $V$ such that $J(u)=v$, and there is no vertex fixed with respect to $J$. The latter means that $1=\rho(u,v)=\min_{x\in M}d(x,I(x))$. This completes the proof of the proposition. Theorem 7 and Proposition 22 raise a natural question. Question 4. What is the minimum number of vertices of a polygon $P$ in Euclidean space whose set of vertices is not Clifford–Wolf homogeneous and has the $\mathrm{MD}$-property? Problem 1. Classify all homogeneous polytopes in Euclidean spaces whose vertex sets have the $\mathrm{MD}$-property. Since the distance between different vertices of the polytope $\operatorname{Goss}_6$ takes only two values, it is natural to consider the following problem. Problem 2. Classify all homogeneous polytopes in Euclidean spaces for which the distance between different vertices takes only two values. If the distance between different vertices of some polytope is the same, then this polytope is a regular simplex (even without the homogeneity assumption). § 8. Seven-dimensional Gosset polytope $\operatorname{Goss}_7$The seven-dimensional Gosset polytope $\operatorname{Goss}_7$ can be implemented in different ways. We define this polytope by the vertex coordinates in $\mathbb{R}^7$, as this was done in [12]. Below, the symbol $\overline{a_1,a_2,\dots,a_m}$ denotes the set of all permutations obtained from the elements $a_1,a_2,\dots,a_m$. Let $a=\sqrt{2}/4$, $b=\sqrt{6}/12$, and $c=\sqrt{3}/6$. We define the points $B_i \in \mathbb{R}^7$, $i=1,\dots,56$, as follows:
$$
\begin{equation*}
\begin{gathered} \, B_1=(0,0,0,0,0,0,3c), \qquad B_2=(0,0,0,0,0,4b,c), \quad B_3=(a,a,a,a,a,b,c), \\ \begin{alignedat}{2} B_4-B_{13} &=(\overline{-a,-a,a,a,a},b,c), &\quad B_{14}-B_{18} &=(\overline{-a,-a,-a,-a,a},b,c), \\ B_{19}-B_{23} &=(\overline{2a,0,0,0,0},-2b,c), &\quad B_{24}-B_{28} &=(\overline{-2a,0,0,0,0},-2b,c), \\ B_{29}-B_{33} &=(\overline{-2a,0,0,0,0},2b,-c), &\quad B_{34}-B_{38} &=(\overline{2a,0,0,0,0},2b,-c), \\ B_{39}-B_{48} &=(\overline{a,a,-a,-a,-a},-b,-c), &\quad B_{49}-B_{53} &=(\overline{a,a,a,a,-a},-b,-c), \\ B_{54} &=(-a,-a,-a,-a,-a,-b,-c), &\quad B_{55} &=(0,0,0,0,0,-4b,-c), \\ B_{56} &=(0,0,0,0,0,0,-3c). &\quad & \end{alignedat} \end{gathered}
\end{equation*}
\notag
$$
The Gosset polytope $\operatorname{Goss}_7$ can be represented as the convex hull of these points. It is clear that the last $28$ points are obtained from the first $28$ points with the help of central symmetry about the origin $O$. Thus, the radius of the hypersphere circumscribed around the polytope under consideration is $3c=\sqrt{3}/2$. It is clear that the convex hull of the points $B_i$ for $2\leqslant i \leqslant 28$ is the Gosset polytope $\operatorname{Goss}_6$ (the vertex figure for the vertex $B_1\in \operatorname{Goss}_7$). The polytope symmetric to it is formed by the points $B_i$ for $29\leqslant i \leqslant 55$. It can readily be seen that $\rho(B_1, B_i)=1$ for $2\leqslant i \leqslant 28$, $\rho(B_1, B_i)=\sqrt{2}$ for $29\leqslant i \leqslant 55$, and $\rho(B_1, B_{56})=\sqrt{3}$. It should be noted that the central symmetry of $\mathbb{R}^7$ with respect to the origin is an isometry of the polytope $\operatorname{Goss}_7$, and it shifts all its vertices by the distance $d_3=\sqrt{3}$ (thus, this is a Clifford–Wolf translation). It can readily be seen that the set of isometries of the space $\mathbb{R}^7$ that take $\operatorname{Goss}_7$ into itself and leave the straight line $Ox_7$ fixed acts transitively on the set of all vertices $B_i$ for $2\leqslant i \leqslant 28$, and also for $29\leqslant i \leqslant 55$ (which in fact means the homogeneity of the Gosset polytope $\operatorname{Goss}_6$). Really, any isometry of $\operatorname{Goss}_6$ can be regarded as an isometry of the vertex figure of $\operatorname{Goss}_7$ with a chosen fixed vertex, and it naturally extends to the isometry of the whole polytope $\operatorname{Goss}_7$. Proposition 23. The set of vertices of the seven-dimensional Gosset polytope $\operatorname{Goss}_7$ is a normal homogeneous metric space. Proof. Let us use the assertion of Proposition 10. As is well known, the Gosset polytope $\operatorname{Goss}_7$ is homogeneous and, as shown above, the distance between its different vertices takes three values ($d_1=1$, $d_2=\sqrt{2}$, and $d_3=\sqrt{3}$ up to similarity).
We claim that (in the above notation) the isotropy group $I(B_1)$ of the vertex $B_1$ acts transitively on the set of points $B_i$, $i=2,\dots, 28$ (the set of vertices of $\operatorname{Goss}_7$ placed at the distance $d_1$ from the vertex $B_1$). This holds, since the corresponding set of points forms a homogeneous Gosset polytope $\operatorname{Goss}_6$. In just the same way, the Gosset polytope $\operatorname{Goss}_6$ is also formed by the set of points $B_i$, $i=29,\dots, 55$ (it is centrally symmetric to the previous one). Thus, the isotropy group $I(B_1)$ of the vertex $B_1$ acts transitively on the set of points $B_i$, $i=29,\dots, 55$ (this is exactly the set of vertices of $\operatorname{Goss}_7$ located at the distance $d_2$ from the the $B_1$). Therefore, condition 3) of Proposition 10 is satisfied. We note that the mapping $\eta$ acting by the rule
$$
\begin{equation*}
(x_1,x_2,x_3,x_4,x_5,x_6,x_7) \mapsto (x_2,x_1,x_3,x_4,x_5,x_6,x_7),
\end{equation*}
\notag
$$
is an isometry of the Gosset polytope $\operatorname{Goss}_7$. Here $\rho (B_i, \eta(B_i)) \leqslant d_1=1$ for all $i=1,\dots,56$. Moreover, $\rho (B, \eta(B))\,{=}\, d_1\,{=}\,1$, where $B=(2a,0,0,0,0,-2b,c)$. Therefore, we have found an isometry of $\operatorname{Goss}_7$ with the maximum vertex shift $d_1=1$.
Now let us consider an isometry $\theta$ of the six-dimensional Gosset polytope $\operatorname{Goss}_6$ with the maximum vertex shift $d_2=\sqrt{2}$ (such an isometry exists due to the homogeneity of the polytope $\operatorname{Goss}_6$). Consider the isometry $\theta'$ of the Gosset polytope $\operatorname{Goss}_7$ defined as follows: $\theta'$ leaves fixed the vertices $B_1$ and $B_{56}$ and acts on the vertex figure related to the vertex $B_1$ by the rule $\theta$ (the vertex figure of $\operatorname{Goss}_7$ is $\operatorname{Goss}_6$ under the natural identification). It is clear that the isometry $\theta'$ is completely determined by these properties, while it acts in agreement with the central symmetry with respect to the point $O=(0,0,0,0,0,0,0)\in \mathbb{R}^7$ on the vertex figure $\operatorname{Goss}_7$ related to $B_{56}$. Thus, we have found an isometry of $\operatorname{Goss}_7$ with the maximum vertex shift $d_2=\sqrt{2}$. Therefore, condition 4) of Proposition 10 also holds. The assertion of our proposition follows now from Proposition 10. This completes the proof of the proposition. Proposition 24. There is no $\mathrm{CW}$-translation $f$ by the distance $\sqrt{2}$ of the set of vertices of the polytope $\operatorname{Goss}_{7}$. Proof. Suppose that such a translation $f$ exists. The facets of the polytope $\operatorname{Goss}_{7}$ are six-dimensional regular simplices and orthoplexes with the edge length $l=1$. Then the radii of their circumscribed spheres are $\sqrt{6/(2\cdot 7)}=\sqrt{3/7}$ and $\sqrt{2}/2$, respectively. The latter radius is greater. Therefore, taking into account the fact that the radius of the circumscribed sphere of the polytope is equal to $\sqrt{3}/2$, we see that the radius of its inscribed sphere and the value $\cos\xi$ for the central angle $\xi$ with which the radius of the circumscribed sphere of the orthoplex is seen, respectively, are equal to
$$
\begin{equation*}
r=\sqrt{\frac{3}{4}-\frac{2}{4}}=\frac{1}{2},\qquad \cos\xi = \frac{1}{2}\cdot \frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}.
\end{equation*}
\notag
$$
The following lemma was proved in the book [23].
Lemma bor (see Lemma 6.5.2 of [23]). Spherical balls of radius $\eta<\pi/2$ on the unit sphere $S^d(0,1)\subset\mathbb{R}^{d+1}$ cover $S^d(0,1)$ if and only if the convex hull of their centres in $\mathbb{R}^{d+1}$ contains the Euclidean ball of radius $\cos\eta$ centred at $0$. By this lemma, the spherical balls on the circumscribed sphere $S^6(0,\sqrt{3}/2)$ of the polytope $\operatorname{Goss}_{7}$ with the angular radius $\eta=\arccos(1/\sqrt{3})$ and centred at the vertices of this polytope form a cover of the sphere $S^6(0,\sqrt{3}/2)$. These vertices are displaced under the action of $f$ by the angular distance $\alpha>0$, where
$$
\begin{equation*}
\cos\alpha=\cos\biggl(2\arcsin\frac{\sqrt{2}}{\sqrt{3}}\biggr)=1-\frac{4}{3}=-\frac{1}{3}.
\end{equation*}
\notag
$$
Since $\arccos(1/\sqrt{3})<\arccos(1/3)$, it follows from the triangle inequality for the angular distance that every point of the sphere $S^6(0,\sqrt{3}/2)$ is shifted by an angular distance not greater than
$$
\begin{equation*}
\arccos\frac1{\sqrt{3}}+\arccos\biggl(-\frac13\biggr)< \pi.
\end{equation*}
\notag
$$
Hence $f$ is induced by a proper orthogonal transformation of the odd-dimensional space $\mathbb{R}^7$ without eigenvalues $-1$ and, therefore, as is well known, having at least one eigenvalue $1$ and the corresponding eigenvector $x\in S^6(0,\sqrt{3}/2)$ such that $f(x)=x$. Therefore, $f$ is identical on the vector subspace $\mathbb{R}^1\subset \mathbb{R}^7$ containing $x$. As was proved, there is a vertex $y$ of the polytope $\operatorname{Goss}_{7}$ such that According to this,
$$
\begin{equation*}
0 < \rho(y,\mathbb{R}^1)=\rho(y,z)\leqslant \frac{\sqrt{2}}2
\end{equation*}
\notag
$$
for some point $z\in\mathbb{R}^1$. Since $f$ is an isometry, it follows that
$$
\begin{equation*}
\rho(f(y),f(z))=\rho(f(y),z)\leqslant \frac{\sqrt{2}}2.
\end{equation*}
\notag
$$
Since $\rho(y,f(y))=\sqrt{2}$, it follows that $z$ is the midpoint of $[y,f(y)]$. It is clear that therefore, $f(a(y-z))=-a(y-z)$ for any $a\in\mathbb{R}$. This contradicts the fact that $f$ shifts all points of the sphere $S^6(0,\sqrt{3}/2)$ by an angular distance smaller than $\pi$. This completes the proof of Proposition 24. Propositions 23 and 24 immediately imply the following assertion. Theorem 8. The set $M$ of vertices of the polytope $\operatorname{Goss}_7$ in $\mathbb{R}^7$ is normally homogeneous and not Clifford–Wolf homogeneous. § 9. Eight-dimensional Gosset polytope $\operatorname{Goss}_8$We recall that the Gosset polytope $\operatorname{Goss}_8$ in $\mathbb{R}^8$ can be realised as the convex hull of some multiplication-closed system of unit octaves called Cayley integers ([10], [17], [24]). This system is not a group under the operation of multiplication, but it is a Moufang loop ([25], [26]). For the explicit expressions for Cayley integers and a discussion of their properties, see [24]. Theorem 9. The set $M$ of vertices of the Gosset polytope $\operatorname{Goss}_8$ in $\mathbb{R}^8$ is Clifford–Wolf homogeneous. Proof. We identify $M$ with the Moufang loop of unit octaves which consists of $240$ elements [25]. However, in fact, more general properties of invertible loops [27] are sufficient for us. Below, $|\,{\cdot}\,|$ stands for the norm in $\mathbb{O}=\mathbb{R}^8$. It should be taken into account that the product of octaves satisfies the equality $|a b|=|a|\cdot |b|$. Among the properties of Moufang loops, we select the presence of a two-sided inverse element for every element of the loop (we use the notation $x^{-1}$ for a two-sided inverse of $x$) and the equality $(yx^{-1})x=y(x^{-1}x)=y$ for any elements $x$ and $y$. Below we consider only elements of $M$ (all of them have unit norm).
Every element $x\in M$ can be translated to any other element $y\in M$ by the multiplication from the left by some element of $M$: The left multiplication is an isometry:
$$
\begin{equation*}
\rho(xy,xz)=|xy-xz|=|x(y-z)|=|x|\,|y-z|=|y-z|=\rho(y,z).
\end{equation*}
\notag
$$
Every left multiplication is a Clifford–Wolf translation:
$$
\begin{equation*}
\rho(xy,y)=|xy-y|=|(xy-y)y^{-1}|=|(xy)y^{-1}-yy^{-1}|=|x-1|=\rho(x,1).
\end{equation*}
\notag
$$
Theorem 9, and Theorem 1 together with it, are completely proved. The authors are grateful to M. Dutour Sikirić for helpful discussions. The authors are grateful to an anonymous reviewer for valuable comments that helped to significantly improve the quality of the text.
Citation in format AMSBIB
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