M. I. Shtogrin, “On a convex polyhedron in a regular point system”, Izv. RAN. Ser. Mat., 86:3 (2022), 187–226; Izv. Math., 86:3 (2022), 586

Abstract: Faceting with a ‘filling’. An ideal crystal structure consists of finitely many equal and parallel translational point lattices. In $\mathbb R^3$ it extends unboundedly in all directions. We distinguish in it a finite part situated in a closed convex polyhedron every face of which contains nodes of a translational point lattice involved in the structure not belonging to the same straight line. Such a polyhedron is called a possible faceting of the ideal crystal structure.
There are 32 well-known crystal classes, or 32 crystallographic point groups. Among them is the symmetry group of the possible faceting calculated taking account of the nodes of the ideal crystal structure belonging to it. A cyclic subgroup $C_n$ of the symmetry group of any possible faceting has order $n\le 4$ or $n=6$.
Faceting without ‘filling’. In this paper we construct two crystal structures in which there are crystal polyhedra whose symmetry groups, calculated without taking account of the nodes of the crystal structure belonging to it, have rotation axes of orders $n=8$ and $n=12$. In both cases, the crystal polyhedron is a right prism of finite height. Without taking account of the internal structure, a possible faceting of a crystal structure in three-dimensional Euclidean space cannot have an axes of rotation of order $n$ satisfying $6<n<\infty$.
The proposed constructions are accompanied by a detailed analysis of ideal crystal structures, as well as Delone sets $S$ of type $(r, R)$ in $\mathbb R^2$ and $\mathbb R^3$. In particular, we produce an expanded proof of one of the theorems stated in 2010 at an international conference dedicated to the 120th anniversary of B. N. Delone.

Introduction

The basic notions and facts in the area of crystallography can be found in [1] or [2]. There are striking revelations on geometric crystallography in [3]–[5]. But those facts and notions that we need are elucidated in detail below.

When speaking of the internal structure of a crystal, we mentally extend it to infinity in all directions in $\mathbb R^3$. As a result we obtain a so-called ideal crystal. Its group of superpositions with itself has a finite fundamental domain and is called a Fëdorov group. In $\mathbb R^3$ there are exactly 230 Fëdorov groups.

The geometry of the global internal organization of a crystal structure is revealed by the following theorem of Schoenflies: every Fëdorov group $G$ contains an abelian subgroup $T$ of finite index. This is the subgroup of translations, or the group of parallel shifts. The quotient group $G/T$ is isomorphic to a finite group of rotations, which belongs to one of 32 crystal classes. It is this quotient group that is called the symmetry group of the external shape of the crystal.

The orbit of a point of the space with respect to the Fëdorov group consists of finitely many equal and parallel translational point lattices; see [6]. In turn, an ideal crystal structure consists of finitely many orbits. Therefore it consists of finitely many equal and parallel translational point lattices.

Generally speaking, the translational point lattices that are parts of an ideal crystal structure do not have equal rights. In contrast, the translational point lattices that are parts of a regular point system, or orbits of a point of the space with respect to the Fëdorov group, always have equal rights. They are equal and parallel to one another. A parallel translation which maps a translational point lattice to itself belongs to the Fëdorov group acting transitively on the points of the regular system. But a parallel translation which maps one translational point lattice in a regular point system to another lattice does not belong to the Fëdorov group; a rigid motion in the Fëdorov group which maps one translational point lattice in a regular point system to another lattice is not a parallel translation.

In $\mathbb R^3$ we fix three non-coplanar vectors $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$ with a common initial point $O\in \mathbb R^3$. All the points of $\mathbb R^3$ with integer coordinates in the frame of reference $O\mathbf{e}_1\mathbf{e}_2\mathbf{e}_3$ comprise a point lattice $\Lambda$. If a plane $\mathbb R^2\subset\mathbb R^3$ intersects the lattice $\Lambda\subset\mathbb R^3$ in a sublattice a basis frame of which consists of two non-collinear vectors issuing from a common node, this sublattice is called a net of the lattice $\Lambda$.

In classical crystallography, a crystal is customarily represented in the form of a finite cutout from an ideal crystal structure. A cutout has the form of a closed convex polyhedron. By the laws of crystallography, the plane of every face of the polyhedron contains a net of a translational point lattice in the crystal structure.

“There is a quite definite connection between the internal structure of a crystal and its external shape: the faces of the crystal correspond to planar nets, the edges to rows of the spatial lattice that have the highest reticular density. The reticular density is the number of nodes per unit area of a planar net, or per unit length of a row of a spatial lattice”; see [7], Ch. 1, § 1.2.2.

The normal to a face of a crystal polyhedron does not necessarily have a rational direction in the translational lattice. But this normal always has a rational direction in the reciprocal lattice of the main lattice. The coordinates of a primitive vector with this direction are coprime integers called the Miller indices. Here, “the indices of a face are not only integers but also small numbers”; see [8], Ch. I, § 12.

A possible faceting is a simple form (see [9]–[12]) or a combination of simple forms. A simple form is defined to be a polyhedron all the faces of which can be obtained from one of them by elements of the symmetry group of the polyhedron. There are 47 simple forms. The number of faces in a simple form and their mutual disposition depend not only on the symmetry group but also on the disposition of the initial face with respect to its elements. The largest number of faces appears in simple forms in which the faces are in the most general position in space with respect to the elements of the symmetry group. They are called general simple forms.

For a sufficiently small change in the position of the initial face, the simple form remains general, but this does not mean that the position of the initial face can be changed continuously. Indeed, a simple form is closely connected with the translational point lattice, metric parameters have fixed values. The normal to the plane of a face of a simple form must be directed along a vector of the reciprocal lattice of the main translational lattice. It has small integer coordinates, the Miller indices.

We now turn to some concrete examples of simple forms; see [12]. A ditetragonal prism cannot be a regular octagonal prism, and a dihexagonal prism cannot be a regular dodecagonal prism, since otherwise there would be normals to faces of the prism with irrational directions in the reciprocal lattice of the main translational point lattice, while in reality these must rational. For the same reason, a pentagondodecahedron cannot be a regular dodecahedron.

In some cases, the symmetry of a faceting can turn out to be richer than the point group of a crystal. For example, this can happen for special parameters of a translational point lattice: a ditrigonal prism may well turn out to be hexagonal, but its symmetry group nevertheless is one of the 32 crystal classes.

The following question arises: is it possible to enlarge the symmetry group of a faceting not by changing the parameters of a translational point lattice, but by changing the mutual disposition of the translational lattices occurring in a given structure under which the symmetry group of the faceting (as a polyhedron) is not one of the 32 crystal classes?

A possible faceting of an ideal crystal is a simple form or a combination of simple forms. The faces that belong to the same simple form have the same physical and chemical properties. The nets situated in the planes of the faces of the same simple form are congruent to one another, and so they have the same reticular density. But we shall not take into account whether or not the densities of the nets in the planes of congruent faces are equal.

The main purpose of this paper is to find out whether a possible faceting, regarded as an ordinary convex polyhedron, can have a richer symmetry group under which the faces of the different simple forms comprising the faceting can become equivalent. Here, a possible faceting can be a combination of simple forms in any number of them, even equal to the number of all the faces of the faceting.

§ 1. The main result

The external shape of an ideal crystal is closely connected with its internal structure. The possible faces of a crystal correspond to planar nets of spatial lattices involved in the crystal structure. These nets have fairly high reticular density, as a rule, the highest. The reticular density of a net means the number of nodes per unit area. A closed convex polyhedron bounded by possible faces is a possible faceting or a possible crystal polyhedron.

In crystallography, when the symmetry group of a faceting is calculated, the positions of the nodes of the crystal structure in it are always taken into account (with ‘filling’).

Below, the symmetry group of a possible faceting is calculated without taking into account the positions of the nodes of the crystal structure in the hypothetical faceting (without ‘filling’).

Under these conditions we construct a special crystal structure which has a possible faceting with an axis of rotation of order $n=8$ or $n=12$.

When determining the order of an axis of rotation of a polyhedron we shall be using only the fact that the normal to a face is a vector of the reciprocal lattice of the main translational lattice in the crystal structure. We shall not be taking into account the positions of its nodes that belong to the possible faceting.

The bases of two such prisms together with the nodes of a multinet are depicted in Figs. 1 and 2, which will be described in detail in § 7 below (see Propositions 10 and 13), in § 10 and in § 12.

§ 2. The spatial translational lattice

We consider an arbitrary parallelepiped in three-dimensional Euclidean space. We construct a partition of the space into parallelepipeds parallel to it in which adjacent parallelepipeds intersect in entire faces. We say that such a partition is normal. Any point of the space belongs to at least one of the parallelepipeds in the partition. A sufficiently small neighbourhood of a point of $\mathbb R^3$ belongs only to:

The vertices of this partition comprise a translational point lattice. The points of the lattice, or nodes, comprise a discrete point system in $\mathbb R^3$.

Suppose that the edges of a parallelepiped issuing from a vertex are unit line segments of the coordinate axes with origin at this vertex. Then all the vertices of the partition and only they have integer coordinates. Thus, a translational point lattice in $\mathbb R^3$ can be defined as the set of points with integer coordinates in a frame of reference, which is called a basis frame of the lattice.

A lattice has the parallel translation property. If we take a vector with initial and terminal points at points of the lattice, then a parallel vector with initial point at a point of the lattice will have its terminal point at a point of the same lattice. The translation of a lattice by a vector that has integer coordinates in the basis frame superimposes the lattice on itself.

Proof. If the nodes are situated on one straight line and the distance between adjacent nodes is rational, then the distances between any nodes of the lattice on this straight line are rational.

Suppose that the nodes are not situated on a straight line. Then there is a parallelogram with vertices at nodes of the lattice. Suppose that the lengths of its sides are rational. Then by the homothety with centre at a vertex of the parallelogram and with homothety coefficient equal to the common denominator of the lengths of its sides we obtain a new parallelogram with vertices at nodes of the lattice the lengths of whose sides are integers. By multiplying the length of every side by the length of an adjacent side we again obtain a parallelogram with vertices at nodes of the lattice, whose sides are equal. Therefore it is a rhombus. The diagonals of a rhombus are orthogonal. Consider the rectangle with vertices at nodes of the lattice whose sides are equal and parallel to the diagonals of the rhombus. Suppose that the lengths of the sides of this rectangle are rational. Then we do the same to the rectangle as we did to the parallelogram. We a obtain a square with vertices at nodes of the lattice. If the length of the side of the square is rational, then the length of the diagonal is irrational. The proposition is proved.

Proof. A triangle and a hexagon occur in a hexagonal lattice. A square occurs in a square lattice. We now prove that in $\mathbb R^d$, $d\geqslant 2$, there is no planar regular convex $n$-gon with vertices at nodes of a given lattice when $n=5$ or $n\geqslant 7$. We argue by contradiction: suppose that there is such an $n$-gon. Then we connect its vertices taking every third (skipping two) by line segments. All such segments are parallel to sides. They bound an $n$-gon homothetic to the original one when $n\geqslant 7$, or centrally symmetric to a homothetic copy of the original one when $n=5$ with homothety coefficient $k<1$. Every vertex of this new $n$-gon, being the fourth vertex of a rhombus constructed on adjacent sides of the original $n$-gon, belongs to the lattice. We can do exactly the same for the new $n$-gon, repeating this again and again. But this process cannot be continued indefinitely, since the original plane $n$-gon contains only finitely many nodes of the lattice. The proposition is proved.

Our original proof of Proposition 3 was based on the following result in [10]: for any translational point lattice in $\mathbb R^3$, the order of an axis of rotation of it is at most $6$ and not equal to $5$. Erokhovets pointed out to the author that the use of the order of an axis of rotation is redundant. Because of this, conversely, by Proposition 3 we obtain a restriction on the order $n$ of an axis of rotation of a lattice: $n\leqslant 6$ and $n\ne 5$.

Following [13], we take $\mathbb R^3$ a rational-faced in particular an integer-polyhedron in $\mathbb R^3$. We focus on the latter. Its vertices have integer coordinates. Therefore its vertices belong to a translational point lattice.

§ 3. Passing from multilattices to Delone systems

A spatial point lattice in $\mathbb R^3$ is centrally symmetric with centre of symmetry at a point of the lattice and at the midpoint of a line segment with endpoints at points of the lattice.

If there is a rotation of the space about an ordinary or helical axis superimposing a point lattice on itself and not fixing any point of it, then there is an ordinary rotation about a parallel axis through the same angle superimposing the lattice on itself and fixing at least one point of it. Indeed, if every point of the lattice takes a new position, then a parallel translation superimposing the lattice on itself can return one of the new positions of a point of the lattice to its old position. The resulting rotation will be about the old position of the point.

A three-dimensional lattice does not have an ordinary axis of rotation of the cyclic group $C_8$ or $C_{12}$; see Proposition 3. Hence it does not have a mirror axis of rotation of the group $S_8$ or $S_{12}$. Indeed, if a point lattice had a mirror axis of rotation of the group $S_8$ or $S_{12}$, then the same kind of axis would be fixing its node $O$. A mirror rotation through the angle $5\pi/4$ or $7\pi/6$, followed by the central symmetry with respect to $O$, would together make up an ordinary rotation of order eighth or twelve, respectively, which is not admitted by a three-dimensional lattice.

Furthermore, a three-dimensional lattice does not have the groups $D_{4d}$ or $D_{6d}$ [17]. This is proved by contradiction. Suppose that a spatial translational point lattice is superimposed on itself by all elements of the group $D_{nd}$ with $n=4$ or $n=6$. Then all the rays emitted from a node of the lattice in the direction of the axes of order two orthogonal to the axis of order $n$ are equivalent with respect to the group $D_{nd}$. These rays contain equivalent nodes of the lattice (see the corollary of Proposition 7 below) that are vertices of a regular $2n$-gon with $n=4$ or $n=6$, which is impossible by Proposition 3.

Based on the main empirical law of Haüy, which reveals the geometry of a faceting of a crystal, all $32$ crystal classes (crystallographic point groups) were found in [10]. Any finite group of isometries of the space superposing a translational lattice on itself is called a crystallographic point group.

As it turned out, Hessel was the first to deduce the $32$ crystallographic point groups in 1830, but by chance his deduction remained unnoticed for a long time.

The group of all rotations of the space around a point of a lattice superimposing the lattice on itself is called the holohedral group. There are $32$ crystallographic point groups in $\mathbb R^3$, among which there are seven different holohedral groups. Some holohedral groups have non-integer-equivalent representations in the form of groups of $(3\times 3)$ integer matrices. If two representations of holohedral groups are integer equivalent, then the lattices are said to belong to the same Bravais type. There are $14$ Bravais types of translational lattices in $\mathbb R^3$. All $230$ Fëdorov groups are contained among the groups of these lattices and their subgroups of finite index.

The presence of the group of parallel translations is a characteristic property of an ideal crystal. A simple ideal crystal can be represented in the form of the orbit of a point of the space with respect to a Fëdorov group, or in the form of a regular point system; see [6]. A Fëdorov group has a subgroup of parallel translations by Schoenflies’ theorem. A simplification of the proof of Schoenflies’ theorem was given in [18], but in the spatial case the proof is not as simple as in the planar case. However, if the central symmetry with respect to a point of $\mathbb R^3$ is an element of the Fëdorov symmetry group, then the proof of Schoenflies’ theorem becomes elementary in the spirit of [19].

Indeed, suppose that the central symmetry with respect to a point $A_0\in\mathbb R^3$ is an element of the Fëdorov symmetry group $G$. Then the central symmetry with respect to any point belonging to the orbit of $A_0$ under $G$ is also an element of $G$. It now remains to use the fact that there is a representative of this orbit in every fundamental domain of $G$. Namely, this orbit contains a point $A_1$ in a fundamental domain that does not contain $A_0$. Next, this orbit contains a point $A_2$ in a fundamental domain that does not intersect the straight line passing through $A_0$ and $A_1$. Finally, this orbit contains a point $A_3$ in a fundamental domain that does not intersect the plane passing through $A_0$, $A_1$ and $A_2$. The product of the central symmetries with respect to the points $A_0$ and $A_i$, $i=1, 2, 3$, is a parallel translation $\mathbf{t}_i$, $|\mathbf{t}_i|=2|A_0A_i|$. The three translations $\mathbf{t}_1$, $\mathbf{t}_2$, $\mathbf{t}_3$ thus obtained are non-coplanar, and their integer combinations comprise the group of parallel translations. All the parallel translations in $G$ form the group of parallel translations. This group has finite index in $G$.

An arbitrary crystal structure consists of the orbits of finitely many points of $\mathbb R^3$ with respect to the same Fëdorov group. A unique generalization of a crystal structure is a widely used Delone set $S\subset\mathbb R^3$ of type $(r, R)$. This is not the only notion that its creator [20] could be proud of.

Dolbilin devoted his studies [19] to Delone point sets in Euclidean space of arbitrary dimension. Here we confine ourselves to dimension $3$. Suppose that every point of some Delone set is a centre of symmetry of this set. Then four centres of symmetry $A_0$, $A_1$, $A_2$, $A_3$ of the Delone set not situated in a plane generate a group of parallel translations superimposing the Delone set on itself. Consequently, this Delone set is a crystal structure. Every straight line containing two of its points intersects it in a one-dimensional point lattice. But, contrary to a conjecture of Kovalëv, such a Delone set does not have to be a translational point lattice.

Indeed, all the vertices of a square lattice in Euclidean plane together with the midpoints of all the sides of the basis squares in this lattice (the midpoints of the sides of a basis square are vertices of a smaller square, and the midpoints of the sides of all basis squares constitute a doubly dense square lattice) define a Delone point set. Every point of this set is a centre of symmetry of this set. Every straight line containing two of its points obviously intersects it in a one-dimensional point lattice. But this set is not a lattice (because it does not contain the centres of the basis squares of the original square lattice). All antipodal Delone sets for Euclidean spaces of arbitrary dimension were found in [21].

In the abstract [22], for a Delone set $S\subset\mathbb R^3$ of type $(r, R)$ with congruent clusters $S_X(2R)=S\cap B_X(2R)$ at all points $X\in S$, it was proved that the order $n$ of a local axis $C_n$ in the group $H_X(2R)$ consisting of isometries of the cluster $S_X(2R)$ is at most $6$. This fundamental bound for the order of an axis was obtained without the presence of a translational point lattice for an uncountable set of such Delone systems.

The author’s theorem on the boundedness of the order of local axes in a Delone set $S\subset\mathbb R^3$ of type $(r, R)$ with equivalent $2R$-clusters was the starting point for the proof of the upper bound $10R$ for the regularity radius for Delone sets in three-dimensional Euclidean space (see [23]). In the process of proof, Dolbilin [24] derived a theorem saying that an absolutely arbitrary Delone set $S\subset\mathbb R^3$ contains at least one point $X\in S$ such that the order of an axis of the group $H_X(2R)$ is at most $6$, which immediately implies the same restriction in the case of a Delone set with equivalent $2R$-clusters. He [25] later strengthened this result by proving that in any Delone set the subset of points $X$ for which the order of an axis in $H_X(2R)$ does not exceed 6 is itself a Delone set with some other value of the parameter $R'$ (greater than or equal to $R$).

A special feature of the papers [24], [25] is that the theorems proved therein for absolutely arbitrary Delone sets imply important substantial assertions for a much narrower class of Delone sets, namely, those with equivalent $2R$-clusters. In the same spirit, Dolbilin [25] made a second conjecture: in any Delone set $S\subset\mathbb R^3$ of type $(r, R)$, the subset of those points at which $H_X(2R)$ is one of the 32 crystal classes is itself a Delone set. This conjecture seems to be the maximal possible generalization of the famous theorem on the impossibility of a pentagonal symmetry in a three-dimensional lattice. Very recently Dolbilin and the author [26] proved the two-dimensional analogue of this conjecture, and also obtained a number of other results concerning local groups in an arbitrary Delone set.

§ 4. Proof of Theorem 2 announced in [22]

The following two theorems were announced in the abstracts of the International Conference dedicated to the 120th anniversary of B. N. Delone [22].

Dolbilin attracted the author’s attention to the absence of a proper proof of Theorem 2 in [22], as well as to the fact that the “locally regular Delone system” taken from § 2.1 in Ch. 2 of [13] can be interpreted incorrectly. In fact, by such a system we mean a Delone set $S\subset\mathbb R^3$ of type $(r, R)$ with congruent spiders $P_A(2R)$ at all $A\in S$.

The completion of the proof of Theorem 2 in fact reduces to the construction in $\mathbb R^3$ of a Delone partition for an $(r, R)$-system $S$ with congruent spiders $P_A(2R)$ such that the order of the axis of rotation $a$ in the group $H_A(2R)$ of isometries of $P_A(2R)$ in $\mathbb R^3$ is equal to $6$.

Note that $R$ is the radius of a covering of $\mathbb R^3$ with equal closed balls $B_A(R)$ with centres at all the points $A\in S$, while $r$ is the diameter of equal open balls $\operatorname{int}B_A(r/2)$ with centres at all $A\in S$ that form a packing in $\mathbb R^3$. Proposition 14, which is known in the theory of finite groups of isometries of $\mathbb R^3$ and proved in the text of this paper, confirms that the local axis of rotation $a$ of order $6$ is unique in $H_A(2R)$.

In the proof of Theorem 1, $a$ has order $n>6$. Therefore the smallest $\rho_*$ for which the vectors of the spider $P_A(\rho_*)$ are non-collinear is greater than $r$.

In Theorem 2, $a$ has order $n=6$, and the smallest $\rho_*$ for which the vectors of $P_A(\rho_*)$ are non-collinear can be either greater than $r$ or equal to $r$.

In any case, in the spider $P_A(\rho_*)$ for a line segment $AB$ non-collinear to $a$, where $A\in a$ and $|AB|=\rho_*\leqslant 2R$, we use the arguments in the proof of Theorem 1. When $|NA|\ne 0$ we obtain a contradiction since the spiders $P_A(\rho_*)$ and $P_B(\rho_*)$ are not congruent.

In what follows, we assume that $|NA|=0$, and so $AB$ is orthogonal to the axis of rotation $a$ of order $6$, which is unique in $H_A(2R)$.

We create copies of $AB$ by applying all the rotations about $a$. In the plane $p$ orthogonal to $a$, we obtain a planar star $\operatorname{st}A$ (in the notation for a planar star we use a lowercase $\mathrm{s}$) consisting of six equilateral triangles that have a common vertex $A$. In $\operatorname{st}A$, we distinguish a triangle $\Delta$ with vertices $A$, $B$, $C$ (in the set $S$) and with sides $|AB|=|BC|=|CA|=\rho_*$.

The star $\operatorname{st}A$ contains two triangles adjacent along $AB$. Therefore three sides of length $\rho_*$ belonging to these two triangles come together at $B$.

Since the local spiders $P_A(2R)$ are congruent at all $A\in S$, at the vertex $B$ situated on the edge of $\operatorname{st}A$ there is a star $\operatorname{st}B$ congruent to $\operatorname{st}A$.

The partition of $p$ into equilateral triangles congruent to $\Delta$ with vertices $A$, $B$, $C$ and with sides $|AB|=|BC|=|CA|=\rho_*$ that are adjacent over entire sides is a Delone partition for the planar hexagonal lattice constructed on the basis triangle $\Delta$. The sides of the equilateral triangles are the edges of the partition. The vertices of the equilateral triangles are the nodes of the lattice.

Proof. In the supposed hexagonal lattice, the radii of $\operatorname{st}A$ form a pre-stable spider $P_A(\rho_*)$. By Lemma 2, the local axis $b$ of order $n=6$ orthogonal to $p$ is uniquely defined at $B$. The two triangles adjacent to $B$ are part of $\operatorname{st}B$; for otherwise a contradiction would be obtained. Rotations about $b$ uniquely complete the two triangles adjacent to $B$ to a star $\operatorname{st}B$. There are exactly the same stars with centres at all the other vertices on the edge of $\operatorname{st}A$. We connect $A$ to all the vertices of these stars. We obtain a stable spider $P_A(\rho_*+2R)$ with $\rho_*=R\sqrt{3}<2R$ in $p$. Here, $R$ is the radius of a covering of the plane $\mathbb R^2=p$ by equal discs with centres at the points of the supposed planar hexagonal lattice, which is uniquely defined by this stable spider.

As a partition $(3^6)$, it is uniquely defined by a pre-stable spider, a part of the local spider $P_A(2R)$. As a result, a hexagonal lattice situated in $p$ is uniquely constructed from $\operatorname{st}A$. This lattice was discovered in [22]. There is a primitive triangle $\Delta$ of this lattice in $p$. It is an equilateral triangle with vertices $A$, $B$, $C$ and sides $|AB|=|BC|=|CA|=\rho_*$. The local axes $a$, $b$, $c$ at $A$, $B$, $C$, respectively, are generators of a group of isometries of $p$ that is transitive on the nodes of the lattice. The lattice itself can now be represented in the form of the intersection $p\cap S$. By Lemma 1 there are no other points of $S$ in $p$. $\square$

Consequently, $\Delta$ is a face of a Delone polyhedron [20]. Moreover, because we proved that the ball in $\mathbb R^3$ is empty, we obtain that $\rho_*/\sqrt{3}<R$, or $\rho_*<R\sqrt{3}$.

A ball in $\mathbb R^3$ of sufficiently small radius that is tangent to $p$ at $A$ contains only one point $A$ of $S$. Keeping $A$ on the surface of the ball, we move its centre away from $p$. With interior not containing points of $S$ and being tangent to $p$ at $A$, the ball, finally, for the first time runs up against at least one ‘new’ point of $S$. At this moment the radius of the ball is at most $R$, and the new point of $S$ (possibly not unique) is not situated in $p$.

Using the same kind of a ball whose centre is situated on the opposite side of $p$, another new point of $S$ is reached. The two new points of $S$ are situated on different sides of $p$ and belong to $B_A(2R)$ and therefore belong to $S_A(2R)$.

Consider the union of the local sets $S_X(2R)$ for all $X\in p\cap S$. We remove from it all the points of $p\cap S$. We obtain a subset of points of $S$ situated on different sides of $p$, which we denote by $\mathcal M$:

Proof. It follows from the fact that $A'\in\mathcal M$ that $A'\in S_X(2R)$, where $X\in p\cap S$, $A'\notin p$, $|A'X|\leqslant 2R$. The ball $B_{A'}(2R)$ contains $X$. Therefore the intersection of the sets $S_{A'}(2R)$ and $p\cap S$ is not empty. This non-empty set contains a point nearest to $A'$, which we denote by $A$. Since $A$ is nearest to $A'$, we have $|A'A|\leqslant|A'X|$. Consequently, $|A'A|\leqslant 2R$ and therefore, $A'\in S_A(2R)$. Therefore the local axis $a$ at $A$ extends its action to $A'$.

Whether $A'$ is a vertex of a polyhedron of the spatial star $\operatorname{St}A$ (the notation for a spatial star uses the capital letter $\mathrm{S}$) will become clear later.

Since among all the points (nodes) of the hexagonal lattice the point $A$ is nearest to the point $A'$, the orthogonal projection of $A'$ onto $p$ belongs to the Dirichlet–Voronoi domain of $A$ in the planar hexagonal lattice $p\cap S$.

Suppose that $A'$ is not situated on $a$. Then all the images of $A'$ under the rotations about $a$ are vertices of a regular convex hexagon. Its orthogonal projection onto $p$ is isometric to it and is contained in the Dirichlet–Voronoi domain of $A$ in the hexagonal lattice. Therefore the side of this hexagon is no longer than the side of the Dirichlet–Voronoi domain equal to $\rho_*/\sqrt{3}$. Since $\rho_*/\sqrt{3}<\rho_*$ and adjacent sides of a hexagon are not situated on the same straight line, we obtain a contradiction to the fact that $\rho_*$ is the smallest number for which the spider $P_A(\rho_*)$ contains non-collinear vectors. Consequently, $A'$ is situated on $a$. $\square$

The point $A$ in Lemma 5 is unique. Otherwise $A'$ would be situated on the local axes orthogonal to $p$ erected at two or three points of $p\cap S$, which is impossible.

Suppose that for some point $A\in p\cap S$ the set $S_A(2R)$ does not have a point on $a$ that is different from $A$. Then there is no point $A'\in\mathcal M$ situated on $a$. Therefore the line segment $AA'$ in Lemma 5 is determined not by $A$ but by $A'$.

Generally speaking, other points of $S$ can be situated inside the line segment $AA'\in a$. Then we redenote by $A'$ the one that is nearest to $A$, and therefore we can assume in what follows that the interior of $AA'$ does not contain points of $S$.

We construct a ball with diameter $AA'\in a$ whose interior does not contain points of $S$. If that ball contains a point $X'\in S$, $X'\ne A$, $X'\ne A'$, then the distance from $X'$ to $p$ is less than $|AA'|$. The point $X'$ is situated on the local axis $x\perp p$ at a point $X\in p\cap S$, $X\ne A$, nearest to $X'$, and $|XX'|<|AA'|$. The point $X'$ is situated on the same side of $p$ as $A'$. The same arguments can be repeated for the line segment $XX'$, and so on.

Every point in $\mathcal M$ has its own line segment situated on the local axis at a point in ${p\,\cap S}$. The length of every line segment situated on a local axis is at most $2R$. Therefore every such line segment is a radius vector of a local spider. All local spiders are congruent to one another. A local spider has only finitely many radii. Consequently, among the line segments on local axes there is a minimal one.

Among all the line segments that are situated on local axes on one side of $p$ (where $A'$ is situated), we denote a minimal one by $XX'$.

Among all the line segments situated on local axes on the other side of the plane $p$ (where there is no point $A'$), we denote a minimal line segment by $YY''$.

The distance from any point of $\mathcal M$ to $p$ is at most $2R$, which follows from the definition of $\mathcal M$. But if the line segment connecting a point in $\mathcal M$ with a nearest point in $p\cap S$ is minimal, then we construct a ball a diameter of which is this line segment. This ball contains only two points of $S$, the endpoints of the diameter. Therefore its length is less than $2R$. Whether this is a global minimum, or is a minimum on one side or the other of $p$ does not matter. For minimal $XX'$ and $YY''$ we have the inequalities $|XX'|<2R$ and $|YY''|<2R$ independently of whether $|XX'|=|YY''|$ or $|XX'|\ne |YY''|$.

All the points of $\mathcal M$ are situated on local axes at some points of $p\cap S$. We divide into three cases the ways in which the minimal line segments are distributed at the nodes of $p\cap S$.

First case. On both sides of $p$, the minimal line segments directed along local axes have the same length, and there is a point of $p\cap S$ at which two minimal line segments come together. The same two minimal line segments come together at every point of $p\cap S$ because local spiders are congruent to one another.

Thus, two oppositely directed line segments $AA'$ and $AA''$ of minimal length $|AA'|=|AA''|$ come together at $A$. Exactly the same two line segments come together at every point of the hexagonal lattice $p\cap S$. The endpoints of all these line segments define hexagonal lattices in the planes $p'$ and $p''$ parallel to $p$ and situated on different sides of $p$.

The hexagonal lattice in $p'$ has a basis triangle $\Delta'$ with vertices $A'$, $B'$, $C'$ which is projected onto $p$ to the triangle $\Delta$ with vertices $A$, $B$, $C$. The triangles $\Delta$ and $\Delta'$ are the bases of the regular triangular prism $ABCA'B'C'$.

On the other side of $p$ there is a similar prism $ABCA''B''C''$ isometric to the prism $ABCA'B'C'$. Any regular triangle of the hexagonal lattice in the plane $p$ is a common base of exactly the same two prisms. All these prisms fill two layers of thickness $|AA'|$ between $p$ and $p'$, and between $p$ and $p''$. There are $12$ regular triangular prisms congruent to the prism $ABCA'B'C'$ coming together at $A$.

Six radii of the planar star $\operatorname{st}A$ together with the line segments $AA'$ and $AA''$ are a part of the local spider $P_A(2R)$, that is, they form a subspider. There is a planar star $\operatorname{st}A'$ at $A'$ congruent to $\operatorname{st}A$ and there is a line segment $A'A$, while the opposite line segment in the subspider is absent. The missing opposite line segment can obviously be reconstructed uniquely. The same happens at any other point of the hexagonal lattice situated in $p'$. What was said above can be repeated for the hexagonal lattice situated in the plane symmetric to $p$ with respect to $p'$, and so on. As a result, we obtain a partition of the space into prisms congruent to $ABCA'B'C'$. The vertices of these prisms form a spatial hexagonal lattice contained in $S$.

Proof. We argue by contradiction. Suppose that a point $X\in S$ is situated between $p$ and $p'$. Then $X$ belongs to the layer of thickness $|AA'|$ between $p$ and $p'$ filled with regular triangular prisms congruent to $ABCA'B'C'$. Consequently, $X$ belongs to some of these prisms. We assume for definiteness that $X$ belongs to $ABCA'B'C'$. But $X$ is not a vertex of it.

The point farthest from all vertices of a prism is its centre. We bound above the distance from $A$ to the centre of the prism. We decompose it into the following summands: the first is the distance from $A$ to the centre of the base and the second is a half of the height of the prism. The first summand is equal to $\rho_*/\sqrt{3}$. It follows from Lemma 4 that $\rho_*/\sqrt{3}<R$. The second summand, which is orthogonal to the first, is equal to $|AA'|/2<R$. As a result, the distance from $A$ to the centre of the prism is less than $R\sqrt{2}<2R$. Therefore $X$ belongs to a local enclosing of some vertex of $ABCA'B'C'$.

The orthogonal projection $\operatorname{Pr}_pX$ of $X$ onto $p$ belongs to the triangle $\Delta$. Therefore the nearest point to $\operatorname{Pr}_pX$ in the planar hexagonal lattice $p\cap S$ is among the vertices $A$, $B$ and $C$ of $\Delta$. The orthogonal projection $\operatorname{Pr}_{p'}X$ of $X$ onto $p'$ belongs to triangle $\Delta'$. Therefore the nearest point to $\operatorname{Pr}_{p'}X$ in the hexagonal lattice $p'\cap S$ is among its vertices $A'$, $B'$ and $C'$. Consequently, the nearest point to $X$ in the spatial hexagonal lattice constructed from triangular prisms congruent to $ABCA'B'C'$ is situated either in $p$ among the vertices $A$, $B$ and $C$ of $ABCA'B'C'$, or in $p'$ among the vertices $A'$, $B'$ and $C'$ of the same prism.

Suppose for definiteness that $A'$ is nearest to $X$. Then $X$ is situated on the local axis $a'\perp p'$ at the point $A'$ belonging to $p'\cap S$. Therefore $X$ is in the interior of the line segment $A'A\in a'$. This contradicts the minimality of the line segment $AA'\in a$, whose interior does not contain points of $S$. Consequently, there are no points of $S$ between $p$ and $p'$. $\square$

Second case. On both sides of $p$, the minimal line segments issuing from points of $p\cap S$ along the local axes have the same length but there is no point of $p\cap S$ at which two of them come together. Thus, by the congruence of the local spiders, at every point of $p\cap S$ there is only one minimal line segment directed along the local axis.

For any point $A\in S$, the group $H_A(2R)$ contains a unique local axis $a$ of order $6$. The unique plane $p$ orthogonal to the axis $a$ passes through $A$. The intersection $p\cap S$ is a planar hexagonal lattice. In this case, at $A$ there is a unique line segment $AA'$ of minimal length directed along $a$. Since the local spiders $P_A(2R)$ are congruent for all points $A\in S$, the minimal line segments directed along the local axes have the same length at all points $A\in S$.

Thus, at every point of the hexagonal lattice $p\cap S$ there is only one minimal line segment directed along the local axis. Among all the minimal line segments, there are line segments whose endpoints are situated on different sides of $p$.

A Delone partition for the planar hexagonal lattice $p\cap S$ consists of equal equilateral triangles. The sides of these triangles are the edges of the partition. Their vertices are the nodes of the planar hexagonal lattice.

Suppose for definiteness that at the vertices $A$ and $B$ of $\Delta$ the minimal line segments directed along the local axes have endpoints situated on different sides of $p$. Then the minimal line segment at the vertex $C$ of the equilateral triangle $\Delta$ will have the same direction as the minimal line segment at either $A$ or $B$.

We assume for definiteness that the minimal line segments at the vertices $A$ and $C$ have the same direction, that is, their endpoints are situated on the same side of $p$. These are the line segments $AA'$ and $CC'$. The line segments $A'C'$ and $AC$ are parallel as they are opposite sides of a rectangle.

At $A$ there is a unique minimal line segment $AA'\in a$. At $A'$ there is a unique minimal line segment $A'U\in a'$. By the congruence of the spiders $P_{A'}(2R)$ and $P_A(2R)$, we have the equation $|A'U|=|AA'|$. We consider all possibilities for the mutual disposition in $\mathbb R^3$ of $A'U$ and $A'A$. They are in general position, they are not situated on the same straight line and are not orthogonal, they are orthogonal, they have opposite directions, or they coincide.

Suppose that $A'U$ is not orthogonal to $A'A$ and is not situated on $a$. Then $|A'U|=|A'A|=|AA'|<2R$. Consequently, $A\in S_{A'}(2R)$. Furthermore, $A$ is not situated in the plane $p'$. There is a point $V'$ in $p'\cap S$ that is nearest to $A$. In the right triangle $V'A'A$ the leg $V'A$ is shorter than the hypotenuse $A'A$. On the local axis $v'$ orthogonal to $p'$ at $V'$, we obtained a line segment $V'A$ which is shorter than the line segment $A'U$ since $|V'A|<|A'A|=|A'U|$. This contradicts the fact that the line segment $A'U$ is minimal. Consequently, $A'U$ and $A'A$ cannot be in general position.

Next, suppose that $A'U$ is orthogonal to $A'A$. Then $A'A$ is situated in $p'$. All the points of $S$ situated in $p'$ form a planar hexagonal lattice $p'\cap S$. Therefore the endpoints of $A'A$ are situated at nodes of the lattice $p'\cap S$. Hence, $|A'A|\geqslant \rho_*$, since the smallest distance between the nodes of $p'\cap S$ is equal to $\rho_*$. It follows from the inequality $|AA'|<2R$ that $A'\in S_A(2R)$. Moreover, in the star $\operatorname{st}A'$ there is a point that is contained in $S_A(2R)$ and is not situated on $a$.

Indeed, $\operatorname{st}A'$ contains six radii of length $\rho_*$. The angle between neighbouring radii is equal to $\pi/3$. The line segment $AA'$ with $|AA'|\geqslant \rho_*$ is situated in the plane of $\operatorname{st}A'$. Consequently, for any mutual disposition of $\operatorname{st}A'$ and $AA'$ in $p'$, the star $\operatorname{st}A'$ contains a point $V$ not situated on $a$ such that the angle between $A'A$ and $A'V$ is at most $\pi/3$. Taking into account the inequality $|AA'|\geqslant \rho_*$, we conclude that $V$ is contained in $S_A(2R)$. Therefore $a$ extends its action to $V$. We create copies of $V$ by all rotations about $a$. We obtain a regular convex hexagon whose sides are smaller than $\rho_*$ since $|A'V|=\rho_*$. Adjacent sides are not situated on the same straight line. This contradicts the fact that $\rho_*$ is the smallest number for which the vectors of the spider $P_A(\rho_*)$ are not collinear. Consequently, $A'U$ is not orthogonal to $A'A$.

Suppose the contrary that $A'U$ and $A'A$ are collinear and have opposite directions. Then the minimality of $AA'$ and the equation $|A'U|=|A'A|$ imply that two minimal line segments come together at $A'$, and this is the first case considered above rather than the second.

Finally, suppose that $A'U$ coincides with $A'A$, or $U$ coincides with $A$. At $A$ we have the minimal line segment $AA'$ directed along $a$. At $A'$ we have the minimal line segment $A'A$ directed along $a'$. Then $p$ and $p'$ are parallel since $p$ is orthogonal to $AA'$ and $p'$ is orthogonal to $A'A$.

The planes $p$ and $p'$ have equal rights. There is a planar hexagonal lattice $p\cap S$ in the planes $p$. There is a planar hexagonal lattice $p'\cap S$ in the planes $p'$. It remains to find what is the mutual disposition of these lattices in $\mathbb R^3$.

Since the plane of $\operatorname{st}A'$ is orthogonal to $A'A$, and this plane already contains the line segment $A'C'$ with $|A'C'|=\rho_*$ and $A'C'\parallel AC$, $\operatorname{st}A'$ is parallel to $\operatorname{st}A$. Therefore the hexagonal lattices $p'\cap S$ and $p\cap S$ are parallel. Consequently, at every point of $p\cap S$ there is a minimal line segment directed along the local axis an endpoint of which is situated in $p'$. At $B$ there are two minimal line segment along $b$ having opposite directions. This is again the first case, rather than the second.

Third case. On one side of $p$ there is a line segment $XX'$, $X'\in S$, issuing from a point $X\in p\cap S$ and directed along a local axis $x\perp p$. On the other side of $p$ there is a line segment $YY''$, $Y''\in S$, issuing from a point $Y\in p\cap S$ and directed along a local axis $y\perp p$. The line segments $XX'$ and $YY''$ are minimal and each is on its own side of $p$. It is assumed that $|XX'|\ne |YY''|$ since both cases where these lengths are equal have already been dealt with in the preceding cases.

The congruence of local spiders means that two line segments $AA'$ and $AA''$ come together at the point $A\in p\cap S$ such that $|AA'|=|XX'|$ and $|AA''|=|YY''|$. For definiteness we assume that $|AA'|<|AA''|$. There are exactly the same two line segments at every point of the hexagonal lattice $p\cap S$. The endpoints of those of length $|AA'|$ are situated on one side of $p$, and those of length $|AA''|$ are situated on the other since the minimal line segments along the local axes on the same side of $p$ have the same length.

The endpoints of the minimal line segments of length $|AA'|$ are situated in $p'$ and form a planar hexagonal lattice $p'\cap S$. The endpoints of those of length $|AA''|$ are situated in $p''$ and form a planar hexagonal lattice $p''\cap S$. The planes $p'$ and $p''$ are parallel to $p$ and are situated on different sides of $p$.

As in the first case, we construct a partition of the layer of thickness $|AA'|$ between $p$ and $p'$ into regular triangular prisms congruent to $ABCA'B'C'$, and a partition of the layer of thickness $|AA''|$ between $p$ and $p''$ into regular triangular prisms congruent to $ABCA''B''C''$. There are $12$ regular triangular prisms that come together at $A$, six of which are congruent to $ABCA'B'C'$ and six are congruent to $ABCA''B''C''$.

In the same way as in the first case, we construct a partition of the space into prisms congruent to $ABCA'B'C'$ and $ABCA''B''C''$. Here, the prisms congruent to $ABCA'B'C'$ are in layers of thickness $|AA'|$, and congruent to $ABCA''B''C''$ are in layers of thickness $|AA''|$. The layers of different thicknesses alternate with one another. The vertices of the prisms of the partition form a spatial hexagonal bilattice contained in $S$.

In Lemma 1 and its strengthening Lemma 4, the triangle $\Delta$ can be replaced by $\Delta'\subset p'$. In Lemma 6, the planes $p$ and $p'$ can be replaced by $p$ and $p''$, since the line segment $AA''$ representing the height of $ABCA''B''C''$ has length $|AA''|<2R$. Recall that in the proof, the minimality of the line segment is used only on one side of $p$ rather than on both sides. For the same reason, in Proposition 5 $ABCA'B'C'$ can be replaced by $ABCA''B''C''$. The corresponding proofs remain exactly the same.

Consequently, the spatial hexagonal bilattice as a whole coincides with $S$. In the bilattice, one hexagonal lattice is shifted with respect to the other along the axis of the group $C_6$. Every local axis of order six is a part of the global axis of the group $C_6$.

Thus, only two cases are possible: the first ($S$ is a spatial hexagonal lattice) and the third ($S$ is a spatial hexagonal bilattice).

Based on the congruence of the local spiders $P_A(2R)$ with an axis of rotation of order $n=6$ in the $(r, R)$-system $S\subseteq\mathbb R^3$, we gave managed to prove that the congruent stars $\operatorname{St}A$, $A\in S$, consist of regular triangular prisms. As a result, the $(r, R)$-system $S\subseteq\mathbb R^3$ as a whole is a spatial hexagonal lattice when $|AA''|=|AA'|$ or a spatial hexagonal bilattice when $|AA''|>|AA'|$. In the case of a bilattice, one lattice is shifted with respect to the other along an axis of order $n=6$.

The proof of this theorem has already been presented as Theorem 4.3 in [17], where the four non-crystallographic point groups $D_{4d}$, $D_{6d}$, $S_8$ and $S_{12}$ were discovered, in each of which an ordinary axis of rotation is crystallographic.

§ 5. Rational directions in a planar lattice

Here we analyse the symmetry group of the set of rational directions in a translational point lattice in $\mathbb R^2$. It contains the group of all isometric maps of the lattice onto itself fixing a node.

A ray emitted from a point of a lattice has rational direction if it contains at least one more point of the lattice. Therefore it contains an infinite set of points of the lattice, which we call its nodes.

A ray of rational direction in a planar lattice can be defined by a vector having coprime integer coordinates in the basis frame of the lattice.

We shall be defining any rational direction in a planar point lattice by a vector of length $1$ collinear to this direction.

Suppose that two lattices are homothetic or situated in parallel, or one of them is a sublattice of the other of full dimension. Then such lattices have the same set of rational directions. We say that all these lattices belong to the same kind.

Thus, we divide all planar lattices into five kinds, choosing one representative in each, as follows:

2) a rectangular lattice, primitive and special, in which there are only two rational directions orthogonal to each other;

3) a rectangular lattice, primitive and special, in which the sides of the rectangle have the same length (this is a tetragonal lattice);

4) a rectangular lattice, primitive and special, in which the diagonal of the rectangle is equal to twice its side (therefore, after adding the centres of all rectangles, it becomes a hexagonal lattice);

5) a rectangular lattice, primitive, general, non-special, not one of the two special ones, and a sublattice not of full dimension of one of the lattices mentioned above.

The symmetry group of the set of rational directions in a lattice of the first kind coincides with the holohedral group $C_2$, and in the lattice of the second kind it coincides with the holohedral group $D_2$. The symmetry group of the set of rational directions in a lattice of the third, fourth or fifth kind is infinite.

A rectangular lattice occurs among lattices of all kinds except for the first. We take adjacent sides of the rectangle for unit line segments of coordinate axes. Let the length of a primitive vector along the abscissa axis be equal to $1$, and the length of a primitive vector along the ordinate axis be equal to $\lambda$. Then the square of the length of a vector with coordinates $x$ and $y$ is equal to $x^2+\lambda^2\,y^2$.

The points with integer coordinates are the nodes of a rectangular lattice. The radius vectors of points $(a, b)$ and $(-\lambda^2b, a)$ are orthogonal to each other. Let $a, b\in\mathbb N$. Then the point $(a, b)$ belongs to the lattice. If $\lambda^2$ is irrational, then only two rational directions in the lattice are orthogonal, and therefore this is a lattice of the second kind with holohedral group $D_2$.

Since $a, b\in\mathbb N$, the point $(-\lambda^2b, a)$ has rational coordinate if and only if $\lambda^2$ is rational. In this case, the radius vector of the node $(a, b)$ is orthogonal to the radius vector of the node $(-\lambda^2bd, ad)$, where $d$ is the denominator of the number $\lambda^2$.

Suppose that $\lambda^2$ is an integer. Then the radius vector of $(-\lambda^2b, a)$ has integer coordinates. It is orthogonal to the radius vector of $(a, b)$. The ratio of the lengths of the two radius vectors is equal to the ratio of the unit line segments on the coordinate axes. The rectangular sublattice constructed on the two radius vectors is similar to the original rectangular lattice with similarity coefficient $\sqrt{a^2+\lambda^2b^2}$ and is rotated with respect to it through the angle $\alpha=\arctan(\lambda b/a)$. Since the set of rational directions in the lattice and in a sublattice of full dimension is the same, this set is superimposed on itself by the rotation through $\alpha$.

By the mirror symmetry of the original rectangular lattice with respect to the abscissa axis, it contains the nodes $(a, -b)$ and $(-\lambda^2b, -a)$. They are situated on the rays emitted from the origin $O(0, 0)$ and equivalent to $Ox$ and $-Oy$ under rotation through $-\alpha$. Therefore the set of rational directions of the original rectangular lattice is superimposed on itself by rotation through $z\alpha$ for all possible values $z\in \mathbb Z$.

Suppose further that the number $\lambda^2$ is an integer but not the perfect square of an integer, and therefore $\lambda$ is irrational. Then rotation through $\pi/2$ superimposing on itself the set of rational directions of the rectangular lattice corresponding to the metric quadratic form $x^2+\lambda^2\,y^2$ does not belong to the holohedral group. This rotation does not superimpose the lattice on itself since the lengths of any two orthogonal primitive vectors of this lattice are not commensurable. Neither does mirror reflection in the bisector of the angle between the orthogonal primitive vectors superimpose the lattice on itself since their lengths are not commensurable. This lattice is superimposed on itself by the group $D_2$, and there are infinitely many such point groups.

Finally, when $\lambda^2=1$ we have a square lattice of the third kind with holohedral group $D_4$. When $\lambda^2=3$, this is a primitive rectangular lattice of the fourth kind with holohedral group $D_2$, and if we add the centres of all the rectangles, then the lattice becomes hexagonal with holohedral group $D_6$. When $\lambda^2=p$, where $p$ is a prime, $p\ne 3$, this is a primitive rectangular lattice of the fifth kind with holohedral group $D_2$.

For lattices of the third, fourth and fifth kinds, the groups of rotations of the set of rational directions are infinite and different for different kinds. In these lattices there is an infinite set of pairs of orthogonal rational directions. Any rational direction occurs in a pair with some other rational direction.

The finite point groups of the sets of rational directions for lattices of the second and fifth kinds are the same. But for lattices of the second kind, there is only one maximal finite point group at a given node, while for lattices of the fifth kind, there are infinitely many.

The set of rational directions of a lattice of the fifth kind has a rotation of order four, although this is not a square lattice in $\mathbb R^2$ or a sublattice of one. But it does not have a rotation of the order eight or twelve, since such a lattice is not a sublattice of a square or hexagonal lattice. (Since the lengths of the orthogonal primitive vectors of a lattice of the fifth kind are incommensurable, the bisector between these vectors has irrational direction, which fact is proved in exactly the same way as Proposition 8 below. This once again confirms that a lattice of the fifth kind does not have a rotation of order eight.)

We are interested in the following question: what are the possible orders of the group of rotations of the set of all rational directions in an arbitrary planar point lattice?

Proof. We connect nodes $O$ and $A$ of the given lattice by a line segment $OA$. Rotation about $O$ through $\varphi=2\pi/n$ takes the line segment $OA$ to a line segment $OB$, $OB$ to $OC$, $OC$ to $OD$, $OD$ to $OE$, $\dots$ . The points $A$, $B$, $C$, $D$, $E$, $\dots$ are the vertices of a regular convex $n$-gon $ABCDE\dots$ . The line segments $OA$, $OB$, $OC$, $OD$, $OE$, $\dots$ are situated in $\mathbb R^2$ and have rational directions in the lattice.

Consider the first three line segments $OA$, $OB$ and $OC$. The line segment $OB$ is directed along the bisector of the angle between $OA$ and $OC$. Since $OB$ has rational direction in the lattice, there is a point $B'$ belonging to the lattice and lying in the interior of $OB$ or an extension of it, or coinciding with $B$. We construct a rhombus with diagonal $OB'$ whose two sides are directed along $OA$ and $OC$. The other two sides pass through the point $B'$. In the basis frame of the lattice, all the sides of the rhombus are defined by linear equations with rational coefficients. Therefore all its vertices have rational coordinates. By multiplying the coordinates of the vertices by the common multiple of their denominators, we obtain points of the lattice equidistant from $O$ on the rays $OA$ and $OC$.

In exactly the same way $OD$ is directed along the bisector of the angle between $OC$ and $OE$. Consequently, the rays $OC$ and $OE$ contain points of the lattice equidistant from $O$. Therefore in this case we have obtained a point of the lattice on the ray $OC$. In the preceding case we have obtained, generally speaking, a quite different point of the lattice on the same ray $OC$. We take a point on $OC$ whose distance from $O$ is a multiple of the distances to $O$ in these two cases. There are points of the lattice at this distance on all the three rays $OA$, $OC$ and $OE$.

Thus, the rays $OA$, $OC$ and $OE$ contain points of the given lattice equidistant from the node $O$. Here we must emphasize that all three rays pass through vertices of the $n$-gon $ABCDE\dots$ taken alternately when going around its perimeter.

By repeating these arguments, we perform a complete circuits of $O$ for even $n$, or two complete circuits for odd $n$. For even $n$, the vertices taken alternately are the vertices of a regular $(n/2)$-gon. For odd $n$, they are the vertices of the same regular $n$-gon. In both cases the vertices of the regular polygon obtained belong to the point lattice. This cannot happen for odd $n>6$. For even $n>6$, or $n/2>3$, the only possibilities are $n/2=4, 6$; see Proposition 3 (which is proved for a planar lattice in exactly the same way as for a spatial one). Thus $n=8$ or $12$. $\square$

In a square lattice, a sublattice does not have to be superimposed on itself by a rotation of order four. In a hexagonal lattice, a sublattice does not have to be superimposed on itself by a rotation of order six. The following proposition holds for the set of rational directions in these lattices.

Thus, the set of rational directions in a planar point lattice, generally speaking, can have a rotation of order eight or twelve. Furthermore, it can have a rotation of infinite order; see [16].

(1) Is the irrational number $(1/\pi)\arctan(b/a)$ transcendental or algebraic?

It is clear that $(1/\pi)\arctan(b/a)$ is not the only irrational number connected with a tetragonal point lattice; there is a countable set of such numbers. We denote every such individual number by $\omega_n$, where $n\in\mathbb N$. All the $\omega_n$ form a countable set $\Omega=\{\omega_n\mid n\in\mathbb N\}$. For $n=\mathrm{const}$ and all possible $z\in\mathbb Z$, $z\ne 0$, the numbers $z\omega_n$ are irrational and form a countable subset of $\Omega$. It is unknown whether $\Omega$ is the union of countably or finitely many such subsets. Suppose that two subsets $\{z\omega_j\mid z\in\mathbb Z$, $z\ne 0\}$ and $\{z\omega_k\mid z\in\mathbb Z$, $z\ne 0\}$ have non-empty intersection. Then $z_j\omega_j=z_k\omega_k$ for some $z_j$ and $z_k$, and therefore $\omega_j$ and $\omega_k$ are linearly dependent over $\mathbb Q$. What the situation is with independent $\omega_n$ is an open question:

how many of the numbers in $\Omega$ are linearly independent over $\mathbb Q$, finitely many or countably many?

These problems can be transferred from lattices of the third kind to lattices of the fourth and fifth kinds.

Are the groups of rotations of the sets of rational directions for different representatives of lattices of the fifth kind isomorphic or not?

§ 6. A regular polygon in a planar multilattice

A group of rigid motions (isometries) of $\mathbb R^2$ is called a Fëdorov group if it has a finite fundamental domain which contains a representative of the orbit of every point of the plane under the group. There are $17$ Fëdorov groups in $\mathbb R^2$. If a Fëdorov group contains rigid motions of the second kind, then there is a subgroup of index $2$ consisting of rigid motions of the first kind and having a finite fundamental domain.

Consider a Fëdorov group consisting of rigid motions of the first kind. The boundary of its fundamental domain contains only finitely many points that are centres of rotation for this group. Consider two centres of rotation that are equivalent under rigid motions of the group. We perform a rotation about one centre clockwise, and then a rotation through the same angle around the other centre anticlockwise. As a result we obtain a parallel translation. An equivalent parallel translation obtained for a rotation of order greater than $2$ is not collinear with it.

Suppose that the order of every rotation in a Fëdorov group is equal to $2$. Then consecutive rotations about two fixed centres produce a parallel translation along the line containing them. We take two other centres belonging to two fundamental domains lying on different sides of this line along which the original translation already constructed acts. The consecutive rotations about these two centres produce a translation that is not collinear with the original one.

The set of all parallel translations in a Fëdorov group form an abelian subgroup of finite index, the group of parallel translations.

The orbit of a point of the plane with respect to a given Fëdorov group is a regular point system. The orbit of a point with respect to the subgroup of parallel translations is a planar lattice. A regular system of points of Euclidean plane consists of finitely many parallel lattices, a so-called multilattice.

We claim that an octagon can be regular if its vertices belong to a planar multilattice consisting of four square lattices; see Fig. 1.

A regular octagon in a planar multilattice is defined in Fig. 1. Each of its horizontal and vertical sides contains two linear rows of nodes of square (tetragonal) lattices. Any other side of the octagon contains a linear row (of seven nodes) belonging to a square lattice in the given multilattice.

For this planar multilattice, the star of Delone polygons contains two squares and two rectangles. If the side of the smaller square with vertices at nodes of the multilattice is equal to $1$, then the side of the other is equal to $\sqrt {1.125}$.

A regular dodecagon in a planar multilattice consisting of six hexagonal lattices is constructed similarly; see Fig. 2.

A side of the dodecagon contains either three nodes of one linear row, or two nodes from each of two linear rows taken from two hexagonal lattices. The star of the Delone partition contains a regular hexagon, an equilateral triangle and two rectangles. If the length of the side of the hexagon is equal to $1$ and the length of the side of the triangle is equal to $2$, then the length of the side of the dodecagon is equal to $4+2\sqrt{3}$.

In both cases there is no frame in which the coordinates of all the vertices of the regular octagon or the regular dodecagon are rational. Every side contains a linear row of nodes belonging to a point lattice that is a part of the planar multilattice. These linear rows do not belong to the same point lattice, although the direction of every linear row is rational. (The lines containing the sides of the regular $n$-gon, where $n=8$ or $n=12$, are defined by linear non-homogeneous equations. If the coefficients of the unknowns are rational, then the free terms must be irrational in at least some of the equations.)

Thus, without taking into account the nodes of the corresponding multilattices, the group of ordinary rotations of the $n$-gon for $n=8$ or $n=12$, see Figs. 1 and 2, has order $n=8$ or $n=12$. Taking account of the nodes of the corresponding multilattices, the group of ordinary rotations of the planar $n$-gon for $n=8$ or $n=12$ (see Figs. 1 and 2) has order $n/2=4$ or $n/2=6$.

Figure 3 depicts a square in a planar bilattice. Each of the horizontal sides of the square contains two linear rows of nodes taken from these lattices. Each of the vertical sides contains one linear row of nodes.

The origin $O(0, 0)$ lies at the centre of the square depicted in Fig. 3. The axis $Ox$ is horizontal and the axis $Oy$ is vertical. The rectangular lattice with the metric quadratic form $x^2+2\,y^2$, that is, a lattice of the fifth kind with $\lambda=\sqrt{2}$, is the original one (none of its nodes in Fig. 3). We construct two rectangular lattices that are equal and parallel to the original one. The first has a node at the point $(3\sqrt{2}-4, 0)$. The second has a node at the point $(-3\sqrt{2}+4, 0)$. We obtain a bilattice whose intersection with the square with sides of length $6\sqrt{2}$ is depicted in Fig. 3. The point $O$ is the centre of symmetry of the bilattice.

The square defined above, as well as the set of rational directions in the original rectangular lattice, has a rotation of order four. But this rotation does not superimpose the original rectangular lattice on itself. Neither do the mirror reflections in the diagonals of the square. The lengths of the primitive vectors with directions equivalent under the rotation of the order four and the mirror reflection in the diagonals are not commensurable.

All is clear in the first two examples: the rotations of order $n=8$ and $n=12$ are not crystallographic. But what can be said about the rotation of order $n=4$ in the third example? It seems to be among the rotations of crystallographic order, but behaves in exactly the same way as the rotations of order $n=8$ and $n=12$ in the first two examples.

§ 7. Rational directions in a spatial lattice

A ray in $\mathbb R^3$ emitted from a node of a lattice has rational direction in this lattice if it contains at least one more node of it. Therefore it contains an infinite set of nodes of the lattice. A ray emitted from a node of a lattice that does not contain any other node has irrational direction.

A ray of rational direction in a given lattice can be defined by a vector having coprime integer coordinates in the basis frame of the lattice.

We are defining any rational direction in an arbitrary point lattice by a vector of length 1 in this direction.

We consider a rotation of $\mathbb R^3$ through the angle $\pi$ about a given axis and take a direction in $\mathbb R^3$, rational or not. The angle between this direction and the axis is arbitrary: it can be a right angle (equal to $\pi/2$), acute or equal to $0$.

Proof. The rotation through $\pi$ about an axis passing through a point $O$ takes a rational direction $OA$ to a rational direction $OB$, where $O$, $A$, $B$ are nodes of the lattice. The line segments $OA$ and $OB$ are commensurable. Denoting their common measure by $e$, we obtain $|OA|=ae$ and $|OB|=be$, where $a, b\in\mathbb N$. Enlarging homothetically the line segment $OA$ with the centre of homothety at $O$ and with the coefficient of homothety $b$, we obtain a line segment $OA'$ with $|OA'|=b|OA|=b(ae)$. Since $b$ is an integer, the point $A'$ belongs to the lattice. In exactly the same way we obtain a point $B'$ on the ray $OB$ that belongs to the lattice, where $|OB'|=a|OB|=a(be)$. The points $A'$ and $B'$ are equidistant from $O$. The line segments $OA'$ and $OB'$ are not collinear since $OA$ and $OB$ are not parallel and not perpendicular to the axis. The diagonal of the rhombus with adjacent sides $OA'$ and $OB'$ is an axis of rotation through $\pi$ and has rational direction. $\square$

Proof. We argue by contradiction. Suppose that the direction of the axis is rational. Then a primitive line segment $OC$, where $O$ and $C$ are points of the lattice, is directed along the axis of order two, while $OA$ and $OB$ are primitive line segments directed along rational directions in the lattice that are equivalent under rotation about this axis through $\pi$. We construct a rhombus with diagonal $OC$ and with sides parallel to the line segments $OA$ and $OB$. Since the points $O$ and $C$ belong to the lattice, while the non-collinear line segments $OA$ and $OB$ have rational directions, the linear equations of the lines containing the sides of the rhombus have rational coefficients in the basis frame of the planar lattice. Consequently, all the vertices of the rhombus have rational coordinates. If we multiply the coordinates of all points by the common denominator of the coordinates, the coordinates become integers. Therefore in the directions of $OA$ and $OB$ there are nodes of the lattice equidistant from $O$. Therefore $OA$ and $OB$ are commensurable, which contradicts the original hypothesis. $\square$

Proof. If two rational directions of general position (not orthogonal and not parallel to the axis) are equivalent under rotation through $\pi$ and the primitive line segments in these two directions are commensurable, then by Proposition 7 the direction of the axis is rational. If two other rational directions of general position (not orthogonal and not parallel to the axis) are equivalent under rotation through $\pi$ about the same axis, then the primitive line segments in these two directions are also commensurable, for otherwise the direction of the axis would be irrational by Proposition 8. But the direction of an axis cannot be simultaneously rational and irrational. Consequently, if the primitive line segments are commensurable for one pair of equivalent directions with respect to this axis, then they are also commensurable for any other pair.

Similarly, if the primitive line segments are incommensurable for one pair of equivalent directions, then they are also incommensurable for any other pair. $\square$

We do not go into further details related to a rotation of order two, and in what follows we study rotations of higher order.

Proof. We consider an ordinary axis of rotation of order $n>6$ that passes through a node $O$. Since axis has only two opposite directions, the set of non-collinear directions contains a direction not directed along the axis. In a rational direction for this lattice, we consider a primitive line segment $OA$ issuing from $O$ and of general position with respect to the axis, that is, not parallel or orthogonal to it. Rotation through $\varphi=2\pi/n$ about the axis takes $OA$ to a line segment $OB$, $OB$ to $OC$, $OC$ to $OD$, $OD$ to $OE$, $\dots$, and all these line segments are rationally directed. Therefore any two of these line segments define a rational plane, which intersects the spatial point lattice in a planar point lattice, that is, a net. We claim that the projections of these line segments onto the plane orthogonal to the axis have rational directions in the spatial point lattice.

Indeed, the points $A$, $B$, $C$, $D$, $E$, $\dots$ are the vertices of a regular $n$-gon situated in the plane not containing $O$ and orthogonal to the axis of rotation.

Suppose that $n$ is even. Then there is a rotation through $\pi$ about the axis. Under rotation through $\pi$, $OA$ and its image are not collinear and lie in the plane containing the axis. Under rotation though $\pi$, the line segment $OB$ and its image are not collinear and they are situated in the plane containing the axis. Therefore the axis has rational direction. As in the proof of Proposition 7, we construct a rhombus with vertices at points of the lattice in which one side is directed along $OA$ and the diagonal is directed along the axis. Then the other diagonal of the rhombus is orthogonal to the axis. Therefore the orthogonal projection of $OA$ onto the plane orthogonal to the axis has rational direction. The line segment $OA$ can be replaced by $OB$, $OC$, $\dots$ . Consequently, the projections of all the line segments $OA$, $OB$, $OC$, $OD$, $OE$, $\dots$ onto the plane orthogonal to the axis have rational directions.

Suppose that $n$ is odd. Then we begin with the trapezium $ABCD$. Its legs $AB$ and $CD$ are equal. Its bases $AD$ and $BC$ are orthogonal to the axis. The planes of the triangles $OAB$ and $OCD$ intersect in a line of rational direction. This line and the line of rational direction passing through $O$ and vertex of the $n$-gon opposite to the side $BC$ define a plane containing the axis. This plane is uniquely defined by the axis and the midpoint of $BC$. The planes of the triangles $OBC$ and $OAD$ intersect in a line of rational direction parallel to $BC$ and situated in a plane orthogonal to the axis. Replacing $BC$ by another side of the $n$-gon $ABCDE\dots$ we verify that the axis has rational direction and that the plane containing $O$ and orthogonal to the axis is rational. By what was said above we conclude that the plane containing the axis and any of the line segments $OA$, $OB$, $OC$, $OD$, $OE$, $\dots$ and the plane containing $O$ and orthogonal to the axis intersect in a line of rational direction. Consequently, the orthogonal projections of $OA$, $OB$, $OC$, $OD$, $OE$, $\dots$ onto the plane passing through $O$ and orthogonal to the axis have rational directions.

Thus, if there exist equivalent line segments $OA$, $OB$, $OC$, $OD$, $OE$, $\dots$ of rational direction not orthogonal to the axis of rotation and not parallel to it, then the axis has rational direction and there exist line segments of the same type orthogonal to it. Due to this fact, in what follows we can (and will, in order not to introduce new notation) assume that all the aforementioned line segments equivalent to $OA$ under rotations about the axis through angles of the form $\varphi=2k\pi/n$, $1\leqslant k\leqslant n$, are orthogonal to the axis and have rational directions.

Thus, the line segments $OA$, $OB$, $OC$, $OD$, $OE$, $\dots$ equivalent to $OA$ with respect to the group $C_n$ of order $n>6$ are situated in the plane $\mathbb R^2$ containing $O$ and have rational directions in the planar point lattice. They satisfy the conclusion of Lemma 8, which was proved earlier for the planar case; see above. $\square$

The set of rational directions in a square or hexagonal lattice does not have a rotation of finite order greater than $8$ or $12$, respectively. The order $8$ or $12$ is the greatest possible; see Propositions 6 and 10.

Proposition 10 applies to the rational directions whose primitive vectors have different lengths. If their lengths are the same, see Proposition 3. The cases $n=8$ and $n=12$ will be made more precise in Propositions 11 and 12.

The set of all rational directions of a square (tetragonal) lattice in $\mathbb R^3$ has a rotation of order four, one of the elements of its holohedral group.

Proof. One of the basis cells of a square (tetragonal) translational point lattice in $\mathbb R^3$ is a square prism. We consider a square point lattice with the metric quadratic form $x^2+y^2+\nu^2z^2$ constructed using a square prism with edges of length $1$ directed along the coordinate axes $Ox$ and $Oy$, and with an edge of length $\nu$ directed along the coordinate axis $Oz$.

The plane $Oxy$ intersects the spatial square lattice in a planar square lattice. The set of rational directions in a planar square lattice has a rotation of order eight; see Proposition 6. Consequently, the set of all those rational directions in the lattice lying in the coordinate plane $Oxy$ has a rotation of order eight. We can add the two opposite directions along the axis $Oz$ to this set.

Any larger subset of the set of rational directions of the lattice does not have a rotation of order eight. Indeed, consider an arbitrary primitive vector in general position. Its coordinates have the form $\{x, y, z\}$, where $x, y, z$ are integers, $x^2+y^2\ne 0$, and $z\ne 0$. After a rotation of order eight about $Oz$ we obtain the vector $\{(x-y)/\sqrt{2}, (x+y)/\sqrt{2}, z\}$. Since $z\ne 0$ and $x-y\ne 0$ and $x+y\ne 0$, the direction of the rotated vector is not rational. $\square$

For transcendental $\nu$, the direction of an axis of order $n=4$ is unique in the lattice whose basis frame is defined by the metric quadratic form $x^2+y^2+\nu^2z^2$. If a subset of the set of rational directions in this square lattice in $\mathbb R^3$ has a rotation of order eight, then it consists of directions orthogonal to the axis and, possibly, parallel to it. All the rational directions orthogonal to the axis or parallel to it form a non-expandable set of rational directions with the group $C_8$.

The set of all rational directions in a hexagonal lattice in $\mathbb R^3$ has a rotation of order six, one of the elements of its holohedral group.

Finally, we use the fact that the normals to the faces of a crystal polyhedron are directed along rational directions in the reciprocal lattice of the main lattice of the crystal structure. Consequently, the analysis of the symmetry group of the polyhedron reduces to an analysis of the symmetry group of the set of normals to its faces.

Such a $kn$-gonal prism has been constructed for $k=1$ (but not for $k\geqslant 2$). A practical application of Proposition 13 will involve the spatial point lattice that is the reciprocal lattice of a translational point lattice of an ideal crystal structure.

§ 8. On axis of rotation of order $n\geqslant 6$ in a finite group

The great Leonhard Euler proved the following remarkable theorem: if a rigid motion of a body has a fixed point, then it is a rotation of the body about some axis.

The following result is known for a finite group of rigid motions in $\mathbb R^3$ superimposing this body with itself.

Proof. We argue by contradiction. Suppose that $G$ contains two subgroups $C_n\subseteq G$ of the same order $n\geqslant 6$. Then the number of all such subgroups is finite because $G$ is finite. The axis of rotation of a group $C_n$ intersects a sphere $\mathbb S^2$ with centre $O\in\mathbb R^3$ in two diametrically opposite points around which $\mathbb S^2$ revolves under rotations about this axis.

Let $A$ denote the point of intersection of $\mathbb S^2$ with one of the axes and $V$ point the nearest to $A$ of intersection of $\mathbb S^2$ with the other. We create copies of $A$ by rotations about the axis $OV$. All the $n\geqslant 6$ images of $A$ are points of intersection of $\mathbb S^2$ with axes of rotations of order $n\geqslant 6$ in $G$. On $\mathbb S^2$, these $n\geqslant 6$ images are the vertices of a regular convex $n$-gon with centre on $OV$. The sides of the $n$-gon are shorter than $|AV|$. This contradicts the fact that $V$ is the nearest point to $A$. Therefore, if $G$ has a subgroup $C_n$ of order $n\geqslant 6$, then it is unique. $\square$

§ 9. A regular point system in space

Recall that the symmetry group of a polyhedron with vertices at nodes of a translational point lattice in $\mathbb R^3$ is a subgroup of the holohedral group of either the lattice itself or of a sublattice of full dimension. Consequently, this group is one of the $32$ point groups, the famous $32$ crystal classes of Gadolin [10].

For a translational point lattice, two faces of a faceting that are equivalent in the geometric sense are equivalent in the physical and chemical sense.

We embark on finding the group of a polyhedron whose faces contain the nodes of nets of the translational point lattices involved in a regular point system. The normal to a face of the polyhedron defines not one plane but a whole family of parallel planes. The same can be said about the corresponding Miller indices. But it is one thing if the planes of the family contain the nets of the same spatial lattice, but quite another if these planes contain the nets of different lattices contained in a regular point system in $\mathbb R^3$, which is what we assume in what follows. In the first case the planes of the faces are defined by linear equations with rational coefficients. In the second case, the coefficients of the equations of the planes of the faces do not have to be rational.

The coefficients of the equations of the planes passing through the point $O(0, 0, 0)$ and parallel to the planes of the faces are rational. They are defined by homogeneous linear equations with rational coefficients in the basis frame of the lattice.

A regular point system in $\mathbb R^3$ is defined to be the orbit of a point of the space under a Fëdorov group. It consists of finitely many equal translational point lattices situated in parallel in $\mathbb R^3$. A regular point system extends unboundedly in $\mathbb R^3$ in all directions. But we are interested only in a finite cutout of a regular point system in the form of a closed convex polyhedron. By definition, the plane of every face of this polyhedron contains a planar net contained in at least one of the spatial lattices involved in the regular point system. The normals to the planes of the faces in a regular point system are vectors of the reciprocal lattice of the main translational point lattice. This choice of a polyhedron in a regular point system is quite natural: it is related to the analysis of the symmetry group of a possible faceting of a crystal (the external shape of the crystal). This choice agrees with the assertions on p. 9 of [7] and p. 94 of [8] mentioned above.

In the next section we use this in the analysis of the symmetry of a crystal polyhedron without taking into account the positions of the nodes of a regular point system.

§ 10. A polyhedron in a regular point system

Here we analyse the symmetry of a possible faceting of a regular point system in $\mathbb R^3$. A closed convex polyhedron on its own, as a possible faceting of a regular point system, can sometimes have an axis of rotation of order $n=8$ or $n=12$. When $6<n<\infty$ such a polyhedron cannot have axes of other orders. For $n=8$ or $n=12$ this polyhedron has the form of a right $kn$-gonal prism of finite height; see Proposition 13.

Indeed, with a regular point system we associate a closed convex polyhedron (a crystallographic polyhedron) every face of which contains nodes of a planar net of at least one of all the point lattices that comprise the system. These lattices are equal and parallel. Therefore their reciprocal lattices are equal and parallel. We can say that they have the same reciprocal lattice since it is not its nodes that are important for us but only the directions in it whose primitive vectors are normals to the faces. We are interested in the symmetry group of the crystallographic polyhedron, more precisely, the symmetry group of the set of normals to its faces.

We are interested only in the external shape of the polyhedron. We do not take into account its internal structure. Thus, we consider an isometric copy of the external shape of a crystallographic polyhedron. It is a closed convex polyhedron, whose faces are closed convex polygons.

In what follows, from an arbitrary crystallographic polyhedron, only its external shape remains, devoid of any internal contents.

Our goal is to find all groups of such polyhedra considered on their own whose symmetry groups contain cyclic subgroups of order $n$, where $6<n<\infty$.

In fact, this problem has already been solved in Propositions 10 and 13 (the latter is applied to the reciprocal lattice of a translational point lattice).

Indeed, examples of regular polygons in $\mathbb R^2$ were constructed having a rotation of order $n=8$ or $n=12$; see Figs. 1 and 2. These two examples give rise to the two claimed regular point systems in $\mathbb R^3$. Namely, we create copies of a planar regular point system by a parallel translation along a line segment orthogonal to the plane in which Fig. 1 or 2 is situated, and by all multiples of it. The direct product of a regular octagon or a regular dodecagon with an orthogonal line segment is a regular octagonal prism of finite height with symmetry group $D_{8h}$ or a regular dodecagonal prism of finite height with symmetry group $D_{12h}$.

A crystallographic polyhedron, as a closed convex polyhedron considered without taking into account its internal structure, that has an axis of rotation $C_n$ of order $n=8$ or $n=12$ is a right $kn$-gonal prism of finite height, which is regular when $k= 1$ with symmetry group $D_{nh}$, and irregular when $k\geqslant 2$, possibly, with symmetry group $D_{nh}$ or $C_{nh}$. We make more precise the supposed form of the symmetry group when $k\geqslant 2$ by taking into account the fact that a right prism has a plane of mirror symmetry parallel to its bases.

We consider an auxiliary list of groups: $S_8$, $C_8$, $C_{8v}$, $D_8$, $D_{4d}$, $D_{8d}$, $S_{16}$, $S_{12}$, $C_{12}$, $C_{12v}$, $D_{12}$, $D_{6d}$, $D_{12d}$, $S_{24}$. The groups $S_8$ and $D_{4d}$ have an axis of rotation $C_4$. The groups $S_{12}$ and $D_{6d}$ have an axis of rotation $C_6$. The order of the axis of rotation of both groups is doubled when in each of them the mirror reflection in the plane orthogonal to the axis is added. The other groups have an ordinary axis of rotation $C_n$ of order $n=8$ or $n=12$ and no higher. We extend each group by adding the mirror reflection in the plane orthogonal to the principal axis.

The groups $S_8$ and $C_8$ obviously extend to $C_{8h}$ and $C_{8v}$, $D_8$, $D_{4d}$ to $D_{8h}$. The extensions of $D_{8d}$, $S_{16}$ contain $C_{16}$. The groups $S_{12}$, $C_{12}$ extend to $C_{12h}$ and $C_{12v}$, $D_{12}$, $D_{6d}$ extend to $D_{12h}$. The extensions of $D_{12d}$ and $S_{24}$ contain $C_{24}$.

The groups $C_{16}$ and $C_{24}$ are not axes in the $kn$-gonal prism under consideration. Only $D_{8h}$ and $C_{8h}$, or $D_{12h}$ and $C_{12h}$, occur in the complete list of groups which can contain the symmetry groups of $kn$-gonal prisms, $k\in\mathbb N$, when $n=8$ or $n=12$, respectively.

We did not undertake the confirmation of the groups $C_{8h}$ and $C_{12h}$. They are already confirmed by the concrete theoretical examples related to Figs. 1 and 2.

Taking account of the nodes of the nets contained in the faces, it turned out that the lateral surface of both $n$-gonal prisms when $n=8$ or $n=12$ is a combination of two simple forms.

§ 11. On actions of a crystallographic point group

The existence of groups with axes of orders 8 and 12 motivated us to have a fresh look at crystallographic point groups since elements of some of them may have unusual actions. For example, in $\mathbb R^3$ a polyhedron was discovered that is a faceting of a bilattice and has a group of rotations not all whose elements superimpose the translational point lattice on itself. Namely, consider the translational point lattice with the metric quadratic form $x^2+2y^2+\nu^2z^2$ constructed using a rectangular parallelepiped with edges of lengths 1, $\sqrt 2$ and $\nu$.

(If $\nu$ is transcendental, then the line segment of length $\nu$ is commensurable neither with the unit line segment of the axis $Ox$ of length $1$, nor with the unit line segment of the axis $Oy$ of length $\sqrt{2}$ nor with any other primitive line segment of the rectangular lattice in the plane $Oxy$.)

Rotation through $\pi/2$ about the axis $Oz$ superimposes on itself the direction of the axis $Oz$ and the set of rational directions of the rectangular lattice in the plane $Oxy$, but it does not superimpose on itself the set of rational directions of a spatial lattice in general position, that is, not in $Oxy$ and not along $Oz$. Indeed, the rotation through $\pi/2$ about $Oz$ takes the node $(1, 0, k)$, $k\in\mathbb N$, belonging to the lattice to the point $(0, 1/\sqrt{2}, k)$ situated on a ray issuing from the node $O(0, 0, 0)$ in an irrational direction.

Rotation through $\pi/2$ about $Oz$ takes a rational direction in this lattice to a rational direction if and only if it is directed along $Oz$ or is orthogonal to it. A polyhedron such that the normals to its faces are rational directions and the rotation through $\pi/2$ about $Oz$ superimposes this polyhedron on itself is a right prism. In particular, if this prism has the minimal number of lateral faces, it is a right square (tetragonal) prism. (Its lateral surface is a combination of two simple forms. An example of such a prism is given by the direct product of the square in Fig. 3 with a line segment of length $\nu$ orthogonal to it.)

This is not the only example of such an action of elements of a crystallographic point group. But we did not set ourselves the goal of describing all the cases that can occur. For example, it is sufficient to take a translational lattice with any non-triclinic holohedral group and a sublattice with a triclinic holohedral group. A possible faceting of this lattice can be chosen in such a way that its symmetry group coincides with the original holohedral group.

§ 12. An arbitrary crystal structure

In classical crystallography, an ideal crystal structure in the general case consists of finitely many regular point systems constructed for the same Fëdorov group. Like a regular point system, it consists of finitely many equal and parallel translational point lattices. Choosing a crystal polyhedron in an arbitrary crystal structure is not in any way different from choosing a crystal polyhedron in a regular point system and does not give rise to anything new. A polyhedron can have an axis of rotation of order 8 or 12, which fact has already been obtained for regular point systems. There are no axes of other orders $n$ if $6<n<\infty$.

The actual existence of a regular octagonal or dodecagonal prism seems unlikely to us. For them to exist the points of a crystal structure must have irrational coordinates. There must be special growth rates of the faces and the growth must stop at a suitable moment. The planes of the faces with different densities of nets must be at the same distance from the centre of the polyhedron. The possibility of all this happening has been confirmed only by artificial theoretical examples.

In every finite group there can only be one axis of order eight or twelve. This well-known fact in the theory of finite point groups confirms Proposition 14. In view of Propositions 10, 13 and 14, the proof of Proposition 1 is complete.

The set of rational directions in a tetragonal or hexagonal point lattice in $\mathbb R^2$ is also superimposed on itself under rotations of infinite order; see [16]. This set is also superimposed on itself in $\mathbb R^3$ under rotations of infinite order about an axis orthogonal to $\mathbb R^2$. But the existence of an axis of infinite order (generated by a rotation through an irrational part of the full angle $2\pi$) still does not mean that this can give rise to a faceting in the form of a right circular cylinder. A possible faceting, as a finite cutout from an infinite crystal structure, must have the form of a closed convex polyhedron with a finite number of faces.

§ 13. Conclusion

A substantial contribution to the development of classical crystallography was made by the study of local properties of regular point systems, or Delone systems, begun in [27], [28] and continued in [13] and [17], [19], [21], [23]–[26]. Conditions on a Delone system were found under which it is a regular point system. It turned out that in a Delone system, all points have the same global surroundings in it if they have the same local surroundings that are not too big but also not too small.

Modern crystallography studies so-called quasicrystals, which can have axes of rotation of orders eight, ten, twelve, and other $n$, where $6<n<\infty$; see [29]. There are many different definitions of a quasicrystal, some of which can be found in Appendix 2 of the book [30]. There is also the survey [31] about quasicrystals, including a general discussion and a comparison of some of the definitions of a quasicrystal.

Quasicrystals are in no way connected with Fëdorov groups. The group of rotational parts of the rigid motions of a Fëdorov group does not contain axes of order $n$ with $6<n<\infty$. In spite of this, a possible faceting of a crystal, without taking into account the nodes of nets, can generally speaking have an axis of order 8 or 12. This can happen in a regular point system if irrational values occur among the coordinates of points. But the irrationality of coordinates of points is not sufficient. They must be closely connected with a faceting.

What was said above is confirmed only by artificial examples; see the regular octagon in Fig. 1 and regular dodecagon in Fig. 2, as well as §§ 10 and 12.

In Fig. 1 the side of a minimal square with vertices at points belonging to a regular point system is equal to $1$. The side of the square next in size is equal to $\sqrt {1.125}$. The sides of such squares are incommensurable and can even be found on the same line. Therefore there is an irrational number among the coordinates of the nodes of this point system.

It is interesting whether in crystallography there exist regular point systems with irrational coordinates in a crystallographic system of coordinates and whether it is technically feasible to confirm or refute this. One would have to look for confirmation among crystals with irrational coordinates of the nodes. This property is enjoyed by structures with icosahedra in [29]. The vertices of an icosahedron cannot have only rational coordinates. Nor can the normals to the faces of an icosahedron. Therefore they are not among the normals to the faces of a faceting of a crystal.

We point out that the problem is not only in the irrationality of coordinates. The choice of a polyhedron in a regular point system is substantially different from its choice in a lattice. A polyhedron in a lattice has a crystallographic symmetry, while a polyhedron in a regular point system can have a non-crystallographic symmetry. The latter is related to the fact that the faces of the polyhedron are contained in the planes of nets in different translational point lattices. Only by chance can these planes turn out to be at the same distance from a point of the space that is fixed by rotations superimposing the polyhedron on itself. This is clearly not a pattern [10] or a rule. It is unknown whether this exception of the rules, improbable in crystallography, is possible or not.

In connection with irrational coordinates, we again mention irrational distances. The rejection of irrational distances between points of some geometric object (see the preface of the editor of the translation in [29]) can result in the rejection of the geometric object itself. Indeed, there are two geometric objects in $\mathbb R^2$, a tetragonal (square) lattice and a hexagonal lattice. One of the basis cells of a tetragonal lattice is a square. If the length of the side of a square is equal to $1$, then the length of the diagonal is equal to $\sqrt 2$. One of the basis cells of a hexagonal lattice is a rhombus composed of two equilateral triangles. If the length of one diagonal of the rhombus is equal to $1$, then the length of the other is equal to $\sqrt 3$. Both planar lattices are nets of the corresponding two lattices in $\mathbb R^3$; see also Proposition 2. In a classical crystal, irrational parameters are always present along with rational ones.

The author is grateful to V. M. Buchstaber, N. P. Dolbilin, N. Yu. Erokhovets, M. D. Kovalëv and A. Yu. Popov for valuable suggestions in the discussion of the results of this paper. The author also thanks the referee for a careful reading of the entire manuscript and substantial comments aimed at the improvement of the text, in particular, at the elimination of gaps.