N. N. Sandakova

PROOF OF ANALOGUES OF THE MINKOWSKI AND VORONOI THEOREMS FOR THREE-DIMENSIONAL ISOGONAL DECOMPOSITIONS ASSOCIATED WITH A GROUP OF PARALLEL TRANSLATIONS

(Presented by Academician I. M. Vinogradov on 26 VII 1963)

A decomposition of space is called normal if its bodies are convex polyhedra and are adjacent along entire faces. On the plane and on the sphere, the most general normal decompositions—both isohedral* and isogonal**—have been considered. In Euclidean space of three or more dimensions, up to now only isohedral decompositions for the group of parallel translations (parallelohedra) have been considered. Isohedral decompositions for an arbitrary group (stereohedra) were first considered in the paper \((^{1})\).

In the present note, general normal isogonal decompositions of three-dimensional Euclidean space associated with a group of parallel translations are considered for the first time.

If a decomposition \(\{Q\}\) is normal and isogonal with respect to a group of parallel translations \(T\), then it satisfies the following conditions:
1) if \(O\) is any vertex of the decomposition, then the totality of all vertices of the decomposition forms an exact lattice \(\Gamma=\{O\tau\}\);
2) under any translations \(T\) the decomposition is transformed into itself.

Theorem 1. The decomposition \(\{Q\}\) is symmetric with respect to any point of the lattice \(\Gamma\) and to the midpoint of any segment joining two points of \(\Gamma\).

Proof. Reflecting the decomposition \(\{Q\}\) in the point \(O\), we obtain another, likewise isogonal, decomposition \(\{Q^*\}\), the totality of whose vertices is the same lattice \(\Gamma\).

Let \(\omega\) and \(\omega^*\) be the totalities of all those edges of the decompositions \(\{Q\}\) and, respectively, \(\{Q^*\}\), which issue from the point \(O\), and suppose that in \(\omega\) there is an edge \(OA\) (\(\overrightarrow{OA}\) is some vector \(\vec t\) from \(T\)). Translating \(\{Q\}\) by the vector \(-\vec t\), we move the point \(A\) to the point \(O\), and the point \(O\) to the point \(A^*\), symmetric to \(A\) with respect to the point \(O\). Therefore each vector in the totality \(\omega\) has in \(\omega\) a vector symmetric to it with respect to the point \(O\); i.e., the totality \(\omega\) is symmetric with respect to the point \(O\), and, consequently, \(\omega=\omega^*\). From the fact that the totality of all edges of the decomposition \(\{Q\}\) is \(\{\omega\tau\}\), it follows that the totalities \(\{\omega\tau\}\) and \(\{\omega^*\tau\}\) of edges of the decompositions \(\{Q\}\) and \(\{Q^*\}\) coincide.

Suppose \(\{Q\}\ne\{Q^*\}\). Then there are in \(\{Q\}\) a polyhedron \(Q\), and in \(\{Q^*\}\) a polyhedron \(Q^*\), such that \(Q\ne Q^*\) and \(Q\cap Q^*=D\), where \(D\) is some three-dimensional polyhedron. In view of \(Q\ne Q^*\), it cannot simultaneously be the case that \(D=Q\) and \(D=Q^*\). Let \(D\ne Q\). Then there exist points \(C\) and \(M\) such that \(C\) is an interior point of the polyhedron \(D\), and \(M\notin D\), but \(M\in Q\). The segment \(CM\) intersects the boundary of the polyhedron \(D\) at a certain point \(F\), which is an interior point of this segment. If the point \(F\) lies on an edge of the polyhedron \(D\), then the point \(C\) can be moved slightly so that \(F\) will be an interior point of some face \(E\) of the polyhedron \(D\); \(E\subset P^*\), where \(P^*\) is some face of the polyhedron \(Q^*\). Denote by \(S\) the intersection of the plane of the face \(P^*\) with the polyhedron \(Q\). It is obvious that \(E=S\cap P^*\)

* Those for which the group of coincidences is transitive with respect to the bodies of the decomposition.
** Those for which the group of coincidences is transitive with respect to the vertices of the decomposition.

The point \(F\) is an interior point also for \(S\), since it is an interior point of the polyhedron \(Q\). From the fact that the point \(F\) is interior both for \(E\) and for \(S\), it follows that \(E\) is a convex polygon which either 1) is contained in \(S\), or 2) coincides with \(S\).

In case 1), there will be points \(C_1\) and \(M_1\) such that \(C_1\) is an interior point of the polygon \(E\), while \(M_1 \notin E\), but \(M_1 \in S\), and such that the segment \(C_1M_1\) intersects some side \(E_1\) of the face \(E\) at a point \(F_1\), which is an interior point of \(E_1\) and of the segment \(C_1M_1\). Consequently, the point \(F_1\) of the edge is at the same time an interior point of \(S\), i.e., of the polyhedron \(Q\), which is impossible.

In the second case, in the polyhedron \(Q\) there will be two vertices \(B_1\) and \(B_2\) such that they lie on different sides of the plane \(S\), since \(S\) is not a face of the polyhedron \(Q\).

In view of the convexity of the polyhedron \(Q\), the segment \(B_1B_2\) intersects the face \(E\), and consequently also the face \(P^*\), at some point \(F_2\), which is not a vertex of \(Q\).

Consider three cases.

A. The point \(F_2\) is an interior point of an edge. This is impossible, since it is simultaneously an interior point of the face \(P^*\).

B. The point \(F_2\) is an interior point of some face \(P\) of the polyhedron \(Q\). Reflect the partition \(\{Q\}\) in the midpoint of the segment \(B_1B_2\). In view of the fact that this reflection is equivalent to reflection in the point \(B_1\) of the lattice \(\Gamma\) and then to translation by the vector \(\overrightarrow{B_1B_2}\) of the lattice \(\Gamma\), the partition \(\{Q\}\) will pass into the partition \(\{Q^*\}\), the polyhedron \(Q\) into some polyhedron \(Q_1^*\), and the face \(P\) will pass into some face \(P_1^*\) of the polyhedron \(Q_1^*\). All interior points of the segment \(B_1B_2\) were interior points of the face \(P\) and, consequently, will pass into interior points of the face \(P_1^*\). The point \(F_2\), consequently, will be simultaneously an interior point both of the face \(P^*\) and of the face \(P_1^*\); but this is impossible, since the faces \(P^*\) and \(P_1^*\) do not coincide, in view of the fact that the vertices \(B_1\) and \(B_2\) do not belong to the face \(P^*\).

C. The point \(F_2\) is an interior point of the polyhedron \(Q\). Having made the reflection in the midpoint of the segment \(B_1B_2\), we shall analogously see that \(F_2\) will be simultaneously an interior point of the polyhedron \(Q_1^*\) and a point of the face \(P^*\), which is impossible. The theorem is thus proved.

Theorem 2. Every partition \(\{Q\}\) can be transformed by an affine transformation into a partition \(\{L\}\) (\(\{L\}\) in the sense of Voronoi).

Proof. We divide all polyhedra \(\{Q\}\) into equivalence classes: in one class we put those polyhedra which are obtained from one another by translations of the group \(T\).

Lemma. There exist such polyhedra of the partition \(\{Q\}\), one taken from each equivalence class, that in sum they form some fundamental parallelepiped of the lattice \(\Gamma\).

Let us prove the lemma. Let \(P\) be any face of the partition \(\{Q\}\), and let \(\Pi\) be its plane.

1. Then the whole plane \(\Pi\) is composed of faces of the partition \(\{Q\}\), since to each face lying in the plane \(\Pi\) one can attach an adjacent face along any of its sides, also lying in this plane, by reflection in the midpoint of this side.

2. Let \(BB_1\) be an edge of the polyhedron \(Q\), issuing from the vertex \(B\) of its face \(P\) and not belonging to this face. Translating the plane \(\Pi\) of the face \(P\) by the vector \(\overrightarrow{BB_1}\), we obtain a plane \(\Pi'\), parallel to the plane \(\Pi\). In the region \(\Phi\) enclosed between the planes \(\Pi\) and \(\Pi'\), there are no points of the lattice \(\Gamma\).

Suppose that in the region \(\Phi\) there were a point \(B_2\) of the lattice \(\Gamma\). Then, translating the plane \(\Pi\) by the vector \(\overrightarrow{BB_2}\), we would obtain some plane \(\Pi_1\), which lies inside the region \(\Phi\) and, consequently, intersects the edge \(BB_1\) at one of its interior points, which is impossible in view of 1.

Consider, besides the face \(P\), two more faces \(P'\) and \(P''\) of this polyhedron.

face \(Q\), issuing from the vertex \(B\). For each of them we construct, analogously, the domains \(\Phi'\), \(\Phi''\). The intersection \(\overline{\Phi}\cap\overline{\Phi'}\cap\overline{\Phi''}\) of the closures of all these three domains is a parallelepiped which, in view of the property of the domains \(\Phi, \Phi', \Phi''\), has inside itself, inside its faces, and inside its edges no points of \(\Gamma\). From consideration of the intersection of the planes of the faces of the tiling \(\{Q\}\) and of the emptiness of the domains \(\Phi, \Phi', \Phi''\), it follows that the edges of the parallelepiped will be edges of the tiling \(\{Q\}\), and its vertices will be vertices of the tiling \(\{Q\}\), i.e., points of \(\Gamma\). Consequently, the parallelepiped obtained is a fundamental one in the lattice \(\Gamma\). The lemma is thus proved.

In order to find all possible tilings (we shall not regard affine images of one another as different), it is enough to find all possible tilings of a parallelepiped into polyhedra \(Q'\) satisfying the following conditions:

1. The vertices of the polyhedra \(Q'\) are the vertices of the parallelepiped, and only these.
2. The polyhedra \(Q'\) are adjacent along whole faces.
3. The tiling \(\{Q'\}\) of the parallelepiped is symmetric with respect to its center \(C\).

Let us consider three cases.

A. The center \(C\) lies inside one of the polyhedra \(Q'_1\) of the tiling \(\{Q'\}\). Then in the polyhedron \(Q'_1\), for each of its vertices there is a vertex symmetric with respect to the center \(C\). In view of the fact that centrally symmetric vertices are the ends of a space diagonal of the parallelepiped, the polyhedron \(Q'_1\) has as vertices the ends of either three such diagonals or four. In the first case \(Q'_1\) is an octahedron, and the remaining parts of the parallelepiped are two tetrahedra. In the second case \(Q'_1\) is the parallelepiped itself.

B. The center \(C\) lies inside one of the faces \(P\) of the tiling \(\{Q'\}\). Arguing analogously, we see that the vertices of this face \(P\) are the ends of two space diagonals. The face \(P\) is a diagonal plane of the parallelepiped and cuts it into two triangular prisms symmetric to one another.

Only two cases are possible:

1. These prisms themselves form a tiling \(\{Q'\}\) of the parallelepiped.
2. If from one of them one cuts off a tetrahedron, and from the other the tetrahedron symmetric to it, then the parallelepiped is divided into two quadrangular pyramids and two tetrahedra.

C. The center \(C\) is an interior point of some edge \(l\) of the tiling \(\{Q'\}\). It is obvious that the edge \(l\), by virtue of the symmetry of the tiling \(\{Q'\}\), is a space diagonal of the parallelepiped. At the edge \(l\) no fewer than three faces meet; but in the present case each such face has a face symmetric to it with respect to the midpoint of the edge \(l\), and therefore four or six faces meet at the edge \(l\). In the first case the parallelepiped is divided into two tetrahedra and two quadrangular pyramids. Each quadrangular pyramid, in turn, can be divided into two tetrahedra by drawing certain diagonals of their quadrangular faces. If the other diagonals of their bases are drawn, a different tiling of the parallelepiped into six tetrahedra is obtained.

The seven cases of tilings \(\{Q'\}\) considered exhaust all possible cases of the required tilings of the parallelepiped. Taking this into account, it is easy to see that all tilings \(\{Q\}\) are affinely equivalent to the five known tilings \(\{L\}\).

I express my profound gratitude to Corresponding Member of the Academy of Sciences of the USSR B. N. Delone.

Mathematical Institute named after V. A. Steklov
Academy of Sciences of the USSR

Received
15 VII 1963

REFERENCES

1. B. N. Delone, N. N. Sandakova. Trudy Mat. Inst. im. V. A. Steklova AN SSSR, 64, 28 (1961).