A Counterexample to Smilansky’s Conjecture

1. Introduction

Minkowski decomposability is governed by the compatibility of edge directions. If a polytope is a sum , its vertices inherit positions in each summand, and the difference of those positions along an edge is parallel to that edge. Conversely, the space of such parallel redrawings records the infinitesimal freedom to change the summands [1]. The resulting local rigidity criteria leave open how much freedom in the facets can survive when they are assembled.

A combinatorial three-cube has a three-dimensional space of parallel redrawings modulo translations. In particular, each cubical facet admits nontrivial deformations. The following construction shows that these deformations can be globally incompatible.

The proof separates the geometry of gluing from the choice of coordinates. Its essential ingredient is a cubical polytope with two adjacent facets that impose different restrictions on its redrawings; one determines the whole polytope, whereas the other receives at most a two-dimensional family of redrawings. Joining three copies cyclically forces both properties onto the same interface. A further gluing along a cube then intersects two deformation spaces whose only common direction is uniform scaling.

Section 3 proves this gluing principle, and Section 4 constructs the required 32-vertex polytope. All convexity arguments are geometric; the coordinates exhibit one block with the stated incidences and seven specified noncoplanar configurations.

2. Parallel redrawings and projective changes

Let have vertices . A parallel redrawing assigns a vector to each vertex, subject to

where the edge scalars are arbitrary real numbers. Write for this vector space modulo translations and . The homothety gives a distinguished line in .

If is a face, or an interface whose edges remain on the boundary after gluing, we say that determines when restriction is injective. Equivalently, a redrawing that vanishes on vanishes everywhere.

We use an explicit form of projective invariance to keep track of restrictions under gluing. The invariance of decomposability itself is due to Kallay [2].

3. A gluing principle for cubical polytopes

The following proposition isolates the properties needed of the auxiliary polytope. We prove it in Section 4, after showing how these properties give the desired construction.

3.1. Three copies around a quadrilateral. Apply the normalization in Proposition 5(iv), choose , and translate to the origin. Let be the direction plane of . Since the affine correspondence fixes pointwise, complementary transverse vectors can be chosen so that corresponding outer vertices have the form

For this splitting, choose a transverse vector in , take its image in , and complement by their span. The whole polytope lies in the wedge between its supporting facets .

In a new transverse plane take , and . For , make an affine copy that fixes and sends to . Formula (3) makes the facet of coincide with the facet of . The three interfaces are cubes, all meeting in .

The pieces occupy three sectors covering the transverse plane. Their union need not yet be convex; a compatible compression makes it so without altering their common interfaces.

The pairwise intersections are cubes and the triple intersection is , whose edges survive. Hence

There are facets and therefore .

3.2. Removing the last deformation. Choose a surviving copy of . We can make it an affine unit cube by a projectivity valid on all of . To check the denominator, choose the desired map on , with positive affine denominator there, and first compress towards by

where vanishes on . The denominator then becomes

which is positive on every vertex for sufficiently large . We may therefore assume .

The three edge-direction scale factors identify with , with homothety line . The restriction image of contains this line and has dimension at most two. If cyclically permutes the cube coordinates, its action on has no invariant real line; an eigenvalue on such a line would be real with cube one, hence one, whereas the quotient has no nonzero fixed vector. Consequently

Place a second copy of on the other side of , using on and reversing the transverse coordinate. Compress the two pieces towards an interior point of . The halfspace argument of Lemma 6, with two sectors, makes their union convex and preserves all noninterface facets; call it .

These compressions preserve the restriction spaces on . Indeed, anchor a redrawing at one vertex of . Its values on lie in the direction space of , where the transverse functional in (2) vanishes. The two restriction images are therefore exactly and .

A redrawing of restricts to both copies of , so its common restriction to is homothetic by (5). After subtracting this homothety and a translation, it vanishes on and hence on both copies, because determines each one. Thus , and Lemma 2 proves indecomposability. Finally,

All changes of basis and projectivities may be chosen rational. The compression parameters may be integers exceeding the finitely many rational bounds in (4), so is rational. This proves Theorem 1 once Proposition 5 is established.

4. The 32-vertex block

Work in homogeneous coordinates , with as affine chart, and put

The rows of the following table define eight further linear forms

Set , , and

Writing , define

4.1. Vertices and supporting facets. Label a vertex by five bits,

and define predicates by

The facet incidences are

Each facet has eight labels, with three bits free, and the graph is the five-cube graph.

The coordinates can be recovered from four linear equations. For a label , let satisfy and satisfy . Its point is the unique solution with of

In all 32 cases these affine equations are independent.

Here is an exact check of the halfspaces. Replace each coordinate vector by its primitive integer homogeneous representative with positive last entry, and each defining form by its primitive integer multiple of the same sign. Substitution in (8) gives zero precisely at the incidences in (7); the least positive evaluations at the remaining vertices are

These inequalities identify the whole polytope. Let be the convex hull of the 32 listed points. The four independent active planes at each point make it a vertex of . Within each proposed facet, the six prescribed quadrilaterals are exposed by neighboring rows and form a cubical boundary. The facet is three-dimensional; a proper exposed face containing four extreme points cannot be a line, and a planar facet could not have such a proper face. The six quadrilaterals therefore form its entire boundary. Every ridge is paired with another proposed facet, so connectedness of the facet-adjacency graph of excludes any additional facet. Its complete facet inequalities are consequently (6), proving and .

The accompanying file verify_block.py reconstructs all coordinates from (8) and checks these integer evaluations, the determinants below, and the two perspectivities by exact rational arithmetic.

Choose

Then , and is disjoint from .

4.2. Two determining facets. Only seven noncoplanarity facts are needed. The following determinants, using the primitive vectors in the indicated order, are nonzero modulo 101

The first six rows are graph four-cycles. The last says that the spoke lines and are skew.

Suppose a redrawing vanishes on . The cycles

each contain a zero edge, so Lemma 3 makes them zero. Thus all vertices and all even vertices are zero. The two-neighbor rule completes the propagation; use for ; then for ; finally for each odd . Hence determines .

If instead the redrawing vanishes on , the first four cycles in the table make all vertices zero. On it is a homothety . The values at 16 and 28 lie on the direction lines of the two skew spokes. If , the point would lie on both spoke lines, which is impossible; if , the vector is parallel to both nonparallel lines and must vanish. That cycle is therefore zero. Next have zero neighbors , and every remaining second-half vertex has zero neighbors . Thus determines as well.

4.3. The restriction on the third facet. Consider

The cycle is nonplanar, whereas is a quadrilateral facet of . Their four corresponding spoke lines concur at

Indeed, the eight forms , , annihilate the homogeneous point . The apex is outside the plane of , since otherwise all vertices of would be coplanar.

Subtract the homothety and translation of a redrawing on . Its values on then have the form . Modulo the direction plane of , the vectors have a common nonzero image, while adjacent redrawing differences have zero image. Thus the scalars agree and the redrawing on is homothetic. By Lemma 3(iii), the inverse image of this line in has dimension two, proving Proposition 5(iii).

4.4. Matching the two interfaces. The perspective center from to is

The map fixes and matches

For each pair , substitution in (8) gives

The projectivity

sends to infinity and fixes pointwise. Since on , its denominator lies in . In the new chart the perspective map is affine, fixes , and maps the whole of onto because it matches all eight vertices. This proves the remaining assertions of Proposition 5 and completes the construction.

References