Explicit nonpolytopal Bier spheres

Caius Folkerts

Abstract. We construct explicit nonpolytopal Bier spheres by bounding the sign rank of a matrix of face-extension data in every convex realization. The bound depends only on a fixed core of labels, and a rectangular Hadamard matrix supplies explicit extension data whose sign rank exceeds it. Finite-field formulas give an example with vertices and dimension , defined by a complex of dimension nine. Because the obstruction uses only selected face memberships, it also gives examples with no ghost labels, with combinatorial central symmetry, and with a defining complex
of dimension three.

1. Introduction

Bier’s construction associates a simplicial sphere to every nonempty proper simplicial complex. Counting shows that most of these spheres are nonpolytopal [1], but does not specify an example. Holleben and Yang [3] proved that every Bier sphere with at most twelve vertices is polytopal and asked for an explicit nonpolytopal sphere. We construct one by encoding a matrix whose sign rank exceeds a bound imposed by every convex realization.

Let be a nonempty proper simplicial complex on a labelled ground set . Singleton labels need not be faces; an absent singleton is called a ghost. The Alexander dual and Bier sphere are

where is a copy of on a disjoint primed ground set. The deleted join is a sphere of dimension [1].

Theorem 1. There is an explicitly specified complex of dimension nine whose Bier sphere is nonpolytopal, has actual vertices, and has dimension . There are also nonpolytopal examples with all ground labels present, with a self-dual input complex, and with an input complex of dimension three.

The obstruction separates a small core of labels from a much larger set of additional labels. For each additional label we record which selected core faces extend to a face containing it, arranging these membership signs as a column of a matrix. Convexity bounds the sign rank of this matrix solely in terms of the number of core labels, independently of how many additional labels we use.

For a sign matrix , define

The precise geometric obstruction is as follows.

Theorem 2. Let be a proper complex on , containing every singleton, where . Choose faces of one fixed positive cardinality, and define

Regard as a complex on , with a ghost. If is polytopal, then

For selected faces of arbitrary cardinalities, the bound holds instead.

Supporting hyperplanes express a matrix with signs through Schur complements of one matrix. The required principal inverses, padded by zeros, satisfy linear identities leaving at most dimensions. This proves the bound using only the selected memberships; the other faces of the complex are unrestricted.

For the principal example, take and retain columns of a Hadamard matrix of order . Forster’s spectral bound [2] and Theorem 2 would give

Section 3 specifies the matrix and the face set by elementary finite-field formulas. The final constructions preserve the obstruction while changing the ghost, symmetry, or input-dimension conditions.

2. The sign-rank obstruction

Put . All unprimed labels in and all primed labels occur in ; the complements defining the primed singletons either contain the ghost or equal the nonface . Thus the sphere has actual vertices and dimension .

In a hypothetical realization, denote the labelled vertices by , for , and . The facets we need are

In each case the positive set is a face, the complement of the primed set is a nonface, and there are vertices. These are therefore facets by the definition of the Bier sphere.

These two kinds of facets distinguish an extension from a nonextension; when is a face, replacing by in gives another facet, whereas a nonextension supplies a facet containing and omitting both and . With as the origin, the two alternatives impose opposite signs on one coordinate of in the basis supplied by . Computing that coordinate gives the following Schur complement.

Lemma 3. If is polytopal, there are a real matrix , row vectors , column vectors , and scalars such that every is invertible and

Proof. The simplex facet and the vertex outside its affine hull form an affine basis. Normalize the realization to

where is column of a real matrix . For a selected face , let have columns for and otherwise. These columns form the simplex facet outside the origin, so is invertible. Expanding in its standard basis columns shows that is invertible as well.

For an anchor , write . Solving first in the rows and then in row gives

We claim that has the prescribed sign .

If is a face, normalize the supporting functionals of and to and . Every column of has -value one, while all except column also have -value one. Since lies outside the first facet and outside the second,

The left side is positive and the parenthesized factor negative, so .

If is a nonface, use the facet . Its supporting linear functional may be chosen zero on the facet and positive at every other vertex. It vanishes on all columns of except column , whence

Thus . The claim follows with

□

To bound the rank, pad every principal inverse to the same matrix size. Multiplication by , on either side, then recovers the indicator of its index set on the diagonal. The resulting commutator identities place all the padded inverses in one linear space of controlled dimension, through which the sign matrix will factor.

Lemma 4. Let be a real matrix and let range over subsets on which is invertible. Pad each inverse by zeros to obtain an matrix , with . Then

Proof. Direct multiplication gives

Join and when or is nonzero, and let be the component sizes of this support graph. All belong to the block-diagonal matrix space of dimension . For every diagonal matrix , they satisfy

The map has kernel the diagonals constant on each support component, and hence rank . The trace pairing on the block-diagonal space is nondegenerate, so these give independent constraints. Therefore

where the last inequality follows from . □

Proof of Theorem 2. Let be the common cardinality of the selected faces. Lemma 3 supplies a matrix with signs and entries

Because , its columns are evaluations of the linear functionals

Lemma 4 bounds its rank by . Indeed, after choosing a basis of the span of the , the matrix factors as the matrix of their coordinates times the matrix of these linear functionals. Fixed cardinality is used precisely to express the constant term as a linear function of . Without the common-cardinality assumption, adjoining the constant feature to this matrix space gives . □

3. An explicit Hadamard example

To encode any sign matrix , choose distinct -subsets of , with and . On , define by the following faces

There are no other faces; every proper subset of a face in the third line occurs in the first two, so the prescription defines a complex. The label is a ghost, and no face contains two anchors.

Forster’s spectral bound [2, Theorem 2.2] states that

where is the Euclidean operator norm. In particular, retaining columns of a Hadamard matrix of order gives sign rank at least , since its columns are orthogonal and have norm . We therefore need

The first inequality defeats the geometric bound; the last leaves enough core faces to encode all rows. We use , and an order , just above the required threshold . We construct the required Hadamard matrix from a quadratic-character matrix and a binary character matrix.

Let , ordered by , and let be the quadratic character of , with . Define

This specifies a sign matrix. If , including its zero diagonal, then

The last identity is the character sum for . One direct proof reduces to and counts the solutions of ; under , , this becomes , with ten solutions. Consequently

Let for , where the dot product is over . Character orthogonality gives . Thus

Index rows and columns of lexicographically by , ordering binary vectors by their integer values, and let consist of its first columns. Take , , and assign its rows to the first nine-subsets of in lexicographic order. This is possible because . Formula (1) now defines the principal example completely.

Principal assertion of Theorem 1. Forster’s bound gives , whereas polytopality would give by Theorem 2. The input has a face of size ten and none larger, so its dimension is nine. There are ground labels. Exactly the unprimed ghost is absent from the Bier sphere, giving vertices and dimension . □

4. Variants of the construction

4.1. Removing the ghost. Making the ghost a vertex changes the sphere by a single facet subdivision. Since this subdivision can be reversed in any convex realization, it preserves the nonpolytopality established above.

For the principal example, put . Then is the stellar subdivision of at the all-primed facet . That facet is replaced by

and all other facets remain unchanged.

A stellar subdivision of a facet can be polytopal only if the original sphere was polytopal. Indeed, in a simplicial -polytope realizing the subdivision, the new vertex has exactly neighbors . Its tangent cone has the independent edge rays , so every point of the polytope has coordinates

Every other vertex except these neighbors satisfies , since otherwise it lies in the corner simplex. Deleting therefore reverses stacking across the hyperplane and realizes the original sphere, a contradiction. Thus is nonpolytopal, has actual vertices, and has dimension . Both copies of every ground label occur.

4.2. A self-dual input. Duality symmetry can be imposed while retaining the original sphere as a vertex link.

Let be the principal input on before adjoining the ghost. Add two labels and set

where the dual is relative to and no face contains both . Checking whether a subset contains neither, one, or both new labels gives . Moreover,

To verify the link identity, its positive part is a face of and cannot contain . Its primed part is either any disjoint subset of , or with . These are exactly the dual conditions for with adjoined as a ghost.

Vertex links in boundaries of convex polytopes are boundaries of vertex figures. Hence is nonpolytopal. It has actual vertices and dimension . Self-duality makes the exchange of each label with its primed copy a simplicial involution taking every nonempty face to a disjoint face. Thus this example is centrally symmetric as a combinatorial sphere.

4.3. Input dimension three. To lower the dimension of the input, we use fewer core labels and many more anchors. Balanced column signs give the required rank obstruction through continuity of the median.

Take the first triples of , and index the anchors by the -subsets . Define

Suppose a real matrix with these signs had rank at most . Its space of row relations would contain a two-dimensional subspace. On the unit circle in that subspace, the median of the coordinates is continuous and odd, so the intermediate value theorem gives a nonzero vector with median zero. This vector has at most positive and at most negative coordinates. Choose a -subset containing all its positive coordinates and none of its negative ones, filling with zero coordinates as needed. Every nonzero summand in the corresponding column’s dependence is positive, a contradiction. Thus . In fact equality holds, since assigning the values and to the positive and negative entries makes every column sum zero.

There are available row sets. Apply (1) with , and . Theorem 2 would give in a convex realization. The resulting nonpolytopal Bier sphere has actual vertices and dimension , with an input of dimension three.

The same argument gives input dimension when .

References

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A. Björner, A. Paffenholz, J. Sjöstrand, and G. M. Ziegler, Bier spheres and posets, Discrete Comput. Geom. 34 (2005), 71–86. DOI 10.1007/s00454-004-1144-0.
[2]
J. Forster, A linear lower bound on the unbounded error probabilistic communication complexity, J. Comput. System Sci. 65 (2002), 612–625. DOI 10.1016/S0022-0000(02)00019-3.
[3]
T. Holleben and Y. Yang, Polytopal Bier spheres and nonrealizable central symmetries, arXiv 2608.07233 (2026).