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 .
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