# The signed Varchenko Determinant for Complexes of Oriented Matroids

Winfried Hochstättler  FernUniversität in Hagen
Fakultät für Mathematik und Informatik
58084 Hagen
Germany<EMAIL_ADDRESS>,  Sophia Keip
FernUniversität in Hagen
Fakultät für Mathematik und Informatik
58084 Hagen
Germany<EMAIL_ADDRESS>and  Kolja Knauer  Departament de
Matemàtiques i Informàtica, Universitat de Barcelona, Spain
<EMAIL_ADDRESS>
###### Abstract.

We generalize the (signed) Varchenko matrix of a hyperplane arrangement to
complexes of oriented matroids and show that it has a nice factorization.

##  1\. Introduction

Schechtman and Varchenko [11] considered a symmetric matrix which may be
viewed as a bilinear form on the vector space of linear forms of the set of
regions of a hyperplane arrangement over some ordered field $\mathbb{K}$. The
value of the product of the characteristic vectors of regions $Q_{i}$ and
$Q_{j}$ is given by a product $\prod_{e\in S(Q_{i},Q_{j})}w_{e}$, where the
$w_{e}$ are weights on the hyperplanes $H_{e}$ of the arrangements and
$S(Q_{i},Q_{j})$ is the set of hyperplanes that have to be crossed on a
shortest path from $Q_{i}$ to $Q_{j}$. The corresponding matrix $B$ is called
the Varchenko Matrix, also see Theorem 2.0. In order to determine when the
bilinear form is degenerate, Varchenko [13] gave an elegant factorization of
the determinant of that matrix, considering the weights as variables.

######  Theorem 1.0 (Varchenko 1993+[13]).

Let $\mathcal{A}$ be a real hyperplane arrangement and $L(\mathcal{A})$ the
geometric lattice formed by the intersections of hyperplanes in $\mathcal{A}$,
then

$\det(B)=\prod_{F\in L(\mathcal{A})}\left(1-w_{F}^{2}\right)^{m_{F}}$

where $w_{F}=\prod_{F\subset H_{e}}w_{e}$ and $m_{F}$ are positive integers
depending only on the geometric lattice $L(\mathcal{A})$.

After the original proof of Varchenko there were several approaches to provide
cleaner proofs of this result. Denham and Henlon [5] sketched an elegant
alternative way to prove the result. Gente [6] provided some more details for
that proof and claimed to have generalized the result to cones, which are also
called topcones or in our notation supertopes, i.e. convex sets of regions.
This method was generalized by Hochstättler and Welker [8] to oriented
matroids, which form a combinatorial model for hyperplane arrangements
reflecting their local linear structure but allowing for some global non-
linearities.

Aguiar and Mahajan [1] generalized the original proof of Varchenko to a signed
version of the matrix and also derived the result for topcones. Here one
considers an oriented hyperplane arrangement and the entries of the signed
Varchenko matrix depend on which side of a hyperplane a cell lies, see
Definition 2.0. Randriamaro [10] generalized their proof to oriented matroids.

Bandelt et al. generalized oriented matroids to complexes of oriented matroids
by relaxing the global symmetry while maintaining convexity and local
symmetry. This framework captures a variety of classes beyond oriented
matroids, e.g., distributive lattices, CAT(0)-cube complexes, lopsided sets,
linear extension graphs, and affine oriented matroids, see [2]. The purpose of
this paper is to show, that the latter requirements are still sufficient for
the factorization formula to hold. The presentation as well as the proof
follow the lines of Hochstättler and Welker [8]. We furthermore achieve a
generalization to the signed version of the Varchenko matrix, thus
generalizing Randriamaro [10].

The paper is organized as follows. In Section 2 we introduce the considered
structures. In Section 3 we present some tools from algebraic topology that we
need for the proof of the main theorem. The latter is presented in Section 4

##  2\. The Varchenko Determinant and Complexes of Oriented Matroids

Before we introduce the Varchenko Determinant, we need to get familiar with
complexes of oriented matroids (COMs). COMs have been introduced in [2] as a
common generalization of oriented matroids, affine oriented matroids, and
lopsided sets. We start with the following definitions and axioms.

######  Definition 2.0.

Let $\mathcal{L}\subseteq\\{0,+,-\\}^{E}$ be a set of sign vectors on a finite
ground set $E$. The _composition_ of two sign vectors $X$ and $Y$ is defined
as

$(X\circ Y)_{e}=\begin{cases}X_{e}&\text{ if }X_{e}\neq 0,\\\ Y_{e}&\text{ if
}X_{e}=0\\\ \end{cases}\forall e\in E.$

The _separator_ of $X$ and $Y$ is defined as

$\displaystyle S(X,Y)=\\{e\in E:X_{e}=-Y_{e}\neq 0\\}.$

The _support_ of X is defined as

$\displaystyle\underline{X}=\\{e\in E:X_{e}\neq 0\\}.$

The _zero-set_ of X is defined as

$\displaystyle z(X)=E\backslash\underline{X}=\\{e\in E:X_{e}=0\\}.$

We introduce three axioms for systems of sign vectors.

(FS):

_Face Symmetry_

$\displaystyle\forall X,Y\in\mathcal{L}:X\circ(-Y)\in\mathcal{L}.$

(SE):

_Strong Elimination_

$\displaystyle\forall X,Y\in\mathcal{L}\,\forall e\in S(X,Y)\,\exists
Z\in\mathcal{L}:$ $\displaystyle Z_{e}=0\text{ and }\forall f\in E\setminus
S(X,Y):Z_{f}=(X\circ Y)_{f}.$

(C):

_Composition_

$\displaystyle\forall X,Y\in\mathcal{L}:X\circ Y\in\mathcal{L}.$

Now we can define the term COM.

######  Definition 2.0 (Complex of Oriented Matroids (COM)).

Let $E$ be a finite set and $\mathcal{L}\subseteq\\{0,+,-\\}^{E}$. The pair
$(E,\mathcal{L})$ is called a COM, if $\mathcal{L}$ satisfies (FS) and (SE).
The elements of $\mathcal{L}$ are called _covectors_.

We can order the covectors of a COM partially by comparing them componentwise
with respect to the order $0<+,-$. Note that (FS) implies (C). Indeed, by (FS)
we first get $X\circ-Y\in\mathcal{L}$ and then $X\circ Y=(X\circ-X)\circ
Y=X\circ-(X\circ-Y)\in\mathcal{L}$ for all $X,Y\in\mathcal{L}$. This
observation allows to define OMs as special COMs:

######  Definition 2.0 (Oriented Matroid (OM)).

Let $E$ be a finite set and $\mathcal{L}\subseteq\\{0,+,-\\}^{E}$. The pair
$(E,\mathcal{L})$ is called a OM, if $\mathcal{L}$ satisfies (FS), (SE) and
the all zeros vector $\mathbf{0}\in\mathcal{L}$.

Let $(E,\mathcal{L})$ be a COM. In the following we assume that
$(E,\mathcal{L})$ is _simple_ , i.e.

$\forall e\in E:\\{X_{e}\mid X\in\mathcal{L}\\}=\\{+,-,0\\}\quad\text{ and
}\quad\forall e\neq f\in E:\\{X_{e}X_{f}\mid X\in\mathcal{L}\\}=\\{+,-,0\\}.$

In this setting the sign-vectors in $\mathcal{L}$ of full support are called
_topes_ and $\mathcal{T}$ is the set of all topes of the COM.

The _restriction_ of a sign-vector $X\in\\{0,+,-\\}^{E}$ to $E\backslash A$,
$A\subseteq E$, denoted by $X\backslash A\in\\{0,+,-\\}^{E\backslash A}$, is
defined by $(X\backslash A)_{e}=X_{e}$ for all $e\in E\backslash A$. We also
write $X|_{E\backslash A}$. The _deletion_ of a COM is defined by
$(E\backslash A,\mathcal{L}\backslash A)$, where $\mathcal{L}\backslash
A=\\{X\backslash A,\;X\in\mathcal{L}\\}$, also written as
$\mathcal{L}|_{E\backslash A}$. It has been shown in [2] and we will make use
of it that the class of COMs is closed under deletion.

######  Note 2.0.

Let $X\in\mathcal{L}$ and $F(X)=\\{X\circ Y\,|\,Y\in\mathcal{L}\\}$. Note that
$(E\backslash\underline{X},F(X)\backslash\underline{X})$ is an OM. This is
easily seen as $(X\circ X)\backslash\underline{X}$ yields the all zeros
vector. The other OM properties are inherited from the COM.

For a covector $X$ we define ${\mathrm{star}}(X)=\\{T\in\mathcal{T}\,|\,X\leq
T\\}$, where the componentwise ordering with respect to $0<+,-$ is used. Note
that ${\mathrm{star}}(X)$ is the set of topes of the oriented matroid $F(X)$.

Let us look at a special OM which we will need in the next chapter. For more
information on OMs and their covectors see [4].

######  Example 2.0 (Graphic OM of a directed cycle).

We look a digraph with three vertices which just consists of a directed cycle,
i.e.

The groundset of this OM corresponds to the arcs. One gets the covectors of
such a OM by looking at the signvectors of directed cuts (indicated with
dotted lines). All possible compositions of those signvectors together with
$\mathbf{0}$ give us our covectors. The signvectors of the directed cuts, also
called _cocircuits_ , are $(+,-,0)$, $(-,+,0)$, $(+,0,-)$, $(-,0,+)$,
$(0,+,-)$ and $(0,-,+)$. Their compositions additionally yield the covectors
$(+,+,-)$, $(+,-,+)$, $(-,+,+)$, $(-,-,+)$, $(-,+,-)$ and $(+,-,-)$. We see
that $\mathcal{T}$ consists here of all possible full support signvectors,
exept $(+,+,+)$ and $(-,-,-)$. It can easily be checked that (FS) and (SE) are
fulfilled, so we have a COM and since this COM contains $\mathbf{0}$ we have
an OM. We call the set of covectors of this particular OM $\mathcal{C}_{3}$.
Analogously we call the covectors of the OM of a directed cycle with $n$
vertices $\mathcal{C}_{n}$.

We define the signed Varchenko matrix for COMs analog to the definition of
this matrix for hyperplane arrangements in [1]. For this purpose we introduce
two variables $x_{e^{+}}$, $x_{e^{-}}$ for each element $e\in E$. Let
$\mathbb{K}$ be a field and let $\mathbb{K}[x_{e^{*}}\,|\,*\in\\{+,-\\},e\in
E]$ the polynomial ring in the set of variables $x_{e^{*}}$, $*\in\\{+,-\\}$,
$e\in E$.

######  Definition 2.0 (Signed Varchenko Matrix of a COM).

Let $(E,\mathcal{L})$ be a COM. The signed Varchenko matrix ${\mathfrak{V}}$
of a COM is defined by a $\\#\mathcal{T}\times\\#\mathcal{T}$-Matrix over
$\mathbb{K}[x_{e^{*}}\,|\,*\in\\{+,-\\},e\in E]$. Its rows and columns are
indexed by the topes $\mathcal{T}$ in a fixed linear order. For
$P,Q\in\mathcal{T}$

$\displaystyle{\mathfrak{V}}_{P,Q}=\prod_{e\in S(P,Q)}x_{e^{P_{e}}}.$

Note that the diagonal entries ${\mathfrak{V}}_{P,P}$ of the matrix are equal
to 1. In this work we will prove the following theorem.

######  Theorem 2.0.

Let ${\mathfrak{V}}$ be the signed Varchenko matrix of the COM
$(E,\mathcal{L})$. Then

$\displaystyle\det({\mathfrak{V}})=\prod_{Y\in\mathcal{L}}(1-a(Y))^{b_{Y}}.$

where $a(Y):=\prod_{e\in z(Y)}x_{e^{+}}x_{e^{-}}$ and $b_{Y}$ are nonnegative
integers.

A corollary of this result, namely the case where $x_{e^{-}}=x_{e^{+}}$, which
is the original version of the Varchenko matrix, has been already proven for
OMs in [8]. We formulate it for COMs.

######  Corollary 2.0.

Let $\mathbf{V}$ be the (unsigned) Varchenko matrix (i.e.
$x_{e^{-}}=x_{e^{+}}=x_{e}$) of the COM $(E,\mathcal{L})$. Then

$\displaystyle\det(\mathbf{V})=\prod_{Y\in\mathcal{L}}(1-c(Y)^{2})^{b_{Y}}.$

where $c(Y):=\prod_{e\in z(Y)}x_{e}$ and $b_{Y}$ are nonnegative integers.

##  3\. Preparation

We start with some basics about partially ordered sets $\mathcal{P}$
(_posets_). For an introduction we recommend [14]. One can associate an
abstract simplicial complex $\Delta(\mathcal{P})$, called _order complex_ , to
every poset. The elements of $\mathcal{P}$ are the vertices of this complex
and the chains (i.e. totally ordered subsets) the faces. Having this
association we can speak about contractible and homotopy equivalent posets. A
poset is called contractible if its order complex is homotopy equivalent to a
point. Clearly a poset is contractible if it has a unique minimal or a unique
maximal element, since this element is contained in every chain. We introduce
now the Möbius function $\mu$ of two poset elements:

$\displaystyle\mu(x,x)=1\text{ for all }x\in\mathcal{P}$
$\displaystyle\mu(x,y)=-\sum_{x\leq z<y}\mu(x,z)\text{ for all
}x<y\in\mathcal{P}.$

The _bounded extension_ $\hat{\mathcal{P}}$ of a poset is the poset together
with a new maximal element $\hat{1}$ and minimal element $\hat{0}$. We denote
by $\mu(\mathcal{P})$ the _Möbius number_ of $\mathcal{P}$. It is defined by

$\displaystyle\mu(\mathcal{P})=\mu(\hat{0},\hat{1}),$

where the right-hand-side is evaluated in $\hat{\mathcal{P}}$.

######  Example 3.0.

Let us look at the poset $\mathcal{P}$ which consists only of one element. In
the following its bounded extension and the value of the Möbius function of
the elements of the bounded extension is depicted.

Hence, the Möbius number of a poset consisting of only one element is

$\displaystyle\mu(\mathcal{P})=\mu(\hat{0},\hat{1})=1+(-1)=0.$

It follows from the following fact that the Möbius number is a topological
invariant with respect to homotopic equivalence.

######  Theorem 3.0.

[14, Philip Hall Theorem] The Möbius number of a poset equals the reduced
Euler characteristic of its order complex, i.e.

$\displaystyle\mu(\mathcal{P})=\chi(\Delta(\mathcal{P}))-1.$

More precisely we have

######  Corollary 3.0.

For two homotopy equivalent posets $\mathcal{P}$ and $\mathcal{Q}$ we have
$\mu(\mathcal{P})=\mu(\mathcal{Q})$. In particular, if $\mathcal{P}$ is
contractible then $\mu(\mathcal{P})=0$.

The second part of Corollary 3.0 follows from the definition of
contractability and Example 3.0. We denote for a poset $\mathcal{P}$ and
$p\in\mathcal{P}$ by $\mathcal{P}_{\leq p}$ the subposet
$\\{q\in\mathcal{P}~{}|~{}q\leq p\\}$.

######  Proposition 3.0 (Quillen Fiber Lemma).

Let ${\mathcal{P}}$ and $\mathcal{Q}$ be posets and
$f:{\mathcal{P}}\rightarrow\mathcal{Q}$ a poset map. If for all $q\in Q$ we
have that $f^{-1}(\mathcal{Q}_{\leq q})$ is contractible, then ${\mathcal{P}}$
and $\mathcal{Q}$ are homotopy equivalent.

We will now associate posets with COMs, so let $(E,\mathcal{L})$ be a COM and
let $R\in\\{+,-\\}^{E}$ be a fixed sign pattern. We consider ${\mathcal{T}}$
as a poset with order relation

$P\preceq_{R}Q\quad\text{ if }\quad S(R,P)\subseteq S(R,Q).$

We write ${\mathcal{T}}_{R}$ if we consider ${\mathcal{T}}$ with this partial
order and we call $R$ the _base pattern_ of the poset.

Now we will introduce a theorem which will help us with our crucial lemma. For
$S^{+},S^{-}\subseteq E,S^{+}\cap S^{-}=\emptyset$, we denote by
${\mathcal{T}}(S^{+},S^{-})$ the set of topes of ${\mathcal{T}}$ that are
positive on $S^{+}$ and negative on $S^{-}$. Note that
${\mathcal{T}}(S^{+},S^{-})$ itself is the set of topes of a COM, called
_topal fiber_ in [2] and restriction $\rho_{(S^{+},S^{-})}({\mathcal{L}})$ in
[9]. Usually, we suppress the redundant coordinates in $(S^{+}\cup S^{-})$, to
obtain a simple COM.

######  Theorem 3.0.

Let $(\mathcal{L},E)$ be a COM such that there is a COM
$(\mathcal{L^{\prime}},E^{\prime})$, with $E=E^{\prime}\setminus(S^{+}\cup
S^{-})$ for some $S^{+},S^{-}\subseteq E^{\prime}$ and
$\mathcal{L}=\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime})$. Let
${\mathcal{T}}$ be the non-empty set of topes of $\mathcal{L}$ and
$R\in\\{+,-\\}^{E}$ be a sign pattern such that $R=R^{\prime}|_{E}$ for some
tope $R^{\prime}$ of $\mathcal{L}^{\prime}$. Then the order complex of
${\mathcal{T}}_{R}$ is contractible.

In order to apply the Quillen Fiber Lemma in the proof of Theorem 3.0 we need
the following lemma.

######  Lemma 3.1.

Let $f\in E$ and $R=\\{+\\}^{E}$. Let ${\mathcal{T}}^{f}$ denote the set of
topes of ${\mathcal{L}}\setminus f$ and ${\mathcal{T}}^{f}_{R\setminus f}$ the
corresponding tope poset with base pattern $R\setminus\\{f\\}$. Consider the
poset map $\pi_{f}:{\mathcal{T}}_{R}\to{\mathcal{T}}^{f}_{R\setminus f}$ given
by restriction. Let $Q\in{\mathcal{T}}^{f}$. Then

$\pi_{f}^{-1}(({\mathcal{T}}^{f}_{R\setminus f})_{\preceq
Q})={\mathcal{T}}(Q^{+},\emptyset).$

###### Proof.

Let $\tilde{Q}\in{\mathcal{T}}^{f}_{\preceq Q}$. As $R\setminus f$ is all
positive, we must have $\tilde{Q}^{-}\subseteq Q^{-}$ and hence
$Q^{+}\setminus f\subseteq\tilde{Q}^{+}$ implying
$\pi_{f}^{-1}(\tilde{Q})\subseteq{\mathcal{T}}(Q^{+},\emptyset)$. If on the
other hand $\hat{Q}\in{\mathcal{T}}(Q^{+},\emptyset)$, then
$\hat{Q}^{-}\subseteq Q^{-}\cup\\{f\\},$
$Q^{+}\subseteq\hat{Q}^{+}\cup\\{f\\}$. Hence
$\pi_{f}(\hat{Q})\preceq_{R\setminus f}Q$. ∎

We need two preparatory results. For the proof of Theorem 3.0, for the first
one also see [7, Lemma 10]. We reprove it here, since in the presentation in
[7] the signs are chosen the opposite way. Recall from Example 2.0 that
$\mathcal{C}_{n}$ is the set of covectors of the OM of the directed cycle on
$n$ vertices.

######  Proposition 3.0.

Let $(E,\mathcal{L})$ be a COM with tope set $\mathcal{T}$, such that for all
$f\in E$ there exists $T^{f}\in\mathcal{T}$ such that

$T^{f}_{g}=\left\\{\begin{array}[h]{rcl}-&\text{ if }&g\in
E\setminus\\{f\\}\\\ +&\text{ if }&g=f.\end{array}\right.$

Let $R=\\{+\\}^{E}$. If $-R\notin\mathcal{T}$, then
$\mathcal{L}=\mathcal{C}_{\lvert E\rvert}$. Hence, if
$\mathcal{L}\neq\mathcal{C}_{n}$, then the poset ${\mathcal{T}}_{R}$ has a
unique maximal element. In particular, it is contractible.

###### Proof.

We will show by induction that all covectors which contain exactly one plus-
entry and at least one minus-entry are in $\mathcal{L}$. Since then in
particular all covectors which contain exactly one minus-entry and one plus-
entry (i.e. the cocircuits) exist in $\mathcal{L}$, we get by (SE) that the
all zero vector is in $\mathcal{L}$. Together, we can conclude that
$\mathcal{L}=\mathcal{C}_{n}$, since we obtain all its covectors by
composition of those vectors. Since $\mathcal{C}_{n}$ is uniform no other
oriented matroid can contain these covectors.

So let $T^{f}\in\mathcal{T}$ for all $f\in E$ and $-R\notin\mathcal{T}$. We
will use induction over the number of zero-entries in the covectors, i.e. we
want to show that for every $n=0,\dots,\lvert E\rvert-2$ all sign-vectors with
$n$ zero entries, one plus-entry and $\lvert E\rvert-(n+1)$ minus-entries are
in $\mathcal{L}$.

$n=0$: By the existence of $T^{f}$ here is nothing to show. We fix $n\geq 0$
and assume that all covectors with $n$ or less zero-entries, exactly one plus-
entry and at least one minus-entry exist in $\mathcal{L}$.

$n\rightarrow n+1\leq\lvert E\rvert-2$: We will show that there exists a
covector with zero-entries in the i-th position, $i\in I\subset E$, $\lvert
I\rvert=n+1$, a plus-entry in the j-th position, $j\notin I$ and $-$
everywhere else. We choose an $\hat{i}\in I$ and consider two covectors with
$0$ in $I\backslash\hat{i}$, where one has a $+$ in the $\hat{i}$-th position
and the other one in the j-th position and both have a $-$ everywhere else.
These do exist by inductive assumption. W.l.o.g. those two covectors look like
this:

$\displaystyle(0,\dots,0,\overbrace{+}^{\hat{i}},\;\,-\,\;,-,\dots,-)$
$\displaystyle(\underbrace{0,\dots,0}_{I\backslash\hat{i}},\;\,-\,\;,\underbrace{+}_{j},-,\dots,-).$

If we now perform strong elimination on $\hat{i}$ with those two covectors we
get the covector

$\displaystyle
X=(\underbrace{0,\dots,0}_{I\backslash\hat{i}},\underbrace{0}_{\hat{i}},\underbrace{*}_{j},-,\dots,-).$

If $*$ was $-$, then $X\circ T^{j}=\\{-\\}^{E}$. Since
$\\{-\\}^{E}\notin\mathcal{T}$ we have $*=+$ and have the covector we were
looking for. ∎

######  Lemma 3.2.

Let $(\mathcal{L},E)$ be a COM such that there is a COM
$(\mathcal{L^{\prime}},E^{\prime})$, with $E\subset E^{\prime}$ and
$\mathcal{L}=\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime})$ for some
$S^{+},S^{-}\subseteq E^{\prime}$. Let ${\mathcal{T}}$ be the non-empty set of
topes of $\mathcal{L}$ and $R\in\\{+,-\\}^{E}$ be a sign pattern such that
$R=R^{\prime}|_{E}$ for some tope $R^{\prime}$ of $\mathcal{L}^{\prime}$. If
$\mathcal{L}=\mathcal{C}_{n}$, then $R\in\mathcal{L}$, in particular
$R\neq\\{+\\}^{E},\\{-\\}^{E}$.

###### Proof.

We saw in Example 2.0 that $\mathbf{0}\in\mathcal{C}_{n}$. By the definition
of $\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime})$ there exists
$Z\in\mathcal{L}^{\prime}$ with

$Z_{e}=\begin{cases}+&\text{ if }e\in S^{+},\\\ -&\text{ if }e\in S^{-}\\\
0&\text{ else. }\end{cases}$

Since the composition of $Z$ with every other covector in
$\mathcal{L}^{\prime}$ is in $\mathcal{L}^{\prime}$, we see that
$\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime})=\mathcal{L}\backslash\\{S^{+}\cup
S^{-}\\}$. So in this case $\mathcal{L}\backslash\\{S^{+}\cup
S^{-}\\}=\mathcal{C}_{n}$, so every tope $R^{\prime}\in\mathcal{L}^{\prime}$
restricted to $E$ has to be in $\mathcal{C}_{n}$. Since
$\\{+\\}^{E},\\{-\\}^{E}\notin\mathcal{C}_{n}$,
$R=R^{\prime}|_{E}\neq\\{+\\}^{E},\\{-\\}^{E}$.

∎

Now we are in position to prove Theorem 3.0.

###### Proof of Theorem 3.0.

First we look at the case $\mathcal{L}=\mathcal{C}_{n}$. From Lemma 3.2 we
know, $R$ is unequal to $\\{+\\}^{E}$ and $\\{-\\}^{E}$ and is a tope of
$\mathcal{C}_{n}$ instead. But since $\mathbf{0}\in\mathcal{C}_{n}$,
$\mathbf{0}\circ-R=-R\in\mathcal{C}_{n}$. So we have a unique maximal element
and $\mathcal{T}_{R}$ is contractible. Now let
$\mathcal{L}\neq\mathcal{C}_{n}$. Possibly reorienting elements we may assume
that $R=\\{+\\}^{E}$. We proceed by induction on $|E|$. If $|E|=1$ then
${\mathcal{T}}_{R}$ either is a singleton or a chain of length 2 and thus
contractible. Hence assume $|E|\geq 2$. If for all $f\in E$ there exists
$T^{f}$ as in Proposition 3.0, then ${\mathcal{T}}_{R}$ is contractible by
Proposition 3.0. Hence we may assume that there exists $f\in E$ such that
$T^{f}\not\in{\mathcal{T}}$. Let ${\mathcal{T}}^{f}_{R\setminus f}$ denote the
tope poset in ${\mathcal{L}}\setminus f$ with base pattern $R\setminus f$.
Since the class of COMs is closed under deletion, we know that
${\mathcal{L}}^{\prime}\setminus f$ is a COM. Since $\mathcal{L}$ evolved from
$\mathcal{L}^{\prime}$ by setting
${\mathcal{L}}=\rho_{(S^{+},S^{-})}{\mathcal{L}}^{\prime}$ for some $S^{+}$
and $S^{-}$, ${\mathcal{L}}\setminus f$ evolves in the same way from
${\mathcal{L}}^{\prime}\setminus f$, i.e. ${\mathcal{L}}\setminus
f=\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime}\setminus f)$ (note that $f$
cannot be in $S^{+}\cup S^{-}$, since $E=E^{\prime}\setminus(S^{+}\cup S^{-})$
and $f\in E$). Also $R\setminus f$ is the restriction to $E$ of the tope
$R^{\prime}\setminus f$ of ${\mathcal{L}}^{\prime}\setminus f$. We see that
${\mathcal{L}}\setminus f$ together with $R\setminus f$ fulfills the
assumtions of the theorem. Furthermore, ${\mathcal{L}}\setminus
f\neq\mathcal{C}_{n}$, this follows from Lemma 3.2. Hence,
${\mathcal{T}}^{f}_{R\setminus f}$ is contractible by inductive assumption. We
now want to show that ${\mathcal{T}}_{R}$ and ${\mathcal{T}}^{f}_{R\setminus
f}$ are homotopy equivalent by using Proposition 3.0. So consider the poset
map $\pi_{f}:{\mathcal{T}}_{R}\to{\mathcal{T}}^{f}_{R\setminus\\{f\\}}$ given
by restriction. Let $Q\in{\mathcal{T}}^{f}_{R\setminus\\{f\\}}$. By Lemma 3.1

$\pi_{f}^{-1}(({\mathcal{T}}^{f}_{R\setminus\\{f\\}})_{\preceq
Q})={\mathcal{T}}(Q^{+},\emptyset).$

${\mathcal{T}}(Q^{+},\emptyset)$ is the set of topes of $\rho_{(S^{+}\cup
Q^{+},S^{-})}{\mathcal{L}}^{\prime}$. If $Q^{+}\neq\emptyset$, then
$\rho_{(S^{+}\cup Q^{+},S^{-})}{\mathcal{L}}^{\prime}$ has fewer elements than
$\mathcal{L}$. Furthermore, by Lemma 3.2, $\rho_{(S^{+}\cup
Q^{+},S^{-})}{\mathcal{L}}^{\prime}\neq\mathcal{C}_{n}$. Hence
$\pi_{f}^{-1}({\mathcal{T}}^{f}_{\preceq Q})$ is contractible by inductive
assumption. If $Q^{+}=\emptyset$ then by the choice of $f$ the preimage
$\pi_{f}^{-1}(Q)$ is the all minus vector. Hence, this is the unique maximal
element in $\pi_{f}^{-1}({\mathcal{T}}^{f}_{\preceq Q})$ and that fiber is
also contractible. So by Proposition 3.0 ${\mathcal{T}}_{R}$ and
${\mathcal{T}}^{f}_{R\setminus f}$ are homotopy equivalent and the claim
follows. ∎

For $e\in E$ and $R\in\mathcal{T}$ we write $\mathcal{T}_{R,e}$ for the poset
$\\{T\in\mathcal{T}~{}|~{}T_{e}=-R_{e}\\}\cup\\{\hat{0}\\}$ with $\hat{0}$ as
its artificial least element and the remaining poset structure induced from
$\mathcal{T}_{R}$. For $P\in\mathcal{T}_{R,e}$ we write $(\hat{0},P)_{R,e}$
for the interval from $\hat{0}$ to $P$ in $\mathcal{T}_{R,e}$.

######  Lemma 3.3.

Let $(E,\mathcal{L})$ be a COM and let $R\in\mathcal{T}$ be the base tope of
the poset $\mathcal{T}_{R}$. Let $P\in\mathcal{T}_{R,e}$ and $e\not\in
S\subseteq E$. Then

(3)
$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S=S(P,Q)\cap
S(Q,R)}}\mu((\hat{0},Q)_{R,e})=\left\\{\begin{array}[]{ccc}-1&\mbox{~{}if~{}}&S=\emptyset\\\
0&\mbox{~{}if~{}}&S\neq\emptyset\end{array}\right.$

and

(6)
$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\\{e\\},\emptyset)}{S=S(P,Q)\cap
S(Q,R)}}\mu((\hat{0},Q)_{R,e})=\left\\{\begin{array}[]{ccc}-1&\mbox{~{}if~{}}&S=\emptyset\\\
0&\mbox{~{}if~{}}&S\neq\emptyset\end{array}\right..$

###### Proof.

In order to prove (3) we assume $R=\\{+\\}^{E}$. We prove the assertion by
induction on $|S|$. If $S=\emptyset$ then

$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S=S(P,Q)\cap
S(Q,R)}}\mu((\hat{0},Q)_{R,e})$ $\displaystyle=$
$\displaystyle\sum_{\hat{0}<_{R,e}Q\leq_{R,e}P}\mu((\hat{0},Q)_{R,e})$

Note that
$\sum_{\hat{0}<_{R,e}Q\leq_{R,e}P}\mu((\hat{0},Q)_{R,e})=\mu\left(\\{Q\in\mathcal{T}_{R,e}\,|\,\hat{0}\leq_{R,e}Q\leq_{R,e}P\\}\right)-\mu((\hat{0},\hat{0})_{R,e})$.
This poset has the maximal element $P$, so it is contractible and has Möbius
number $0$. Therefore we have

$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S=S(P,Q)\cap
S(Q,R)}}\mu((\hat{0},Q)_{R,e})=-\mu((\hat{0},\hat{0})_{R,e})=-1.$

Assume $|S|>0$. Set

$T^{+}=\\{f\in E\setminus(S\cup\\{e\\})~{}|~{}P_{f}=+\\}.$

Then

(7)
$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S(P,Q)\cap
S(Q,R)\subseteq S}}\mu((\hat{0},Q)_{R,e})$ $\displaystyle=$
$\displaystyle\sum_{Q\in\mathcal{T}(T^{+},\\{e\\})}\mu((\hat{0},Q)_{R,e}).$

$\mathcal{T}(T^{+},\\{e\\})$ is isomorphic to the set of topes of the COM
$\rho_{(T^{+},\\{e\\})}({\mathcal{L}})$ which is contractible by Theorem 3.0.
So the right hand side of (7) ranges over the elements of a contractible
poset. By the same argument as above it is $-\mu(\hat{0},\hat{0})=-1$ minus
the Möbius number of the poset. Since the poset is contractible its Möbius
number is $0$ and we have shown that

(8)
$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S(P,Q)\cap
S(Q,R)\subseteq S}}\mu((\hat{0},Q)_{R,e})$ $\displaystyle=$ $\displaystyle-1.$

Now rewrite the left hand side of (8) as

(9)
$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S(P,Q)\cap
S(Q,R)\subseteq S}}\mu((\hat{0},Q)_{R,e})$ $\displaystyle=$
$\displaystyle\sum_{T\subseteq
S}\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S(P,Q)\cap
S(Q,R)=T}}\mu((\hat{0},Q)_{R,e})$

By induction the summand
$\displaystyle{\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S(P,Q)\cap
S(Q,R)=T}}\mu((\hat{0},Q)_{R,e})}$ is $0$ for $T\neq S,\emptyset$ and $-1$ for
$T=\emptyset$. Thus combining (8) and (9) we obtain:

$\displaystyle-1$ $\displaystyle=$
$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S(P,Q)\cap
S(Q,R)\subseteq S}}\mu((\hat{0},Q)_{R,e})$ $\displaystyle=$
$\displaystyle-1+\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S(P,Q)\cap
S(Q,R)=S}}\mu((\hat{0},Q)_{R,e})$

From this we conclude

$\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S(P,Q)\cap
S(Q,R)=S}}\mu((\hat{0},Q)_{R,e})=0.$

The second claim follows analogously by reorienting all the signs. ∎

######  Lemma 3.4.

Let $X\in\mathcal{L}$ and $P\in{\mathcal{T}}$. The tope $Q=X\circ
P\in{\mathrm{star}}(X)$ is the only tope in ${\mathrm{star}}(X)$ such that for
all $O\in{\mathrm{star}}(X)$ we have

(10) $\displaystyle S(P,O)$ $\displaystyle=S(P,Q)\cup S(Q,O)$ (11)
$\displaystyle\emptyset$ $\displaystyle=S(P,Q)\cap S(Q,O).$

###### Proof.

It is easy to see that $Q$ fulfills (10) and (11). Let us assume there is
another tope $Q^{*}\neq Q$ in ${\mathrm{star}}(X)$ which has this property. By
the definition of $Q$ we have

$S(Q,O)=S(P,O)\cap z(X)\text{ and }S(P,Q)=(P,O)\backslash z(X)\text{ for all
}O\in{\mathrm{star}}(X).$

Since $Q^{*}\neq Q$ and $S(Q^{*},O)$ can only contain elements from $z(X)$,
$S(P,Q^{*})$ has to contain at least one element from $z(X)$. Now considering
$O^{*}=(X\circ-Q^{*})\in{\mathrm{star}}(X)$ we see that $S(P,Q^{*})\cap
S(Q^{*},O^{*})\neq\emptyset$, so $Q^{*}$ does not fulfill the property and we
have a contradiction. ∎

For $e\in E$ and $P\in\mathcal{T}$ we say that $e$ _does define a proper face_
of $P$ if there is a covector $X\in\mathcal{L}$ with $X\leq P$ and $X_{e}=0$
with $X\neq\mathbf{0}$. Note that in this case there is a unique maximal such
covector, namely the composition of all of them. Otherwise, we say that $e$
_does not define a proper face_ of $P$.

######  Theorem 3.0.

Let $R\in{\mathcal{T}}$ and let $e\in E$ define a proper face of $R$. Let
$Y\in\mathcal{L}$ be the maximal covector such that $Y\leq R$ and $Y_{e}=0$
and choose $P_{top}\in{\mathcal{T}}_{R,e}\setminus{\mathrm{star}}(Y)$. Then
$(\hat{0},P_{top})_{R,e}$ is contractible. In particular,
$\mu((\hat{0},P_{top})_{R,e})=0$.

###### Proof.

Let $P\in(\hat{0},P_{top})_{R,e}$. Then by Lemma 3.4 the tope $Q=Y\circ
P\in{\mathrm{star}}(Y)$ is the unique tope in ${\mathrm{star}}(Y)$ such that
for all $O\in{\mathrm{star}}(Y)$ we have

$\displaystyle S(P,O)$ $\displaystyle=$ $\displaystyle S(P,Q)\cup S(Q,O)$
$\displaystyle\emptyset$ $\displaystyle=$ $\displaystyle S(P,Q)\cap S(Q,O).$

Since $Y_{e}=0$ and $P\in\mathcal{T}_{R,e}$ it also follows that $Q_{e}=-$.
Since $Y\leq R$, clearly $S(R,Q)=S(R,Y\circ P)\subseteq S(R,P)$ and hence
$Q\preceq_{R}P$. This shows $Q\in(\hat{0},P_{top})_{R,e}$. We now define the
map

$\displaystyle\circ_{Y}:(\hat{0},P_{top})_{R,e}\rightarrow(\hat{0},P_{top})_{R,e}$
$\displaystyle\circ_{Y}(P)=Y\circ P$

and prove that it is a closure operator by showing that it is order preserving
and idempotent (i.e. $\circ_{Y}(\circ_{Y}(P))=\circ_{Y}(P)$). So let
$Q\preceq_{R}Q^{\prime}$. Then $Y\circ Q\preceq_{R}Y\circ Q^{\prime}$. Since
$Y\leq R$ it follows that $Y\circ Q\preceq_{R}Q$. Obviously $Y\circ(Y\circ
Q)=Y\circ Q$. So $\circ_{Y}$ is a closure operator and it follows that
$(\hat{0},P_{top})_{R,e}$ is homotopy equivalent to its image (see e.g, [3,
Corollary 10.12]).

Since $P_{top}\not\in{\mathrm{star}}(Y)$ and $Y\circ
P_{top}\in{\mathrm{star}}(Y)\cap(\hat{0},P_{top})_{R,e}$, it also follows that
$Y\circ Q\preceq_{R}Y\circ P_{top}$ for all $Q\in(\hat{0},P_{top})_{R,e}$.
Hence the image of $\circ_{Y}$ has a unique maximal element and hence is
contractible. ∎

##  4\. Main Proof

Recall, that we assume $\mathcal{T}$ to be linearly ordered. Note however that
swapping two topes leads to a row swap and a columns swap at the same time, so
we do not change the sign of our determinant. Hence, in this section we will
rearrange the ordering on $\mathcal{T}$, whenever convenient for the proof.
Moreover, for the proof we also fix a linear ordering on $E$, i.e.,
$E=\\{e_{1}\prec\cdots\prec e_{r}\\}$.

For any sign pattern $\epsilon=(\epsilon_{1},\epsilon_{2})\in\\{+,-\\}^{2}$
let ${\mathfrak{V}}^{e,\epsilon}$ be a matrix with rows indexed by
$\mathcal{T}(\\{e\\},\emptyset)$ for $\epsilon_{1}=+$,
$\mathcal{T}(\emptyset,\\{e\\})$ for $\epsilon_{1}=-$ and columns indexed by
$\mathcal{T}(\\{e\\},\emptyset)$ for $\epsilon_{2}=+$,
$\mathcal{T}(\emptyset,\\{e\\})$ for $\epsilon_{2}=-$. For a tope $R$ indexing
a row and a tope $Q$ indexing a column we set
${\mathfrak{V}}^{e,\epsilon}_{R,Q}={\mathfrak{V}}_{R,Q}$. After reordering
$\mathcal{T}$ this yields a block decomposition of ${\mathfrak{V}}$ as

(14)
$\displaystyle{\mathfrak{V}}=\left(\begin{array}[]{cc}{\mathfrak{V}}^{e,(-,-)}&{\mathfrak{V}}^{e,(-,+)}\\\
{\mathfrak{V}}^{e,(+,-)}&{\mathfrak{V}}^{e,(+,+)}\end{array}\right).$

We fix such a linear ordering on $\mathcal{T}$ and set $\mathcal{M}^{e}$ to

$\displaystyle\mathcal{M}^{e}_{Q,R}=\left\\{\begin{array}[]{ccc}1&\mbox{~{}if~{}}&Q=R\\\
-\mu((\hat{0},Q)_{R,e})\,{\mathfrak{V}}_{Q,R}&\mbox{~{}if~{}}&e\text{ is the
maximal element of }S(Q,R),\\\ 0&\mbox{~{}otherwise}&\end{array}\right..$

Note that this matrix has the following form

$\displaystyle\mathcal{M}^{e}=\left(\begin{array}[]{cc}\mathcal{I}^{e}_{\ell}&U^{e}\\\
L^{e}&\mathcal{I}^{e}_{m}\end{array}\right),$

where

$\displaystyle
U^{e}_{Q,R}=-\mu((\hat{0},Q)_{R,e})\,{\mathfrak{V}}_{Q,R},\;\;\;$
$\displaystyle e\text{ is the maximal element of S}(Q,R),$ $\displaystyle
Q\in\mathcal{T}(\emptyset,\\{e\\}),R\in\mathcal{T}(\\{e\\},\emptyset),$
$\displaystyle
L^{e}_{Q,R}=-\mu((\hat{0},Q)_{R,e})\,{\mathfrak{V}}_{Q,R},\;\;\;$
$\displaystyle e\text{ is the maximal element of S}(Q,R),$ $\displaystyle
Q\in\mathcal{T}(\\{e\\},\emptyset),R\in\mathcal{T}(\emptyset,\\{e\\})$

and $\mathcal{I}$ the identity matrix with
$\ell=\\#\mathcal{T}(\emptyset,\\{e\\})$ and
$m=\\#\mathcal{T}(\\{e\\},\emptyset)$.

######  Lemma 4.1.

Let $e$ be the maximal element of $E$. Then ${\mathfrak{V}}^{e,(-,+)}$ factors
as

(15) $\displaystyle{\mathfrak{V}}^{e,(-,+)}={\mathfrak{V}}^{e,(-,-)}\cdot
U^{e}$

and ${\mathfrak{V}}^{e,(+,-)}$ as

(16) $\displaystyle{\mathfrak{V}}^{e,(+,-)}={\mathfrak{V}}^{e,(+,+)}\cdot
L^{e}.$

###### Proof.

Let us prove (15) first. For $P\in\mathcal{T}(\emptyset,\\{e\\})$ and
$R\in\mathcal{T}(\\{e\\},\emptyset)$ the entry in row $P$ and column $R$ on
the left hand side of (15) is ${\mathfrak{V}}_{P,R}$. On the right hand side
the corresponding entry is:

$\displaystyle\sum_{Q\in\mathcal{T}(\emptyset,\\{e\\})}{\mathfrak{V}}_{P,Q}\cdot
U^{e}_{Q,R}$ $\displaystyle=$
$\displaystyle-\sum_{Q\in\mathcal{T}(\emptyset,\\{e\\})}\mu((\hat{0},Q)_{R,e})\cdot{\mathfrak{V}}_{P,Q}\cdot{\mathfrak{V}}_{Q,R}$

This follows from the fact that $e$ is the maximal element of any separator of
the topes indexing $U^{e}$. By definition we have for
$Q\in\mathcal{T}(\emptyset,\\{e\\})$

$\displaystyle{\mathfrak{V}}_{P,Q}\cdot{\mathfrak{V}}_{Q,R}$ $\displaystyle=$
$\displaystyle{\mathfrak{V}}_{P,R}\cdot\prod_{f\in S(P,Q)\cap
S(Q,R)}x_{f^{+}}x_{f^{-}}.$

We see that
${\mathfrak{V}}_{P,Q}\cdot{\mathfrak{V}}_{Q,R}={\mathfrak{V}}_{P,R}$ if
$S(P,Q)\cap S(Q,R)=\emptyset$. Thus the claim of the lemma is proved once we
have shown that for a fixed subset $S\subseteq E$ and fixed $P,R$ we have:

(19)
$\displaystyle\sum_{\genfrac{}{}{0.0pt}{}{Q\in\mathcal{T}(\emptyset,\\{e\\})}{S=S(P,Q)\cap
S(Q,R)}}\mu((\hat{0},Q)_{R,e})=\left\\{\begin{array}[]{cc}0&\mbox{~{}if~{}}S\neq\emptyset\\\
-1&\mbox{~{}otherwise.}\end{array}\right..$

But this is the content of Lemma 3.3 and we are done. For (16) the right hand
side is

$\displaystyle\sum_{Q\in\mathcal{T}(\\{e\\},\emptyset)}{\mathfrak{V}}_{P,Q}\cdot
L^{e}_{Q,R}$ $\displaystyle=$
$\displaystyle-\sum_{Q\in\mathcal{T}(\\{e\\},\emptyset)}\mu((\hat{0},Q)_{R,e})\cdot{\mathfrak{V}}_{P,Q}\cdot{\mathfrak{V}}_{Q,R}$

and we can proceed analogous to the proof above. ∎

Next we use the matrices $\mathcal{M}^{e}$ to factorize ${\mathfrak{V}}$. The
following lemma yields the base case for the inductive step in the
factorization.

######  Lemma 4.2.

Let $e$ be the maximal element of $E$ and let ${\mathfrak{V}}_{x_{e}=0}$ be
the matrix ${\mathfrak{V}}$ after evaluating $x_{e^{+}}$ and $x_{e^{-}}$ to
$0$. Then

${\mathfrak{V}}={\mathfrak{V}}_{x_{e}=0}\cdot\mathcal{M}^{e}$

###### Proof.

Let $\mathcal{T}$ be in that order, that we get the block decomposition (14)
of ${\mathfrak{V}}$. Using lemma Lemma 4.1, we see that

(26)
$\displaystyle{\mathfrak{V}}=\left(\begin{array}[]{cc}{\mathfrak{V}}^{e,(-,-)}&{\mathfrak{V}}^{e,(-,+)}\\\
{\mathfrak{V}}^{e,(+,-)}&{\mathfrak{V}}^{e,(+,+)}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}{\mathfrak{V}}^{e,(-,-)}&0\\\
0&{\mathfrak{V}}^{e,(+,+)}\end{array}\right)\cdot\left(\begin{array}[]{cc}\mathcal{I}^{e}_{\ell}&U^{e}\\\
L^{e}&\mathcal{I}^{e}_{m}\end{array}\right)$ (29)
$\displaystyle=\left(\begin{array}[]{cc}{\mathfrak{V}}^{e,(-,-)}&0\\\
0&{\mathfrak{V}}^{e,(+,+)}\end{array}\right)\cdot\mathcal{M}^{e}.$

Now the monomial ${\mathfrak{V}}_{P,Q}$ has a factor $x_{e^{+}}$ or
$x_{e^{-}}$ if and only if $P\in\mathcal{T}(\emptyset,\\{e\\})$ and
$Q\in\mathcal{T}(\\{e\\},\emptyset)$ or $P\in\mathcal{T}(\\{e\\},\emptyset)$
and $Q\in\mathcal{T}(\emptyset,\\{e\\})$. Hence

(32) $\displaystyle{\mathfrak{V}}_{x_{e}=0}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}{\mathfrak{V}}^{e,(-,-)}&0\\\
0&{\mathfrak{V}}^{e,(+,+)}\end{array}\right).$

Combining (26) and (32) yields the claim. ∎

Now we are in position to state and prove the crucial factorization.

######  Proposition 4.0.

Let $E=\\{e_{1}\prec\cdots\prec e_{r}\\}$ be a fixed ordering. Then

${\mathfrak{V}}=\mathcal{M}^{e_{1}}\cdots\mathcal{M}^{e_{r}}.$

###### Proof.

We will prove by downward induction on $i$ that

(33) $\displaystyle{\mathfrak{V}}$ $\displaystyle=$
$\displaystyle{\mathfrak{V}}_{x_{i}=\cdots=x_{r}=0}\cdot\mathcal{M}^{e_{i}}\cdots\mathcal{M}^{e_{r}}.$

For $i=r$ the assertion follows directly from Lemma 4.2. For the inductive
step assume $i>1$ and (33) holds for $i$. We know from Lemma 4.2 that if we
choose a linear ordering on $E$ for which $e_{i-1}$ is the largest element
then

(34) $\displaystyle{\mathfrak{V}}$ $\displaystyle=$
$\displaystyle{\mathfrak{V}}_{x_{i-1}=0}\cdot\mathcal{N},$

where $\mathcal{N}=(N_{Q,R})_{Q,R\in\mathcal{T}}$ is defined as

$N_{Q,R}=\left\\{\begin{array}[]{ccc}1&\mbox{~{}if~{}}&Q=R\\\
-\mu((\hat{0},Q)_{R,e_{i-1}})\,{\mathfrak{V}}_{Q,R}&\mbox{~{}if~{}}&e_{i-1}\in\text{Sep}(Q,R)\\\
0&\mbox{~{}otherwise}&\end{array}\right.,$

Since $\mathcal{N}=\mathcal{M}^{e_{i-1}}$ for this particular ordering. Now we
go back to the ordering in the assumption and set $x_{i}=\cdots=x_{r}=0$ in
$\mathcal{N}$. We see that

$(N_{Q,R})_{x_{i}=\cdots=x_{r}=0}\left\\{{\begin{array}[]{cl}1&\mbox{~{}if~{}}Q=R\\\
-\mu((\hat{0},Q)_{R,e_{i-1}})\,{\mathfrak{V}}_{Q,R}&\mbox{~{}if~{}}e_{i-1}\mbox{~{}is
the largest element in~{}}S(Q,R)\\\ 0&\mbox{~{}otherwise}\end{array}}\right..$

But then $\mathcal{N}_{x_{i}=\cdots=x_{r}=0}=\mathcal{M}^{e_{i-1}}$.

Now (34) implies

$\displaystyle{\mathfrak{V}}_{x_{i}=\cdots=x_{r}=0}$ $\displaystyle=$
$\displaystyle{\mathfrak{V}}_{x_{i-1}=\cdots=x_{r}=0}\cdot\mathcal{N}_{x_{i}=\cdots=x_{r}=0}$
$\displaystyle=$
$\displaystyle{\mathfrak{V}}_{x_{i-1}=\cdots=x_{r}=0}\cdot\mathcal{M}^{e_{i-1}}$

With the induction hypothesis this completes the induction step by

$\displaystyle{\mathfrak{V}}$ $\displaystyle=$
$\displaystyle{\mathfrak{V}}_{x_{i}=\cdots=x_{r}=0}\cdot\mathcal{M}^{e_{i}}\cdots\mathcal{M}^{e_{r}}$
$\displaystyle=$
$\displaystyle{\mathfrak{V}}_{x_{i-1}=x_{i}=\cdots=x_{r}=0}\cdot\mathcal{M}^{e_{i-1}}\cdots\mathcal{M}^{e_{r}}.$

For $i=1$ the matrix ${\mathfrak{V}}_{x_{1}=\cdots=x_{r}=0}$ is the identity
matrix. Thus (33) yields:

${\mathfrak{V}}=\mathcal{M}^{e_{1}}\cdots\mathcal{M}^{e_{r}}.$

∎

Before we prove the following proposition, we quote [8, Corollary 3], which is
a result for oriented matroids.

######  Lemma 4.3.

Let $\mathbf{0}\in\mathcal{L}$ and let $P\in\mathcal{T}_{R,e}$ such that $e$
does not define a proper face of $P$. Then the Möbius number
$\mu((\hat{0},P)_{R,e})$ is $0$ if $-R\neq P$ and
$(-1)^{{\mathrm{rank}}(\mathcal{L})}$ if $-R=P$.

Now let $Y\in\mathcal{L}$ and $e\in z(Y)$ be the maximal element of $z(Y)$.
Define $\mathcal{T}^{Y,e}$ as the set of topes $P\in\mathcal{T}$ such that $Y$
is the maximal element of $\mathcal{L}$ for which $Y_{e}=0$ and $Y\leq P$.

######  Proposition 4.0.

For any pair of topes $Q,R\in\mathcal{T}^{Y,e}$ we have

$\displaystyle\mu((\hat{0},Q)_{R,e})=\left\\{\begin{array}[]{ccc}(-1)^{{\mathrm{rank}}(\mathcal{L}|_{z(Y)})}&\mbox{~{}if~{}}&Q_{z(Y)}=-R_{z(Y)}\\\
0&\mbox{~{}otherwise~{}}&\end{array}\right..$

###### Proof.

By the definition of $\mathcal{T}^{Y,e}$ we have $Y\leq Q,R$. Thus, if we
consider the poset $\mathcal{T}_{R|_{z(Y)},e}$ in the restriction
$\mathcal{L}|_{z(Y)}$ we find that the interval $(\hat{0},Q)_{R,e}$ is
isomorphic to $(\hat{0},Q|_{z(Y)})_{R|_{z(Y)},e}$. We saw in Note 2.0, that
$(E\backslash\underline{Y},F(Y)\backslash\underline{Y})$ is an OM. Further,
$\mathcal{T}_{R|_{z(Y)},e}$ is a poset and $(\hat{0},Q|_{z(Y)})_{R|_{z(Y)},e}$
is an interval in this particular OM. Furthermore, since $Y$ is the maximal
element satisfying $Y_{e}=0$ and $Y\leq Q$, $e$ does not define a proper face
of $Q|_{z(Y)}$. Since our interval is in an OM, we can use Lemma 4.3 and the
claim follows. ∎

We define $b_{Y,e}=0$ if $e$ is not the maximal element of $z(Y)$ and
$\frac{1}{2}\\#\mathcal{T}^{Y,e}$ otherwise. Since $P\mapsto Y\circ(-P)$ is a
perfect pairing on $\mathcal{T}^{Y,e}$ it follows that $\mathcal{T}^{Y,e}$
contains an even number of topes. In particular, $b_{Y,e}$ is a nonnegative
integer. We denote by $\mathcal{M}^{Y,e}$ the submatrix of $\mathcal{M}^{e}$
obtained by selecting rows and columns indexed by $\mathcal{T}^{Y,e}$.

######  Lemma 4.4.

Let $Y\in\mathcal{L}$ and $e\in z(Y)$. If $\mathcal{T}^{Y,e}\neq\emptyset$.
then

$\det(\mathcal{M}^{Y,e})=(1-a(Y))^{b_{Y,e}}$

where $a(Y):=\prod_{e\in z(Y)}x_{e^{+}}x_{e^{-}}$.

###### Proof.

If $Q_{z(Y)}=-R_{z(Y)}$ then ${\mathfrak{V}}_{Q,R}=\prod_{e\in
z(Y),Q_{e}=*}x_{e^{*}}$. Using the definition of $\mathcal{M}^{e}$ and
Proposition 4.0 we find

$\displaystyle\mathcal{M}^{Y,e}_{Q,R}=\left\\{\begin{array}[]{ccc}1&\mbox{~{}if~{}}&Q=R\\\
-(-1)^{{\mathrm{rank}}(\mathcal{L}|_{z(Y)})}\prod_{e\in
z(Y),Q_{e}=*}x_{e^{*}}&\mbox{~{}if~{}}&Q=Y\circ(-R)\\\ &&e\mbox{ largest
element of }S(Q,R)\\\ 0&\mbox{~{}otherwise~{}}&\end{array}\right..$

We order rows and columns of $\mathcal{M}^{Y,e}$ so that the elements $R$ and
$Y\circ(-R)$ are paired in consecutive rows and columns. With this ordering
$\mathcal{M}^{Y,e}$ is a block diagonal matrix having along its diagonal
$b_{Y,e}$ two by two matrices

$\left(\begin{array}[]{cc}1&-(-1)^{{\mathrm{rank}}(\mathcal{L}|_{z(Y)})}\prod_{e\in
z(Y),R_{e}=*}x_{e^{*}}\\\
-(-1)^{{\mathrm{rank}}(\mathcal{L}|_{z(Y)})}\prod_{e\in
z(Y),-R_{e}=*}x_{e^{*}}&1\end{array}\right)$

if $e$ is the maximal element of $z(Y)$ and identity matrices otherwise. In
any case we find $\det(\mathcal{M}^{Y,e})=(1-a(Y))^{b_{Y,e}}$ as desired. ∎

######  Lemma 4.5.

After suitably ordering $\mathcal{T}$ the matrix $\mathcal{M}^{e}$ is the
block lower triangular matrix with the matrices $\mathcal{M}^{Y,e}$ for
$Y\in\mathcal{L}$ with $Y_{e}=0$ and $\mathcal{T}^{Y,e}\neq\emptyset$ on the
main diagonal.

###### Proof.

If $Y=\mathbf{0}$, $\mathcal{T}^{Y,e}$ would be empty, so we assume that this
is not the case. We fix a linear ordering of $\mathcal{T}$ such that for fixed
$e\in E$ and $Y\in\mathcal{L}$ the topes from $\mathcal{T}^{Y,e}$ form an
interval and such that the topes from $\mathcal{T}^{Y,e}$ precede those of
$\mathcal{T}^{Y^{\prime},e}$ if $Y<Y^{\prime}$.

For this order the claim follows if we show that the entry
$(\mathcal{M}^{e})_{Q,R}$ is zero whenever $Q\in\mathcal{T}^{Y^{\prime},e}$,
$R\in\mathcal{T}^{Y,e}$ and $Y^{\prime}<Y$.

If $Q_{e}=R_{e}$ then by $Q\neq R$ we have $(\mathcal{M}^{e})_{Q,R}=0$. Hence
it suffices to consider the case $Q_{e}\neq R_{e}$. Since $Y\neq\mathbf{0}$,
$e$ is a proper face of $R$.

If $Q\not\in{\mathrm{star}}(Y),\,Q\in\mathcal{T}(\emptyset,\\{e\\})$ and
$R\in\mathcal{T}(\\{e\\},\emptyset)$ then it follows from Theorem 3.0 that
$\mu((\hat{0},Q)_{R,e})=0$ and therefore $(\mathcal{M}^{e})_{Q,R}=0$.
Analogously if
$Q\not\in{\mathrm{star}}(Y),\,Q\in\mathcal{T}(\\{e\\},\emptyset)$ and
$R\in\mathcal{T}(\emptyset,\\{e\\})$ then $\mu((\hat{0},Q)_{R,e})=0$ and
therefore $(\mathcal{M}^{e})_{Q,R}=0$.

On the other hand, if $Q\in{\mathrm{star}}(Y)$, then in particular $Y\leq Q$.
Since by definition of $\mathcal{T}^{Y^{\prime},e}$ we have that $Y^{\prime}$
is the maximal covector such that $Y^{\prime}\leq Q$ and $Y^{\prime}_{e}=0$ it
follows that $Y\leq Y^{\prime}$. Since $Y\neq Y^{\prime}$ we must have that
$Y<Y^{\prime}$, i.e. $(\mathcal{M}^{e})_{Q,R}$ is an entry below the diagonal
and we are done.

∎

###### Proof of Theorem 2.0.

After fixing a linear order on $E$ it follows from Proposition 4.0 that
$\det{\mathfrak{V}}$ is the product of the determinants of $\mathcal{M}^{e}$
for $e\in E$. By Lemma 4.5 the determinant of each $\mathcal{M}^{e}$ is a
product of determinants of $\mathcal{M}^{Y,e}$ for $e\in E$ and
$Y\in\mathcal{L}$ for which $\mathcal{T}^{Y,e}\neq\emptyset$. Then Lemma 4.4
completes the proof. ∎

######  Note 4.0 (Description of $b_{Y}$).

In our theorem we just describe $b_{Y}$ as a nonnegative integer, but this can
be made more precise. Fix any linear order on $E$ and let $e_{Y}$ be the
maximal element of $z(Y)$. From Lemma 4.4 we deduce, that $b_{Y}=b_{Y,e_{Y}}$.
Thus $2b_{Y}$ counts the topes $P\in\mathcal{T}$ such that $Y$ is the maximal
element of $\mathcal{L}$ for which $Y_{e_{Y}}=0$ and $Y\leq P$. Note in
particular, that as a consequence $b_{Y}$ does not depend on the choice of the
linear ordering on $E$.

##  5\. Applications

We give two applications of our formula for the Varchenko determinant on two
COMs associated to a poset $\mathcal{P}$, its lattice of ideals and its set of
linear extensions. As an example we will use the poset $\mathcal{Q}$ in Figure
1.

Figure 1. A poset $\mathcal{Q}$, its lattice $L(\mathcal{Q})$ of ideals and
its set $X(\mathcal{Q})$ of linear extensions. Edges in the graphs in the
middle and on the right are drawn if endpoints correspond to topes with
separator consisting of a single element. Edges corresponding to the same
element are parallel.

###  5.1. Distributive Lattices

By the Fundamental Theorem of Finite Distributive Lattices, for every
distributive lattice $L$ there exists a poset $\mathcal{P}$, such that
ordering the ideals (downward closed sets) of $\mathcal{P}$ by inclusion
yields a lattice isomorphic to $L$. The topes of the COM associated to $L$
correspond to the ideals of $\mathcal{P}$, the empty set can be seen as the
all-plus vector, the ground set $E$ of this COM is the ground set of
$\mathcal{P}$, and the separator of two ideals $I,I^{\prime}$ is the symmetric
difference $I\Delta I^{\prime}$. So this allows, to quickly write down the
(unsigned) Varchenko matrix $\mathbf{V}_{L}$ of $L$. In our example we
indicate $\mathbf{V}_{L}(\mathcal{P})$ in the following way, where we just
display the elements of the symmetric difference of two ideals to make it
easier to read. Note that in order to get the Varchenko matrix itself one has
to exchange a string $s_{1}\ldots s_{k}$ for the product $\prod_{s\in
S}x_{s}$. The $\emptyset$ translates therefore to the empty product, which is
1.

$\setcounter{MaxMatrixCols}{13}\begin{pmatrix}\emptyset&a&b&ab&be&abc&abd&abe&abcd&abce&abde&abcde\\\
a&\emptyset&ab&b&abe&bc&bd&be&bcd&bce&bde&bcde&\\\
b&ab&\emptyset&a&e&ac&ad&ae&acd&ace&ade&acde\\\
ab&b&a&\emptyset&ae&c&d&e&cd&ce&de&cde\\\
be&abe&e&ae&\emptyset&ace&ade&a&acde&ac&ad&acd\\\
abc&bc&ac&c&ace&\emptyset&cd&ce&d&e&cde&de\\\
abd&bd&ad&d&ade&cd&\emptyset&de&c&cde&e&ce\\\
abe&be&ae&e&a&ce&de&\emptyset&cde&c&d&cd\\\
abcd&bcd&acd&cd&acde&d&c&cde&\emptyset&de&ce&e\\\
abce&bce&ace&ce&ac&e&ced&c&de&\emptyset&cd&d\\\
abde&bde&ade&de&ad&cde&e&d&ce&cd&\emptyset&c\\\
abcde&bcde&acde&cde&acd&de&ce&cd&e&d&c&\emptyset\end{pmatrix}$

More generally, given two antichains $A^{\prime}\subseteq A$, the set of
ideals $\\{\downarrow A^{\prime\prime}\mid A^{\prime}\subseteq
A^{\prime\prime}\subseteq A\\}$ corresponds to the covector $Y(A^{\prime},A)$,
that is $0$ on $A\setminus A^{\prime}$, $-$ on all elements in or below
$A^{\prime}$, and $+$ on all elements above $A$. In particular, when
$A^{\prime}=A$ we get a tope corresponding to the ideal $\downarrow
A^{\prime}$ and the all $-$-tope corresponds to the empty ideal. Now, if we
pick a linear ordering on $E$, let $e_{Y}$ be the largest element of
$A\setminus A^{\prime}$, then $2b_{Y(A^{\prime},A)}$ counts those ideals $I$
such that for the antichain of maxima $\mathrm{Max}(I)$ we have

  * •

$A^{\prime}\subseteq\mathrm{Max}(I)\subseteq A$,

  * •

if $B^{\prime}\subseteq\mathrm{Max}(I)\subseteq B$ and $e_{Y}$ is the largest
element of $B\setminus B^{\prime}$, then $A\setminus A^{\prime}\subsetneq
B\setminus B^{\prime}$.

But note that this condition is only satisfied if $A^{\prime}=\mathrm{Max}(I)$
and $A=A^{\prime}\cup\\{e\\}$ for some $e\in\mathcal{P}$ or
$A=\mathrm{Max}(I)$ and $A^{\prime}=A\setminus\\{e\\}$ for some
$e\in\mathcal{P}$. Indeed, if otherwise
$A^{\prime}\subsetneq\mathrm{Max}(I)\subsetneq A$ and $e_{Y}$ is the largest
element of $A\setminus A^{\prime}$ one can set
$B^{\prime}=\mathrm{Max}(I)\setminus\\{e_{Y}\\}$ and
$B=\mathrm{Max}(I)\cup\\{e_{Y}\\}$, contradiction the above condition. Hence
$b_{Y(A^{\prime},A)}$ is $1$ if $|A\setminus A^{\prime}|=1$ and $0$ otherwise.
Thus, Corollary 2.0 and Note 4.0 yield that

$\displaystyle\det(\mathbf{V}_{L})=\prod_{A\in\mathcal{A}}\prod_{p\in
A}(1-x_{p}^{2})=\prod_{p\in\mathcal{P}}(1-x_{p}^{2})^{m_{p}},$

where $\mathcal{A}$ denotes the set of antichains of $\mathcal{P}$ and $m_{p}$
denotes the number of antichains containing $p$. In our example we get the
following formula for $\det(\mathbf{V}_{L(\mathcal{Q})})$:

$(1-x_{a}^{2})\cdot(1-x_{b}^{2})\cdot((1-x_{a}^{2})(1-x_{b}^{2}))\cdot(1-x_{e}^{2})\cdot(1-x_{c}^{2})\cdot(1-x_{d}^{2})\cdot((1-x_{a}^{2})(1-x_{e}^{2}))\cdot((1-x_{c}^{2})(1-x_{d}^{2}))\cdot((1-x_{c}^{2})(1-x_{e}^{2}))\cdot((1-x_{d}^{2})(1-x_{e}^{2}))\cdot((1-x_{c}^{2})(1-x_{d}^{2})(1-x_{e}^{2}))=$

$(1-x_{a}^{2})^{3}(1-x_{b}^{2})^{2}(1-x_{c}^{2})^{4}(1-x_{d}^{2})^{4}(1-x_{e}^{2})^{5}.$

###  5.2. Linear extensions

Another instance is the ranking COM of a poset $\mathcal{P}$, that was
described in [2]. The topes are the linear extensions of $\mathcal{P}$, and
the separator of two linear extensions $L,L^{\prime}$ is the set of pairs of
elements of $\mathcal{P}$ that are ordered differently in $L$ and
$L^{\prime}$. In particular, the ground set of this COM consists of the set
$\mathrm{Inc}(\mathcal{P})$ of incomparable pairs of $\mathcal{P}$, e.g.,
$\mathrm{Inc}(\mathcal{Q})=\\{ab,ae,cd,ce,de\\}$. We can thus define the
(unsigned) Varchenko matrix $\mathbf{V}_{X(\mathcal{P})}$. We get a
description of $\mathbf{V}_{X(\mathcal{Q})}$. We deem it too large to display
it entirely, but for example the entry corresponding to extensions
$abcde,beadc$ is $x_{ab}x_{ae}x_{cd}x_{ce}x_{de}$.

The covectors of the ranking COM are the weak extensions of $\mathcal{P}$,
i.e., those poset extensions of $\mathcal{P}$ that are chains of antichains.
The set $z(Y)$ of such an extension $Y$ corresponds to its set of incomparable
pairs $\mathrm{Inc}(Y)$. In order to properly define the signs of the
covectors, one can pick an arbitrary linear extension $L_{0}$ of
$\mathcal{P}$, and set at non-zero coordinate of $Y$ to $+$ if the
corresponding incomparable pair of $\mathcal{P}$ is ordered the same way in
$L_{0}$ and $Y$ and to $-$ otherwise. To define $b_{Y}$ we can fix an
arbitrary linear order on the set $\mathrm{Inc}(\mathcal{P})$ and let
$e_{Y}=\\{p,q\\}$ be the largest element of $\mathrm{Inc}(Y)$. Then $2b_{Y}$
counts linear extensions $L$ of $\mathcal{P}$ such that

  * •

$L$ is a linear extension of $Y$,

  * •

if another weak extension $Z$ of $\mathcal{P}$ has $e_{Y}$ as largest
incomparable pair, then either $L$ is not an extension of $Z$ or $Z$ is not an
extension of $Y$.

In this setting one can see that not such $Z$ can exist if and only if $Y$ is
a chain of antichains only one which - say $A$ \- has size larger than $1$. In
this cases the feasible $L$ are extensions of $Y$ that extend $A$ by starting
and ending with an element among $\\{p,q\\}$. Hence, there are $2(|A|-2)!$
such linear extensions. By Corollary 2.0 and Note 4.0 we have

$\displaystyle\det(\mathbf{V}_{\mathcal{P}})=\prod_{A\in\mathcal{A}_{\geq
2}}(1-\prod_{p\neq q\in A}x_{p,q}^{2})^{(|A|-2)!},$

where $\mathcal{A}_{\geq 2}$ denotes the set of antichains of size at least
$2$ of $\mathcal{P}$.

##  6\. Conclusion

One might wonder, to which extent our result could be further generalized to
other classes. A natural next class are partial cubes, i.e., isometric
subgraphs of the hypercube $Q_{d}$. These generalize (tope graphs of) COMs and
allow for ana analogous definition of the Varchenko Matrix, where the $(u,v)$
entry contains a product of monomials indexed by those coordinates in
$\\{1,\ldots,d\\}$ where $u$ and $v$ differ. The smallest partial cube that is
not the tope graph of a COM is the full subdivision of $K_{4}$. In this case
the Varchenko matrix looks like the following

$\displaystyle\scriptsize\left(\begin{array}[]{cccccccccc}1&x_{1}&x_{2}&x_{3}&x_{1}x_{4}&x_{1}x_{3}x_{4}&x_{3}x_{4}&x_{2}x_{3}x_{4}&x_{2}x_{4}&x_{1}x_{2}x_{4}\\\
x_{1}&1&x_{1}x_{2}&x_{1}x_{3}&x_{4}&x_{3}x_{4}&x_{1}x_{3}x_{4}&x_{1}x_{2}x_{3}x_{4}&x_{1}x_{2}x_{4}&x_{2}x_{4}\\\
x_{2}&x_{1}x_{2}&1&x_{2}x_{3}&x_{1}x_{2}x_{4}&x_{1}x_{2}x_{3}x_{4}&x_{2}x_{3}x_{4}&x_{3}x_{4}&x_{4}&x_{1}x_{4}\\\
x_{3}&x_{1}x_{3}&x_{2}x_{3}&1&x_{1}x_{3}x_{4}&x_{1}x_{4}&x_{4}&x_{2}x_{4}&x_{2}x_{3}x_{4}&x_{1}x_{2}x_{3}x_{4}\\\
x_{1}x_{4}&x_{4}&x_{1}x_{2}x_{4}&x_{1}x_{3}x_{4}&1&x_{3}&x_{1}x_{3}&x_{1}x_{2}x_{3}&x_{1}x_{2}&x_{2}\\\
x_{1}x_{3}x_{4}&x_{3}x_{4}&x_{1}x_{2}x_{3}x_{4}&x_{1}x_{4}&x_{3}&1&x_{1}&x_{1}x_{2}&x_{1}x_{2}x_{3}&x_{2}x_{3}\\\
x_{3}x_{4}&x_{1}x_{3}x_{4}&x_{2}x_{3}x_{4}&x_{4}&x_{1}x_{3}&x_{1}&1&x_{2}&x_{2}x_{3}&x_{1}x_{2}x_{3}\\\
x_{2}x_{3}x_{4}&x_{1}x_{2}x_{3}x_{4}&x_{3}x_{4}&x_{2}x_{4}&x_{1}x_{2}x_{3}&x_{1}x_{2}&x_{2}&1&x_{3}&x_{1}x_{3}\\\
x_{2}x_{4}&x_{1}x_{2}x_{4}&x_{4}&x_{2}x_{3}x_{4}&x_{1}x_{2}&x_{1}x_{2}x_{3}&x_{2}x_{3}&x_{3}&1&x_{1}\\\
x_{1}x_{2}x_{4}&x_{2}x_{4}&x_{1}x_{4}&x_{1}x_{2}x_{3}x_{4}&x_{2}&x_{2}x_{3}&x_{1}x_{2}x_{3}&x_{1}x_{3}&x_{1}&1\end{array}\right)$

and its determinant is of the following form:

$\displaystyle(x_{4}-1)^{3}(x_{4}+1)^{3}(x_{3}-1)^{3}(x_{3}+1)^{3}(x_{2}-1)^{3}(x_{2}+1)^{3}(x_{1}-1)^{3}(x_{1}+1)^{3}$
$\displaystyle(3x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{2}-x_{1}^{2}x_{2}^{2}x_{3}^{2}-x_{1}^{2}x_{2}^{2}x_{4}^{2}-x_{1}^{2}x_{3}^{2}x_{4}^{2}-x_{2}^{2}x_{3}^{2}x_{4}^{2}+1)$

Thus, in this case there is no nice factorization.

######  Problem 6.0.

Are there classes of partial cubes beyond COMs, that allow for a factorization
theorem of the Varchenko Matrix?

Hochstättler and Welker proved the factorization formula not only for the full
oriented matroid but also for supertopes, i.e. convex sets of topes in an
oriented matroid. The main motivation for the work in the present paper is
that COMs seem to capture convexity in oriented matroids. At the moment we do
not know an example of a COM which cannot be extended to become the supertope
of an oriented matroid. This supports the the suspicion that they might not
exist. Therefore we pose the following problem in our language, that is
equivalent to previous conjectures [2, Conjecture 1] and [9, Conjecture 1]:

######  Problem 6.0.

Are supertopes of oriented matroids a proper subclass of the class of
Complexes of Oriented Matroids?

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