This paper was converted on www.awesomepapers.org from LaTeX by an anonymous user.
Want to know more? Visit the Converter page.

The signed Varchenko Determinant for Complexes of Oriented Matroids

Winfried Hochstättler FernUniversität in Hagen
Fakultät für Mathematik und Informatik
58084 Hagen
Germany
winfried.hochstaettler@fernuni-hagen.de
Sophia Keip FernUniversität in Hagen
Fakultät für Mathematik und Informatik
58084 Hagen
Germany
sophia.keip@fernuni-hagen.de
 and  Kolja Knauer Departament de Matemàtiques i Informàtica, Universitat de Barcelona, Spain kolja.knauer@ub.edu
Abstract.

We generalize the (signed) Varchenko matrix of a hyperplane arrangement to complexes of oriented matroids and show that its determinant has a nice factorization. This extends previous results on hyperplane arrangements and oriented matroids.

1. Introduction

Varchenko [17] 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 𝒜\mathcal{A} over some ordered field 𝕂\mathbb{K}. The value of the product of the characteristic vectors of regions QiQ_{i} and QjQ_{j} is given by a product eS(Qi,Qj)we\prod_{e\in S(Q_{i},Q_{j})}w_{e}, where the wew_{e} are weights on the hyperplanes HeH_{e} of the arrangements and S(Qi,Qj)S(Q_{i},Q_{j}) is the set of hyperplanes that have to be crossed on a shortest path from QiQ_{i} to QjQ_{j}. The corresponding Varchenko Matrix B𝒜B_{\mathcal{A}} has entries of the form eS(Qi,Qj)we\prod_{e\in S(Q_{i},Q_{j})}w_{e} for any pair of regions QiQ_{i} and QjQ_{j}, also see Theorem 2.0. In order to determine when the bilinear form is degenerate, Varchenko [17] gave an elegant factorization of the determinant of that matrix, considering the weights as variables.

Theorem 1.0 (Varchenko 1993 [17]).

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

det(B𝒜)=FL(𝒜)(1wF2)mF\det(B_{\mathcal{A}})=\prod_{F\in L(\mathcal{A})}\left(1-w_{F}^{2}\right)^{m_{F}}

where wF=FHewew_{F}=\prod_{F\subset H_{e}}w_{e} and mFm_{F} are positive integers depending only on L(𝒜)L(\mathcal{A}).

After the original proof of Varchenko there were several approaches to provide cleaner proofs of this result. Denham and Henlon [6] sketched an elegant alternative way to prove the result. Gente [7] provided some more details for that proof and generalized their approach 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 [9] 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. In the signed case 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 [12] generalized their proof to oriented matroids and proved the factorization formula also for topoplane arrangements [13]. In his habilitation thesis he furthermore showed very recently that topoplane arrangements form a topological representation for complexes of oriented matroids (COMs) [14], leading to our result from a different perspective. Note that Varchenko and Brylawski [5] already generalized another, simpler bilinear form from hyperplane arrangements to matroids.

Bandelt et al. generalized oriented matroids to COMs 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]. Hochstättler and Welker proved the factorization formula not only for the full oriented matroid but also for supertopes, i.e. topal fibers in oriented matroids. Every supertope is a COM and it has even been conjectured in [2, Conjecture 1] and [10, Conjecture 1] that any COM can be realized as a supertope of an oriented matroid. Also see Problem 6.0. However, that conjecture seems out of reach at the moment. A big part of the motivation for the work in the present paper is to support that conjecture by explicitly proving the factorization formula for general COMs. The presentation as well as the proof follow the lines of Hochstättler and Welker [9]. We furthermore achieve a generalization to the signed version of the Varchenko matrix, thus generalizing Randriamaro [12].

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. We give some examples and applications in Section 5 and conclude the paper with some further remarks in Section 6.

2. The Varchenko Determinant and Complexes of Oriented Matroids

Before we introduce the Varchenko Determinant, we need to get familiar with COMs. COMs have been introduced in [2] as a common generalization of oriented matroids, affine oriented matroids, and lopsided sets. We will use the notation from [2] and [4]. Note that the symbols +,+,- and 0 act like 1,11,-1 and 0 when it comes to negation and multiplication. We start with the following definitions and axioms.

Definition 2.0.

We consider sign vectors on a finite ground set EE, i.e., elements of {0,+,}E\{0,+,-\}^{E}. The composition of two sign vectors XX and YY is defined as the sign-vector

(XY)e={Xe if Xe0,Ye if Xe=0eE.(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 reorientation of XX with respect to AEA\subseteq E is defined as the sign-vector

XeA={Xe if eA,Xe if eAeE.{}_{A}X_{e}=\begin{cases}-X_{e}&\text{ if }e\in A,\\ X_{e}&\text{ if }e\notin A\\ \end{cases}\forall e\in E.

The separator of XX and YY is defined as

S(X,Y)={eE:Xe=Ye0}.\displaystyle S(X,Y)=\{e\in E:X_{e}=-Y_{e}\neq 0\}.

The support of XX is defined as

X¯={eE:Xe0}.\displaystyle\underline{X}=\{e\in E:X_{e}\neq 0\}.

The zero-set of XX is defined as

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

For a set {0,+,}E\mathcal{L}\subseteq\{0,+,-\}^{E} we introduce five axioms:

(FS):

Face Symmetry

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

Strong Elimination

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

Composition

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

Zero

The all zeros vector 𝟎.\displaystyle\text{The all zeros vector }\mathbf{0}\in\mathcal{L}.
(Sym):

Symmetry

X:X.\displaystyle\forall X\in\mathcal{L}:-X\in\mathcal{L}.

Now we can define the term COM.

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

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

Let us first present OMs as special COMs.

Definition 2.0 (Oriented Matroid (OM)).

Let EE be a finite set and {0,+,}E\mathcal{L}\subseteq\{0,+,-\}^{E}. The pair =(E,){\mathcal{M}}=(E,\mathcal{L}) is called an OM, if it is a COM that satisfies (Z).

Remark 2.0.

Usually OMs are defined satisfying (C),(Sym),(SE). But note that (FS) implies (C). Indeed, by (FS) we first get XYX\circ-Y\in\mathcal{L} and then XY=(XX)Y=X(XY)X\circ Y=(X\circ-X)\circ Y=X\circ-(X\circ-Y)\in\mathcal{L} for all X,YX,Y\in\mathcal{L}. Further, (Z) together with (FS) clearly implies (Sym). Conversely, (Sym) and (C) imply (FS) while (Sym) and (SE) imply (Z).

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

eE:{XeX}={+,,0} and efE:{XeXfX}={+,,0}.\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 their collection is denoted by 𝒯{\mathcal{T}}.

The restriction of a sign-vector X{0,+,}EX\in\{0,+,-\}^{E} to E\AE\backslash A, AEA\subseteq E, denoted by X\A{0,+,}E\AX\backslash A\in\{0,+,-\}^{E\backslash A}, is defined by (X\A)e=Xe(X\backslash A)_{e}=X_{e} for all eE\Ae\in E\backslash A. We also write X|E\AX|_{E\backslash A}. The deletion of a COM is defined by (E\A,\A)(E\backslash A,\mathcal{L}\backslash A), where \A={X\A,X}\mathcal{L}\backslash A=\{X\backslash A,\;X\in\mathcal{L}\}, also written as |E\A\mathcal{L}|_{E\backslash A}.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a COM and S+,SES^{+},S^{-}\subseteq E such that there exists a tope T𝒯T\in{\mathcal{T}} where S+S^{+} is a subset of the positive elements of TT and SS^{-} is part of the negative elements of TT. The topal fiber ρ(S+,S)()\rho_{(S^{+},S^{-})}({\mathcal{L}}) is defined by the covectors {XX,Xe=+eS+,Xe=eS}\{X\mid X\in\mathcal{L},X_{e}=+\,\forall\,e\in S^{+},X_{e}=-\,\forall\,e\in S^{-}\}. We denote by 𝒯(S+,S){\mathcal{T}}(S^{+},S^{-}) the set of topes of ρ(S+,S)()\rho_{(S^{+},S^{-})}({\mathcal{L}}). Since all covectors of a topal fiber have the same entries on S+SS^{+}\cup S^{-}, we usually suppress the redundant coordinates in S+SS^{+}\cup S^{-}, to obtain a simple COM on the groundset E\(S+S)E\backslash(S^{+}\cup S^{-}). We will make use of the fact (shown in [2]) that the class of simple COMs is closed under deletion and under taking topal fibers.

For a covector XX\in\mathcal{L}, the set F(X)={XYY}F(X)=\{X\circ Y\mid Y\in\mathcal{L}\} is usually called the face of XX. We define star(X)={T𝒯|XT}{\mathrm{star}}(X)=\{T\in{\mathcal{T}}\,|\,X\leq T\}, where the componentwise ordering with respect to 0<+,0<+,- is used. Note that star(X)\X¯{\mathrm{star}}(X)\backslash\underline{X} is the set of topes of (E\X¯,F(X)\X¯)(E\backslash\underline{X},F(X)\backslash\underline{X}), which is well-known and easily seen to be an oriented matroid.

Let us look at a special OM which we will need in the next chapter.

Definition 2.0 (Graphic OM of a directed nn-cycle).

This OM has a ground set EE of size nn and its set of covectors 𝒞n\mathcal{C}_{n} consists of 𝟎\mathbf{0} and all compositions of sign-vectors from {0,+,}E\{0,+,-\}^{E} with exactly one positive and exactly one negative entry. Those generating sign-vectors are called the cocircuits of 𝒞n\mathcal{C}_{n}.

It can easily be checked that 𝒞n\mathcal{C}_{n} is the set of covectors of an OM. We use 𝒞3\mathcal{C}_{3} as an example:

Example 2.0 (Graphic OM of a directed triangle).

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

[Uncaptioned image]

The ground set EE of this OM corresponds to the three arcs. One gets the covectors of such an OM by looking at the sign vectors of directed cuts (indicated with dotted lines). These sign vectors are (+,,0)(+,-,0), (,+,0)(-,+,0), (+,0,)(+,0,-), (,0,+)(-,0,+), (0,+,)(0,+,-) and (0,,+)(0,-,+). Their compositions additionally yield the covectors (+,+,)(+,+,-), (+,,+)(+,-,+), (,+,+)(-,+,+), (,,+)(-,-,+), (,+,)(-,+,-) and (+,,)(+,-,-). We see that 𝒯{\mathcal{T}} consists of all full support sign vectors, except (+,+,+)(+,+,+) and (,,)(-,-,-).

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

Definition 2.0 (Signed Varchenko Matrix of a COM).

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a COM. The signed Varchenko matrix 𝔙{\mathfrak{V}} of \mathcal{M} is defined by a #𝒯×#𝒯\#{\mathcal{T}}\times\#{\mathcal{T}}-Matrix over

𝕂[xe|{+,},eE].\displaystyle\mathbb{K}[x_{e}^{*}\,|\,*\in\{+,-\},e\in E].

Its rows and columns are indexed by the topes 𝒯{\mathcal{T}} in a fixed linear order. For P,Q𝒯P,Q\in{\mathcal{T}}

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

Note that the diagonal entries 𝔙P,P{\mathfrak{V}}_{P,P} of the matrix are equal to 1. Let us illustrate this definition with the graphic OM of a directed triangle.

Example 2.0 (continued).

The signed Varchenko Matrix of the graphic OM of a directed triangle is

𝔙=(1x2+x3x1+x3x1+x2+x3x1+x2+x2x3+1x1+x2x1+x1+x2x3+x3+x1x3+x1x2+1x2+x3+x1x2+x3+x1x2x3+x1x21x2x3+x1x3+x1x1x2+x3x3x2+x31x1x2+x2x3x1+x2x3x1+x3x1+x21)\displaystyle{\mathfrak{V}}=\left(\begin{array}[]{cccccc}1&x_{2}^{+}x_{3}^{-}&x_{1}^{+}x_{3}^{-}&x_{1}^{+}x_{2}^{+}x_{3}^{-}&x_{1}^{+}&x_{2}^{+}\\ x_{2}^{-}x_{3}^{+}&1&x_{1}^{+}x_{2}^{-}&x_{1}^{+}&x_{1}^{+}x_{2}^{-}x_{3}^{+}&x_{3}^{+}\\ x_{1}^{-}x_{3}^{+}&x_{1}^{-}x_{2}^{+}&1&x_{2}^{+}&x_{3}^{+}&x_{1}^{-}x_{2}^{+}x_{3}^{+}\\ x_{1}^{-}x_{2}^{-}x_{3}^{+}&x_{1}^{-}&x_{2}^{-}&1&x_{2}^{-}x_{3}^{+}&x_{1}^{-}x_{3}^{+}\\ x_{1}^{-}&x_{1}^{-}x_{2}^{+}x_{3}^{-}&x_{3}^{-}&x_{2}^{+}x_{3}^{-}&1&x_{1}^{-}x_{2}^{+}\\ x_{2}^{-}&x_{3}^{-}&x_{1}^{+}x_{2}^{-}x_{3}^{-}&x_{1}^{+}x_{3}^{-}&x_{1}^{+}x_{2}^{-}&1\end{array}\right)

In this work we will prove the following theorem.

Theorem 2.0.

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

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

where a(Y):=ez(Y)xe+xea(Y):=\prod_{e\in z(Y)}x_{e}^{+}x_{e}^{-} and bYb_{Y} are nonnegative integers that can be explicitly computed, see Remark 4.0.

Example 2.0 (continued).

For our example the determinant of the signed Varchenko matrix factorizes to

det(𝔙)=(1x1+x1)2(1x2+x2)2(1x3+x3)2(1x1+x1x2+x2x3+x3).\displaystyle\det({\mathfrak{V}})=(1-x_{1}^{+}x_{1}^{-})^{2}(1-x_{2}^{+}x_{2}^{-})^{2}(1-x_{3}^{+}x_{3}^{-})^{2}(1-x_{1}^{+}x_{1}^{-}x_{2}^{+}x_{2}^{-}x_{3}^{+}x_{3}^{-}).

A corollary of this result, namely the case where xe=xe+x_{e}^{-}=x_{e}^{+}, which is the original version of the Varchenko matrix, has been already proven for OMs in [9]. We formulate it for COMs.

Corollary 2.0.

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

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

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

Example 2.0 (continued).

For our example the determinant of the (unsigned) Varchenko matrix factorizes to

det(𝔙)=(1x12)2(1x22)2(1x32)2(1x12x22x32).\displaystyle\det({\mathfrak{V}})=(1-x_{1}^{2})^{2}(1-x_{2}^{2})^{2}(1-x_{3}^{2})^{2}(1-x_{1}^{2}x_{2}^{2}x_{3}^{2}).

3. Preparation

We start with some basics about partially ordered sets 𝒫\mathcal{P} (posets). For an introduction we recommend [18]. One can associate an abstract simplicial complex Δ(𝒫)\Delta(\mathcal{P}), called order complex, to every poset. The elements of 𝒫\mathcal{P} are the vertices of this complex and the chains (i.e. totally ordered subsets) the faces. Two posets are homotopy equivalent if their order complexes are homotopy equivalent. 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 maximal chain and consequently in every maximal face of the order complex. For details see [18]. We introduce now the Möbius function μ\mu of a poset:

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

The bounded extension 𝒫^\hat{\mathcal{P}} of a poset is the poset together with a new maximal element 1^\hat{1} and a new minimal element 0^\hat{0}. The Möbius number of 𝒫\mathcal{P} is defined by

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

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

Example 3.0.

Let us look at the poset 𝒫\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 are depicted.

[Uncaptioned image]

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

μ(𝒫)=μ(0^,1^)=1+(1)=0.\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.

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

μ(𝒫)=χ(Δ(𝒫))1.\displaystyle\mu(\mathcal{P})=\chi(\Delta(\mathcal{P}))-1.

In particular we get the following corollary, whose second part follows from the definition of contractability and Example 3.0.

Corollary 3.0.

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

For a poset 𝒫\mathcal{P} and p𝒫p\in\mathcal{P} we denote by 𝒫p\mathcal{P}_{\leq p} the subposet {q𝒫|qp}\{q\in\mathcal{P}~|~q\leq p\}.

Proposition 3.0.

[11, Quillen Fiber Lemma] Let 𝒫{\mathcal{P}} and 𝒬\mathcal{Q} be posets and f:𝒫𝒬f:{\mathcal{P}}\rightarrow\mathcal{Q} order preserving. If for all qQq\in Q we have that f1(𝒬q)f^{-1}(\mathcal{Q}_{\leq q}) is contractible, then 𝒫{\mathcal{P}} and 𝒬\mathcal{Q} are homotopy equivalent.

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

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

We write 𝒯R{\mathcal{T}}_{R} if we consider 𝒯{\mathcal{T}} with this partial order and we call RR the base pattern of the poset.

Now we will introduce a theorem which will help us with our crucial Lemma 3.3.

Theorem 3.0.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a topal fiber of a COM =(E,){\mathcal{M}}^{\prime}=(E^{\prime},\mathcal{L^{\prime}}), R𝒯R^{\prime}\in{\mathcal{T}}^{\prime} a tope of {\mathcal{M}}^{\prime} and R=R|ER=R^{\prime}_{|E} its restriction to EE. Then the order complex of 𝒯R{\mathcal{T}}_{R} is contractible.

Note that the restriction RR in the statement of Theorem 3.0 is not necessarily a tope of {\mathcal{M}}. In order to apply the Quillen Fiber Lemma in the proof of Theorem 3.0 we need the following lemma.

Lemma 3.1.

Let fEf\in E and R={+}ER=\{+\}^{E}. Let 𝒯f{\mathcal{T}}\setminus f denote the set of topes of f{\mathcal{M}}\setminus f and 𝒯fRf{\mathcal{T}}\hskip-2.84544pt\setminus\hskip-2.84544ptf_{R\setminus f} the corresponding tope poset with base pattern R{f}R\setminus\{f\}. Consider the order-preserving map πf:𝒯R𝒯fRf\pi_{f}:{\mathcal{T}}_{R}\to{\mathcal{T}}\hskip-2.84544pt\setminus\hskip-2.84544ptf_{R\setminus f} given by restriction. Let Q𝒯fQ\in{\mathcal{T}}\hskip-2.84544pt\setminus\hskip-2.84544ptf. Then

πf1((𝒯fRf)Q)=𝒯(Q+,).\pi_{f}^{-1}(({\mathcal{T}}\hskip-2.84544pt\setminus\hskip-2.84544ptf_{R\setminus f})_{\preceq Q})={\mathcal{T}}(Q^{+},\emptyset).
Proof.

Let Q~(𝒯fR\f)Q\tilde{Q}\in({\mathcal{T}}\setminus f_{R\backslash f})_{\preceq Q}. As RfR\setminus f is all positive, we must have Q~Q\tilde{Q}^{-}\subseteq Q^{-} and hence Q+Q~+Q^{+}\subseteq\tilde{Q}^{+} implying πf1(Q~)𝒯(Q+,)\pi_{f}^{-1}(\tilde{Q})\subseteq{\mathcal{T}}(Q^{+},\emptyset). If on the other hand Q^𝒯(Q+,)\hat{Q}\in{\mathcal{T}}(Q^{+},\emptyset), then Q^Q{f},\hat{Q}^{-}\subseteq Q^{-}\cup\{f\}, Q+Q^+{f}Q^{+}\subseteq\hat{Q}^{+}\cup\{f\}. Hence πf(Q^)RfQ\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 [8, Lemma 10]. We reprove that result here, since in the presentation in [8] the signs are chosen the opposite way. Recall from Example 2.0 that 𝒞n\mathcal{C}_{n} is the set of covectors of the OM of the directed cycle on nn vertices.

Proposition 3.0.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}), 𝒞n\mathcal{L}\neq\mathcal{C}_{n}, be a COM with tope set 𝒯{\mathcal{T}} and let R={+}ER=\{+\}^{E}. If for all fEf\in E we have fR𝒯-_{f}R\in{\mathcal{T}}, then the poset 𝒯R{\mathcal{T}}_{R} is contractible.

Proof.

We will show by induction that if R𝒯-R\notin\mathcal{T}, then 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 of 𝒞n\mathcal{C}_{n}) exist in \mathcal{L}, we get by (SE) that the all zero vector is in \mathcal{L}. Together, we can conclude that 𝒞n\mathcal{C}_{n}\subseteq\mathcal{L}, since we obtain all its covectors by composition of those vectors. Since 𝒞n\mathcal{C}_{n} is uniform and contains 𝟎\mathbf{0}, every other COM containing it has to be a OM which is free. This means that every possible sign-vector vector is a covector and is a consequence of the strong elimination axiom. Therefore 𝒞n=\mathcal{C}_{n}=\mathcal{L}, since R𝒯-R\notin\mathcal{T}. Taking the contraposition we see that if 𝒞n\mathcal{L}\neq\mathcal{C}_{n}, then R𝒯-R\in{\mathcal{T}}, so the poset 𝒯R{\mathcal{T}}_{R} has a unique maximal element. In particular, it is contractible.

So let fR𝒯-_{f}R\in{\mathcal{T}} for all fEf\in E and R𝒯-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,,|E|2n=0,\dots,\lvert E\rvert-2 all sign-vectors with nn zero entries, one plus-entry and |E|(n+1)\lvert E\rvert-(n+1) minus-entries are in \mathcal{L}.

n=0n=0: By the existence of fR-_{f}R there is nothing to show. We fix n0n\geq 0 and assume that all covectors with nn or fewer zero-entries, exactly one plus-entry and at least one minus-entry exist in \mathcal{L}.

nn+1|E|2n\rightarrow n+1\leq\lvert E\rvert-2: We will show that there exists a covector with zero-entries in the i-th position, iIEi\in I\subset E, |I|=n+1\lvert I\rvert=n+1, a plus-entry in the j-th position, jIj\notin I and - everywhere else. We choose an i^I\hat{i}\in I and consider two covectors with 0 in I\i^I\backslash\hat{i}, where one has a ++ in the i^\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:

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

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

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

If * was -, then XjR={}EX\circ-_{j}R=\{-\}^{E}. Since {}E=R𝒯\{-\}^{E}=-R\notin{\mathcal{T}} we have =+*=+ and have the covector we were looking for, so =𝒞n\mathcal{L}=\mathcal{C}_{n}.

Lemma 3.2.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a topal fiber of a COM =(E,){\mathcal{M}}^{\prime}=(E^{\prime},\mathcal{L^{\prime}}), R𝒯R^{\prime}\in{\mathcal{T}}^{\prime} a tope of {\mathcal{M}}^{\prime} and R=R|ER=R^{\prime}_{|E} its restriction to EE. If =𝒞n\mathcal{L}=\mathcal{C}_{n}, then RR\in\mathcal{L}, in particular R{+}E,{}ER\neq\{+\}^{E},\{-\}^{E}.

Proof.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a COM such that there is a COM =(E,){\mathcal{M}}=(E^{\prime},\mathcal{L^{\prime}}), with EEE\subset E^{\prime} and =ρ(S+,S)()\mathcal{L}=\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime}) for some S+,SES^{+},S^{-}\subseteq E^{\prime} and =𝒞n\mathcal{L}=\mathcal{C}_{n}. We saw in Definition 2.0 that 𝟎𝒞n\mathbf{0}\in\mathcal{C}_{n}. By the definition of ρ(S+,S)()\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime}) there exists ZZ\in\mathcal{L}^{\prime} with

Ze={+ if eS+, if eS0 else. Z_{e}=\begin{cases}+&\text{ if }e\in S^{+},\\ -&\text{ if }e\in S^{-}\\ 0&\text{ else. }\end{cases}

Since the composition of ZZ with every other covector in \mathcal{L}^{\prime} is in \mathcal{L}^{\prime} (see Remark 2.0), we see that ρ(S+,S)()=\{S+S}\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime})=\mathcal{L}^{\prime}\backslash\{S^{+}\cup S^{-}\}. So in this case \{S+S}=𝒞n\mathcal{L}^{\prime}\backslash\{S^{+}\cup S^{-}\}=\mathcal{C}_{n}, so every tope RR^{\prime}\in\mathcal{L}^{\prime} restricted to EE has to be in 𝒞n\mathcal{C}_{n}. Since {+}E,{}E𝒞n\{+\}^{E},\{-\}^{E}\notin\mathcal{C}_{n}, R=R|E{+}E,{}ER=R^{\prime}|_{E}\neq\{+\}^{E},\{-\}^{E}.

Now we are in position to prove Theorem 3.0.

Proof of Theorem 3.0.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a COM such that there is a COM =(E,){\mathcal{M}}^{\prime}=(E^{\prime},\mathcal{L^{\prime}}), with EEE\subset E^{\prime} and =ρ(S+,S)()\mathcal{L}=\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime}) for some S+,SES^{+},S^{-}\subseteq E^{\prime}, R𝒯R^{\prime}\in{\mathcal{T}}^{\prime} a tope of {\mathcal{M}}^{\prime} and R=R|ER=R^{\prime}_{|E} its restriction to EE. First we look at the case =𝒞n\mathcal{L}=\mathcal{C}_{n}. From Lemma 3.2 we know, RR is a tope of 𝒞n\mathcal{C}_{n}. So we have a unique minimal element and 𝒯R{\mathcal{T}}_{R} is contractible. Now let 𝒞n\mathcal{L}\neq\mathcal{C}_{n}. Possibly reorienting elements we may assume that R={+}ER=\{+\}^{E}. We proceed by induction on |E||E|. If |E|=1|E|=1 then 𝒯R{\mathcal{T}}_{R} either is a singleton or a chain of length 2 and thus contractible. Hence assume |E|2|E|\geq 2. If for all fEf\in E there exists fR-_{f}R as in Proposition 3.0, then 𝒯R{\mathcal{T}}_{R} is contractible by Proposition 3.0. Hence we may assume that there exists fEf\in E such that fR𝒯-_{f}R\not\in{\mathcal{T}}. Let (𝒯f)Rf({\mathcal{T}}\hskip-2.84544pt\setminus\hskip-2.84544ptf)_{R\setminus f} denote the tope poset in f{\mathcal{L}}\setminus f with base pattern RfR\setminus f. Since the class of COMs is closed under deletion, we know that f{\mathcal{L}}^{\prime}\setminus f is a COM. Since \mathcal{L} evolved from \mathcal{L}^{\prime} by setting =ρ(S+,S){\mathcal{L}}=\rho_{(S^{+},S^{-})}{\mathcal{L}}^{\prime} for some S+S^{+} and SS^{-}, f{\mathcal{L}}\setminus f evolves in the same way from f{\mathcal{L}}^{\prime}\setminus f, i.e. f=ρ(S+,S)(f){\mathcal{L}}\setminus f=\rho_{(S^{+},S^{-})}({\mathcal{L}}^{\prime}\setminus f) (note that ff cannot be in S+SS^{+}\cup S^{-}, since E=E(S+S)E=E^{\prime}\setminus(S^{+}\cup S^{-}) and fEf\in E). Also RfR\setminus f is the restriction to EE of the tope RfR^{\prime}\setminus f of f{\mathcal{L}}^{\prime}\setminus f. We see that f{\mathcal{L}}\setminus f together with RfR\setminus f fulfills the assumptions of Theorem 3.0. Furthermore, f𝒞n{\mathcal{L}}\setminus f\neq\mathcal{C}_{n}, this follows from Lemma 3.2. Hence, (𝒯f)Rf({\mathcal{T}}\hskip-2.84544pt\setminus\hskip-2.84544ptf)_{R\setminus f} is contractible by inductive assumption. We now want to show that 𝒯R{\mathcal{T}}_{R} and (𝒯f)Rf({\mathcal{T}}\setminus f)_{R\setminus f} are homotopy equivalent by using Proposition 3.0. So consider the order-preserving map πf:𝒯R(𝒯f)Rf\pi_{f}:{\mathcal{T}}_{R}\to({\mathcal{T}}\hskip-2.84544pt\setminus\hskip-2.84544ptf)_{R\setminus f} given by restriction. Let Q(𝒯f)RfQ\in({\mathcal{T}}\hskip-2.84544pt\setminus\hskip-2.84544ptf)_{R\setminus f}. By Lemma 3.1

πf1(((𝒯f)Rf)Q)=𝒯(Q+,).\pi_{f}^{-1}((({\mathcal{T}}\setminus f)_{R\setminus f})_{\preceq Q})={\mathcal{T}}(Q^{+},\emptyset).

𝒯(Q+,){\mathcal{T}}(Q^{+},\emptyset) is the set of topes of ρ(S+Q+,S)\rho_{(S^{+}\cup Q^{+},S^{-})}{\mathcal{L}}^{\prime}. If Q+Q^{+}\neq\emptyset, then ρ(S+Q+,S)\rho_{(S^{+}\cup Q^{+},S^{-})}{\mathcal{L}}^{\prime} has fewer elements than \mathcal{L}. Furthermore, by Lemma 3.2, ρ(S+Q+,S)𝒞n\rho_{(S^{+}\cup Q^{+},S^{-})}{\mathcal{L}}^{\prime}\neq\mathcal{C}_{n}. Hence πf1((𝒯f)Q)\pi_{f}^{-1}(({\mathcal{T}}\setminus f)_{\preceq Q}) is contractible by inductive assumption. If Q+=Q^{+}=\emptyset then by the choice of ff the preimage πf1(Q)\pi_{f}^{-1}(Q) is the all minus vector. Hence, this is the unique maximal element in πf1((𝒯f)Q)\pi_{f}^{-1}(({\mathcal{T}}\setminus f)_{\preceq Q}) and that fiber is also contractible. So by Proposition 3.0 𝒯R{\mathcal{T}}_{R} and (𝒯f)Rf({\mathcal{T}}\setminus f)_{R\setminus f} are homotopy equivalent and the claim follows. ∎

We now introduce 𝒯R,e{\mathcal{T}}_{R,e}, which is a truncated version of 𝒯R{\mathcal{T}}_{R}. For eEe\in E and R𝒯R\in{\mathcal{T}} we define 𝒯R,e{\mathcal{T}}_{R,e} as the poset {T𝒯|Te=Re}{0^}\{T\in{\mathcal{T}}~|~T_{e}=-R_{e}\}\cup\{\hat{0}\} with 0^\hat{0} as its artificial least element and the remaining poset structure induced from 𝒯R{\mathcal{T}}_{R}. For P𝒯R,eP\in{\mathcal{T}}_{R,e} we write (0^,P)R,e(\hat{0},P)_{R,e} for the interval from 0^\hat{0} to PP in 𝒯R,e{\mathcal{T}}_{R,e}. The following result will help us later in the main proof to obtain a factorization of the Varchenko matrix.

Lemma 3.3.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a COM, R𝒯R\in{\mathcal{T}} a tope, eEe\in E an element, 0^P𝒯R,e\hat{0}\neq P\in{\mathcal{T}}_{R,e} and SS such that eSEe\not\in S\subseteq E. Then

(3) Q𝒯(,{e})S=S(P,Q)S(Q,R)μ((0^,Q)R,e)={1 if S=0 if S\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) Q𝒯({e},)S=S(P,Q)S(Q,R)μ((0^,Q)R,e)={1 if S=0 if S.\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={+}ER=\{+\}^{E}. We prove the assertion by induction on |S||S|. If S=S=\emptyset then

Q𝒯(,{e})S=S(P,Q)S(Q,R)μ((0^,Q)R,e)\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= 0^<R,eQR,ePμ((0^,Q)R,e)\displaystyle\sum_{\hat{0}<_{R,e}Q\leq_{R,e}P}\mu((\hat{0},Q)_{R,e})

Note that 0^<R,eQR,ePμ((0^,Q)R,e)=μ({Q𝒯R,e|0^<R,eQR,eP})μ((0^,0^)R,e)\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}<_{R,e}Q\leq_{R,e}P\}\right)-\mu((\hat{0},\hat{0})_{R,e}). The poset {Q𝒯R,e|0^<R,eQR,eP}\{Q\in{\mathcal{T}}_{R,e}\,|\,\hat{0}<_{R,e}Q\leq_{R,e}P\} has the maximal element PP, so it is contractible and has Möbius number 0. Therefore we have

Q𝒯(,{e})S=S(P,Q)S(Q,R)μ((0^,Q)R,e)=μ((0^,0^)R,e)=1.\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|S|>0. Set

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

Then

(7) Q𝒯(,{e})S(P,Q)S(Q,R)Sμ((0^,Q)R,e)\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= Q𝒯(T+,{e})μ((0^,Q)R,e).\displaystyle\sum_{Q\in{\mathcal{T}}(T^{+},\{e\})}\mu((\hat{0},Q)_{R,e}).

𝒯(T+,{e}){\mathcal{T}}(T^{+},\{e\}) is the set of topes of the COM ρ(T+,{e})\rho_{(T^{+},\{e\})}{\mathcal{L}}, where the associated poset 𝒯(T+,{e})R{\mathcal{T}}(T^{+},\{e\})_{R} is contractible by Theorem 3.0 (note that we did not suppress the redundant coordinates (T+{e})(T^{+}\cup\{e\}) here to keep notation simple). Since the right hand side of (7) ranges over the elements of a contractible poset, we have again

Q𝒯(T+,{e})μ((0^,Q)R,e)=μ(𝒯(T+,{e})R)μ(0^,0^)=μ(0^,0^)=1.\displaystyle\sum_{Q\in{\mathcal{T}}(T^{+},\{e\})}\mu((\hat{0},Q)_{R,e})=-\mu({\mathcal{T}}(T^{+},\{e\})_{R})-\mu(\hat{0},\hat{0})=-\mu(\hat{0},\hat{0})=-1.

and showed

(8) Q𝒯(,{e})S(P,Q)S(Q,R)Sμ((0^,Q)R,e)\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= 1.\displaystyle-1.

Now rewrite the left hand side of (8) as

(9) Q𝒯(,{e})S(P,Q)S(Q,R)Sμ((0^,Q)R,e)\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= TSQ𝒯(,{e})S(P,Q)S(Q,R)=Tμ((0^,Q)R,e)\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 Q𝒯(,{e})S(P,Q)S(Q,R)=Tμ((0^,Q)R,e)\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 TS,T\neq S,\emptyset and 1-1 for T=T=\emptyset. Thus combining (8) and (9) we obtain:

1\displaystyle-1 =\displaystyle= Q𝒯(,{e})S(P,Q)S(Q,R)Sμ((0^,Q)R,e)\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= 1+Q𝒯(,{e})S(P,Q)S(Q,R)=Sμ((0^,Q)R,e)\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

Q𝒯(,{e})S(P,Q)S(Q,R)=Sμ((0^,Q)R,e)=0.\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. ∎

We conclude this section with another result on contractability needed in the proof of our main theorem. We start with a lemma:

Lemma 3.4.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a COM, XX\in\mathcal{L} and P𝒯P\in{\mathcal{T}}. The tope Q=XPstar(X)Q=X\circ P\in{\mathrm{star}}(X) is the only tope in star(X){\mathrm{star}}(X) such that for all Ostar(X)O\in{\mathrm{star}}(X) we have

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

In the case where XX is a tope there is nothing to show, so we assume z(X)z(X)\neq\emptyset. It is easy to see that QQ fulfills (10) and (11). Let us assume there is another tope QQQ^{*}\neq Q in star(X){\mathrm{star}}(X) which has this property. By the definition of QQ we have

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

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

For eEe\in E and P𝒯P\in{\mathcal{T}} we say that ee defines a proper face of PP if there is a covector XX\in\mathcal{L} with XPX\leq P and Xe=0X_{e}=0 with X𝟎X\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 ee does not define a proper face of PP.

Theorem 3.0.

Let =(E,){\mathcal{M}}=(E,\mathcal{L}) be a COM, R𝒯R\in{\mathcal{T}} a tope, and let eEe\in E define a proper face of RR. Let YY\in\mathcal{L} be the maximal covector such that YRY\leq R and Ye=0Y_{e}=0 and choose Ptop𝒯R,estar(Y)P_{top}\in{\mathcal{T}}_{R,e}\setminus{\mathrm{star}}(Y). Then (0^,Ptop)R,e(\hat{0},P_{top})_{R,e} is contractible. In particular, μ((0^,Ptop)R,e)=0\mu((\hat{0},P_{top})_{R,e})=0.

Proof.

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

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

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

Y:(0^,Ptop)R,e(0^,Ptop)R,e\displaystyle\circ_{Y}:(\hat{0},P_{top})_{R,e}\rightarrow(\hat{0},P_{top})_{R,e}
Y(P)=YP\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. Y(Y(P))=Y(P)\circ_{Y}(\circ_{Y}(P))=\circ_{Y}(P)). So let QRQQ\preceq_{R}Q^{\prime}. Then YQRYQY\circ Q\preceq_{R}Y\circ Q^{\prime}. Since YRY\leq R it follows that YQRQY\circ Q\preceq_{R}Q. Obviously Y(YQ)=YQY\circ(Y\circ Q)=Y\circ Q. So Y\circ_{Y} is a closure operator and it follows that (0^,Ptop)R,e(\hat{0},P_{top})_{R,e} is homotopy equivalent to its image (see e.g, [3, Corollary 10.12]).

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

4. Main Proof

In this Section we assume that =(E,){\mathcal{M}}=(E,\mathcal{L}) is a COM with topes 𝒯{\mathcal{T}} and signed Varchenko matrix 𝔙{\mathfrak{V}}. Recall that we assume 𝒯{\mathcal{T}} to be linearly ordered. Note however that swapping two topes leads to a row swap and a column 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 𝒯{\mathcal{T}}, whenever it is convenient for the proof. Moreover, for the proof we also fix a linear ordering on EE, i.e., E={e1er}E=\{e_{1}\prec\cdots\prec e_{r}\}.

For any sign vector ϵ=(ϵ1,ϵ2){+,}2\epsilon=(\epsilon_{1},\epsilon_{2})\in\{+,-\}^{2} let 𝔙e,ϵ{\mathfrak{V}}^{e,\epsilon} be a matrix with rows indexed by 𝒯({e},){\mathcal{T}}(\{e\},\emptyset) for ϵ1=+\epsilon_{1}=+, 𝒯(,{e}){\mathcal{T}}(\emptyset,\{e\}) for ϵ1=\epsilon_{1}=- and columns indexed by 𝒯({e},){\mathcal{T}}(\{e\},\emptyset) for ϵ2=+\epsilon_{2}=+, 𝒯(,{e}){\mathcal{T}}(\emptyset,\{e\}) for ϵ2=\epsilon_{2}=-. For a tope RR indexing a row and a tope QQ indexing a column we set 𝔙R,Qe,ϵ=𝔙R,Q{\mathfrak{V}}^{e,\epsilon}_{R,Q}={\mathfrak{V}}_{R,Q}. After reordering 𝒯{\mathcal{T}} this yields a block decomposition of 𝔙{\mathfrak{V}} as

(14) 𝔙=(𝔙e,(,)𝔙e,(,+)𝔙e,(+,)𝔙e,(+,+)).\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 𝒯{\mathcal{T}} and set e\mathcal{M}^{e} to

Q,Re={1 if Q=Rμ((0^,Q)R,e)𝔙Q,R if e is the maximal element of S(Q,R),0 otherwise.\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

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

where

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

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

Lemma 4.1.

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

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

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

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

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

Q𝒯(,{e})𝔙P,QUQ,Re\displaystyle\sum_{Q\in{\mathcal{T}}(\emptyset,\{e\})}{\mathfrak{V}}_{P,Q}\cdot U^{e}_{Q,R} =\displaystyle= Q𝒯(,{e})μ((0^,Q)R,e)𝔙P,Q𝔙Q,R\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 ee is the maximal element of any separator of the topes indexing UeU^{e}. By definition we have for Q𝒯(,{e})Q\in{\mathcal{T}}(\emptyset,\{e\})

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

We see that 𝔙P,Q𝔙Q,R=𝔙P,R{\mathfrak{V}}_{P,Q}\cdot{\mathfrak{V}}_{Q,R}={\mathfrak{V}}_{P,R} if S(P,Q)S(Q,R)=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 SES\subseteq E and fixed P,RP,R we have:

(19) Q𝒯(,{e})S=S(P,Q)S(Q,R)μ((0^,Q)R,e)={0 if S1 otherwise..\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

Q𝒯({e},)𝔙P,QLQ,Re\displaystyle\sum_{Q\in{\mathcal{T}}(\{e\},\emptyset)}{\mathfrak{V}}_{P,Q}\cdot L^{e}_{Q,R} =\displaystyle= Q𝒯({e},)μ((0^,Q)R,e)𝔙P,Q𝔙Q,R\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 e\mathcal{M}^{e} to factorize 𝔙{\mathfrak{V}}. The following lemma yields the base case for the inductive step in the factorization.

Lemma 4.2.

Let ee be the maximal element of EE and let 𝔙xe=0{\mathfrak{V}}_{x_{e}=0} be the matrix 𝔙{\mathfrak{V}} after evaluating xe+x_{e}^{+} and xex_{e}^{-} to 0. Then

𝔙=𝔙xe=0e{\mathfrak{V}}={\mathfrak{V}}_{x_{e}=0}\cdot\mathcal{M}^{e}
Proof.

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

(26) 𝔙=(𝔙e,(,)𝔙e,(,+)𝔙e,(+,)𝔙e,(+,+))\displaystyle{\mathfrak{V}}=\left(\begin{array}[]{cc}{\mathfrak{V}}^{e,(-,-)}&{\mathfrak{V}}^{e,(-,+)}\\ {\mathfrak{V}}^{e,(+,-)}&{\mathfrak{V}}^{e,(+,+)}\end{array}\right) =(𝔙e,(,)00𝔙e,(+,+))(eUeLeme)\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) =(𝔙e,(,)00𝔙e,(+,+))e.\displaystyle=\left(\begin{array}[]{cc}{\mathfrak{V}}^{e,(-,-)}&0\\ 0&{\mathfrak{V}}^{e,(+,+)}\end{array}\right)\cdot\mathcal{M}^{e}.

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

(32) 𝔙xe=0\displaystyle{\mathfrak{V}}_{x_{e}=0} =\displaystyle= (𝔙e,(,)00𝔙e,(+,+)).\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={e1er}E=\{e_{1}\prec\cdots\prec e_{r}\} be a fixed ordering. Then

𝔙=e1er.{\mathfrak{V}}=\mathcal{M}^{e_{1}}\cdots\mathcal{M}^{e_{r}}.
Proof.

We will prove by downward induction on ii that

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

For i=ri=r the assertion follows directly from Lemma 4.2. For the inductive step assume i>1i>1 and (33) holds for ii. We know from Lemma 4.2 that if we choose a linear ordering on EE for which ei1e_{i-1} is the largest element then

(34) 𝔙\displaystyle{\mathfrak{V}} =\displaystyle= 𝔙xi1=0𝒩,\displaystyle{\mathfrak{V}}_{x_{i-1}=0}\cdot\mathcal{N},

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

NQ,R={1 if Q=Rμ((0^,Q)R,ei1)𝔙Q,R if ei1Sep(Q,R)0 otherwise,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 𝒩=ei1\mathcal{N}=\mathcal{M}^{e_{i-1}} for this particular ordering. Now we go back to the ordering in the assumption and set xi==xr=0x_{i}=\cdots=x_{r}=0 in 𝒩\mathcal{N}. We see that

(NQ,R)xi==xr=0{1 if Q=Rμ((0^,Q)R,ei1)𝔙Q,R if ei1 is the largest element in S(Q,R)0 otherwise.(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 𝒩xi==xr=0=ei1\mathcal{N}_{x_{i}=\cdots=x_{r}=0}=\mathcal{M}^{e_{i-1}}.

Now (34) implies

𝔙xi==xr=0\displaystyle{\mathfrak{V}}_{x_{i}=\cdots=x_{r}=0} =\displaystyle= 𝔙xi1==xr=0𝒩xi==xr=0\displaystyle{\mathfrak{V}}_{x_{i-1}=\cdots=x_{r}=0}\cdot\mathcal{N}_{x_{i}=\cdots=x_{r}=0}
=\displaystyle= 𝔙xi1==xr=0ei1\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

𝔙\displaystyle{\mathfrak{V}} =\displaystyle= 𝔙xi==xr=0eier\displaystyle{\mathfrak{V}}_{x_{i}=\cdots=x_{r}=0}\cdot\mathcal{M}^{e_{i}}\cdots\mathcal{M}^{e_{r}}
=\displaystyle= 𝔙xi1=xi==xr=0ei1er.\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=1i=1 the matrix 𝔙x1==xr=0{\mathfrak{V}}_{x_{1}=\cdots=x_{r}=0} is the identity matrix. Thus (33) yields:

𝔙=e1er.{\mathfrak{V}}=\mathcal{M}^{e_{1}}\cdots\mathcal{M}^{e_{r}}.

Before we prove the following proposition, we quote [9, Corollary 3], which is a result for oriented matroids.

Lemma 4.3.

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

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

Proposition 4.0.

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

μ((0^,Q)R,e)={(1)rank(|z(Y)) if Qz(Y)=Rz(Y)0 otherwise .\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 𝒯Y,e{\mathcal{T}}^{Y,e} we have YQ,RY\leq Q,R, so Ye=Qe=ReY_{e}=Q_{e}=R_{e} for all eE\z(Y)e\in E\backslash z(Y). Thus, if we consider the poset 𝒯R|z(Y),e{\mathcal{T}}_{R|_{z(Y)},e} in the restriction |z(Y)\mathcal{L}|_{z(Y)} we find that the interval (0^,Q)R,e(\hat{0},Q)_{R,e} is isomorphic to (0^,Q|z(Y))R|z(Y),e(\hat{0},Q|_{z(Y)})_{R|_{z(Y)},e}, since the elements in (0^,Q)R,e(\hat{0},Q)_{R,e} only differ in z(Y)z(Y). We saw in Section 2, that (E\Y¯,F(Y)\Y¯)(E\backslash\underline{Y},F(Y)\backslash\underline{Y}) is an OM. Further, 𝒯R|z(Y),e{\mathcal{T}}_{R|_{z(Y)},e} is a poset and (0^,Q|z(Y))R|z(Y),e(\hat{0},Q|_{z(Y)})_{R|_{z(Y)},e} is an interval in this particular OM. Furthermore, since YY is the maximal element satisfying Ye=0Y_{e}=0 and YQY\leq Q, ee does not define a proper face of Q|z(Y)Q|_{z(Y)}. Since our interval is in an OM, we can use Lemma 4.3 and the claim follows. ∎

We define bY,e=0b_{Y,e}=0 if ee is not the maximal element of z(Y)z(Y) and 12#𝒯Y,e\frac{1}{2}\#{\mathcal{T}}^{Y,e} otherwise. Since PP together with Y(P)Y\circ(-P) is a perfect pairing on 𝒯Y,e{\mathcal{T}}^{Y,e} it follows that 𝒯Y,e{\mathcal{T}}^{Y,e} contains an even number of topes. In particular, bY,eb_{Y,e} is a nonnegative integer. We denote by Y,e\mathcal{M}^{Y,e} the submatrix of e\mathcal{M}^{e} obtained by selecting rows and columns indexed by 𝒯Y,e{\mathcal{T}}^{Y,e}.

Lemma 4.4.

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

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

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

Proof.

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

Q,RY,e={1 if Q=R(1)rank(|z(Y))ez(Y),Qe=xe if Q=Y(R)e largest element of S(Q,R)0 otherwise .\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 Y,e\mathcal{M}^{Y,e} so that the elements RR and Y(R)Y\circ(-R) are paired in consecutive rows and columns. With this ordering Y,e\mathcal{M}^{Y,e} is a block diagonal matrix having along its diagonal bY,eb_{Y,e} two by two matrices

(1(1)rank(|z(Y))ez(Y),Re=xe(1)rank(|z(Y))ez(Y),Re=xe1)\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 ee is the maximal element of z(Y)z(Y) and identity matrices otherwise. In any case we find det(Y,e)=(1a(Y))bY,e\det(\mathcal{M}^{Y,e})=(1-a(Y))^{b_{Y,e}} as desired. ∎

Lemma 4.5.

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

Proof.

Note that for each tope TT there is exactly one YY with T𝒯Y,eT\in\mathcal{T}^{Y,e}, and that 𝒯0,e=\mathcal{T}^{0,e}=\emptyset. We fix a linear ordering of 𝒯{\mathcal{T}} such that for each YY\in\mathcal{L} the topes from 𝒯Y,e{\mathcal{T}}^{Y,e} form an interval and such that the topes from 𝒯Y,e{\mathcal{T}}^{Y,e} precede those of 𝒯Y,e{\mathcal{T}}^{Y^{\prime},e} if Y<YY<Y^{\prime}.

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

If Qe=ReQ_{e}=R_{e} then by QRQ\neq R we have (e)Q,R=0(\mathcal{M}^{e})_{Q,R}=0. Hence it suffices to consider the case QeReQ_{e}\neq R_{e}. Since Y𝟎Y\neq\mathbf{0}, ee is a proper face of RR.

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

On the other hand, if Qstar(Y)Q\in{\mathrm{star}}(Y), then in particular YQY\leq Q. Since by definition of 𝒯Y,e{\mathcal{T}}^{Y^{\prime},e} we have that YY^{\prime} is the maximal covector such that YQY^{\prime}\leq Q and Ye=0Y^{\prime}_{e}=0 it follows that YYY\leq Y^{\prime}. Since YYY\neq Y^{\prime} we must have that Y<YY<Y^{\prime}, i.e. (e)Q,R(\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 EE it follows from Proposition 4.0 that det𝔙\det{\mathfrak{V}} is the product of the determinants of e\mathcal{M}^{e} for eEe\in E. By Lemma 4.5 the determinant of each e\mathcal{M}^{e} is a product of determinants of Y,e\mathcal{M}^{Y,e} for eEe\in E and YY\in\mathcal{L} for which 𝒯Y,e{\mathcal{T}}^{Y,e}\neq\emptyset. Then Lemma 4.4 completes the proof. ∎

Remark 4.0 (Description of bYb_{Y}).

In Theorem 2.0 we describe bYb_{Y} as a nonnegative integer, but this can be made more precise: Fix any linear order on EE and let eYe_{Y} be the maximal element of z(Y)z(Y). From Lemma 4.4 we deduce, that bY=bY,eYb_{Y}=b_{Y,e_{Y}}. Thus 2bY2b_{Y} counts the topes P𝒯P\in{\mathcal{T}} such that YY is the maximal element of \mathcal{L} for which YeY=0Y_{e_{Y}}=0 and YPY\leq P. In particular, bYb_{Y} does not depend on the choice of the linear ordering on EE.

5. Applications

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

Refer to caption
Figure 1. A poset 𝒬\mathcal{Q}, its lattice L(𝒬)L(\mathcal{Q}) of ideals and its set X(𝒬)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 LL there exists a poset 𝒫\mathcal{P}, such that ordering the ideals (downward closed sets) of 𝒫\mathcal{P} by inclusion yields a lattice isomorphic to LL. The topes of the COM associated to LL correspond to the ideals of 𝒫\mathcal{P}, the empty set can be seen as the all-plus vector, the ground set EE of this COM is the ground set of 𝒫\mathcal{P}, and the separator of two ideals I,II,I^{\prime} is the symmetric difference IΔII\Delta I^{\prime}. So this allows, to quickly write down the (unsigned) Varchenko matrix 𝐕L\mathbf{V}_{L} of LL. In our example we indicate 𝐕L(𝒫)\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 s1sks_{1}\ldots s_{k} for the product sSxs\prod_{s\in S}x_{s}. The \emptyset translates therefore to the empty product, which is 1.

(ababbeabcabdabeabcdabceabdeabcdeaabbabebcbdbebcdbcebdebcdebabaeacadaeacdaceadeacdeabbaaecdecdcedecdebeabeeaeaceadeaacdeacadacdabcbcaccacecdcedecdedeabdbdaddadecddeccdeeceabebeaeeacedecdecdcdabcdbcdacdcdacdedccdedeceeabcebceaceceacecedcdecddabdebdeadedeadcdeedcecdcabcdebcdeacdecdeacddececdedc)\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}

Let us define the covectors of that COM: Let III\subseteq I^{\prime} be two ideals such that I\II^{\prime}\backslash I forms an antichain. Then these two ideals define a covector YY by setting

Y(I,I)e={ if eI0 if eI\I+ otherwise. \displaystyle Y(I,I^{\prime})_{e}=\begin{cases}-\text{ if }e\in I\\ 0\text{ if }e\in I^{\prime}\backslash I\\ +\text{ otherwise. }\\ \end{cases}

In particular, when I=II=I^{\prime}, we get a tope corresponding to the ideal II and the all-plus tope corresponds to the empty ideal.

Now, if we pick a linear ordering on EE, let eYe_{Y} be the largest element of III^{\prime}\setminus I, then 2bY(I,I)2b_{Y(I,I^{\prime})} counts those ideals KK such that we have

  • IKII\subseteq K\subseteq I^{\prime},

  • if JKJJ\subseteq K\subseteq J^{\prime} and eYe_{Y} is the largest element of JJJ^{\prime}\setminus J, then IIJJI^{\prime}\setminus I\subsetneq J^{\prime}\setminus J.

But note that this condition is only satisfied if I=KI=K and I=I{e}I^{\prime}=I\cup\{e\} for some e𝒫e\in\mathcal{P} or I=KI^{\prime}=K and I=I{e}I=I^{\prime}\setminus\{e\} for some e𝒫e\in\mathcal{P}. Indeed, if otherwise IKII\subsetneq K\subsetneq I^{\prime} and eYe_{Y} is the largest element of III^{\prime}\setminus I one can set J=K{eY}J=K\setminus\{e_{Y}\} and J=K{eY}J^{\prime}=K\cup\{e_{Y}\}, a contradiction to the above condition. Hence bY(I,I)b_{Y(I,I^{\prime})} is 11 if |II|=1|I^{\prime}\setminus I|=1 and 0 otherwise. All pairs I,II,I^{\prime} with |II|=1|I^{\prime}\setminus I|=1 look like I=I\{p}I=I^{\prime}\backslash\{p\}, where pp is a maximal element of II^{\prime}.

Thus, Corollary 2.0 and Remark 4.0 yield that

det(𝐕L)=Ipmax(I)(1xp2)=p𝒫(1xp2)mp,\displaystyle\det(\mathbf{V}_{L})=\prod_{I\in\mathcal{I}}\prod_{p\in\max(I)}(1-x_{p}^{2})=\prod_{p\in\mathcal{P}}(1-x_{p}^{2})^{m_{p}},

where \mathcal{I} denotes the set of ideals of 𝒫\mathcal{P}, max(I)\max(I) the set of maximal elements of an ideal II and mpm_{p} denotes the number of ideals having pp as maximal element. In our example we get the following formula for det(𝐕L(𝒬))\det(\mathbf{V}_{L(\mathcal{Q})}):

(1xa2)(1xb2)((1xa2)(1xb2))(1xe2)(1xc2)(1xd2)((1xa2)(1xe2))((1xc2)(1xd2))((1xc2)(1xe2))((1xd2)(1xe2))((1xc2)(1xd2)(1xe2))=(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}))=

(1xa2)3(1xb2)2(1xc2)4(1xd2)4(1xe2)5.(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 𝒫\mathcal{P}, that was described in [2]. The topes are the linear extensions of 𝒫\mathcal{P}, and the separator of two linear extensions L,LL,L^{\prime} is the set of pairs of elements of 𝒫\mathcal{P} that are ordered differently in LL and LL^{\prime}. In particular, the ground set of this COM consists of the set Inc(𝒫)\mathrm{Inc}(\mathcal{P}) of incomparable pairs of 𝒫\mathcal{P}, e.g., Inc(𝒬)={ab,ae,cd,ce,de}\mathrm{Inc}(\mathcal{Q})=\{ab,ae,cd,ce,de\}. We can thus define the (unsigned) Varchenko matrix 𝐕X(𝒫)\mathbf{V}_{X(\mathcal{P})}. We get a description of 𝐕X(𝒬)\mathbf{V}_{X(\mathcal{Q})}. We deem it too large to display it entirely, but for example the entry corresponding to extensions abcde,beadcabcde,beadc is xabxaexcdxcexdex_{ab}x_{ae}x_{cd}x_{ce}x_{de}.

The covectors of the ranking COM are the weak extensions of 𝒫\mathcal{P}, i.e., those poset extensions of 𝒫\mathcal{P} that are chains of antichains. The set z(Y)z(Y) of such an extension YY corresponds to its set of incomparable pairs Inc(Y)\mathrm{Inc}(Y). In order to properly define the signs of the covectors, one can pick an arbitrary linear extension L0L_{0} of 𝒫\mathcal{P}, and set non-zero coordinates of YY to ++ if the corresponding incomparable pair of 𝒫\mathcal{P} is ordered the same way in L0L_{0} and YY and to - otherwise. To define bYb_{Y} we can fix an arbitrary linear order on the set Inc(𝒫)\mathrm{Inc}(\mathcal{P}) and let eY={p,q}e_{Y}=\{p,q\} be the largest element of Inc(Y)\mathrm{Inc}(Y). Then 2bY2b_{Y} counts linear extensions LL of 𝒫\mathcal{P} such that

  • LL is a linear extension of YY,

  • if another weak extension ZZ of 𝒫\mathcal{P} has eYe_{Y} as largest incomparable pair, then either LL is not an extension of ZZ or ZZ is not an extension of YY.

In this setting one can see that no such ZZ can exist if and only if YY is a chain of antichains only one of which - say AA - has size larger than 11. In this case the feasible LL are extensions of YY that extend AA by starting and ending with an element among {p,q}\{p,q\}. Hence, there are 2(|A|2)!2(|A|-2)! such linear extensions. By Corollary 2.0 and Remark 4.0 we have

det(𝐕𝒫)=A𝒜2(1pqAxp,q2)(|A|2)!,\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 𝒜2\mathcal{A}_{\geq 2} denotes the set of antichains of size at least 22 of 𝒫\mathcal{P}.

6. Conclusion

One might wonder to what extent our result could be further generalized to other classes. A natural next class are partial cubes, i.e., isometric subgraphs of the hypercube QdQ_{d}. These generalize (tope graphs of) COMs and allow for an analogous definition of the Varchenko matrix, where the (u,v)(u,v) entry contains a product of monomials indexed by those coordinates in {1,,d}\{1,\ldots,d\} where uu and vv differ. The smallest partial cube that is not the tope graph of a COM is the full subdivision of K4K_{4}, see [10]. In this case the Varchenko matrix looks like the following

(1x1x2x3x1x4x1x3x4x3x4x2x3x4x2x4x1x2x4x11x1x2x1x3x4x3x4x1x3x4x1x2x3x4x1x2x4x2x4x2x1x21x2x3x1x2x4x1x2x3x4x2x3x4x3x4x4x1x4x3x1x3x2x31x1x3x4x1x4x4x2x4x2x3x4x1x2x3x4x1x4x4x1x2x4x1x3x41x3x1x3x1x2x3x1x2x2x1x3x4x3x4x1x2x3x4x1x4x31x1x1x2x1x2x3x2x3x3x4x1x3x4x2x3x4x4x1x3x11x2x2x3x1x2x3x2x3x4x1x2x3x4x3x4x2x4x1x2x3x1x2x21x3x1x3x2x4x1x2x4x4x2x3x4x1x2x1x2x3x2x3x31x1x1x2x4x2x4x1x4x1x2x3x4x2x2x3x1x2x3x1x3x11)\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:

(x41)3(x4+1)3(x31)3(x3+1)3(x21)3(x2+1)3(x11)3(x1+1)3\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}
(3x12x22x32x42x12x22x32x12x22x42x12x32x42x22x32x42+1)\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?

As mentioned in the introduction, we are not aware of an example of a COM which cannot be extended to become the supertope of an oriented matroid. The conjectures from [2, Conjecture 1] and [10, Conjecture 1] in our language are equivalent to the following:

Problem 6.0.

Are supertopes of oriented matroids a proper subclass of the class of complexes of oriented matroids?

Acknowledgements:

We sincerely thank the referees for their attentive review and valuable comments, which have significantly improved the quality of this work. KK was partially supported by the French Agence nationale de la recherche through project ANR-17-CE40-0015 and by the Spanish Ministerio de Economía, Industria y Competitividad through grant RYC-2017-22701, grant PID2019-104844GB-I00 and grant PID2022-137283NB-C22.

References

  • [1] M. Aguiar, S. Mahajan, Topics in hyperplane arrangements, Mathematical Surveys and Monographs 226, American Mathematical Society, Providence, RI, 2017.
  • [2] H.-J. Bandelt, V. Chepoi , K. Knauer, COMs: complexes of oriented matroids, J. Combin. Theory Ser. A, 156 (2018) 195–237.
  • [3] A. Björner, Topological methods, in: Handbook of combinatorics, Volume 2, 1819–1872, Elsevier Sci. B. V., Amsterdam, 1995.
  • [4] A. Björner, M. Las Vergnas, B. Sturmfels, W. White, G.M. Ziegler, Oriented matroids. Second edition. Encyclopedia of Mathematics and its Applications 46. Cambridge University Press, Cambridge, 1999.
  • [5] T. Bryławski, A. Varchenko, The determinant formula for a matroid bilinear form, Adv. Math. 129 (1997) 1–24.
  • [6] G. Denham, P. Hanlon, Some algebraic properties of the Schechtman-Varchenko bilinear forms, in: New perspectives in algebraic combinatorics (Berkeley, CA, 1996–97), 149–176, Math. Sci. Res. Inst. Publ., 38, Cambridge University Press, Cambridge, 1999.
  • [7] R. Gente, The Varchenko Matrix for Cones, PhD-Thesis, Philipps-Universität Marburg, 2013.
  • [8] W. Hochstättler, S. Keip, and K.Knauer. ”Kirchberger’s theorem for complexes of oriented matroids.” Linear Algebra and its Applications 693 (2024): 288-296.
  • [9] W. Hochstättler, V. Welker. ”The Varchenko determinant for oriented matroids.” Mathematische Zeitschrift 293.3 (2019): 1415-1430.
  • [10] K. Knauer, T. Marc, On tope graphs of complexes of oriented matroids, Discrete Comput. Geom. 63 (2020): 377-417.
  • [11] D. Quillen, ”Homotopy properties of the poset of nontrivial p-subgroups of a group.” Advances in Mathematics 28.2 (1978): 101-128.
  • [12] H. Randriamaro, The Varchenko Determinant of an Oriented Matroid, Trans. Comb. (10) 4 (2021), 7–18.
  • [13] H. Randriamaro. ”The Varchenko matrix for topoplane arrangements.” Communications in Contemporary Mathematics 24.10 (2022): 2150086.
  • [14] H. Randriamaro, Computer Algebra of Conditional Oriented Matroids, Habilitation, (2023) Universität Kassel.
  • [15] V.V. Schechtman, A.N. Varchenko, Arrangements of hyperplanes and Lie algebra homology, Invent. Math. 106 (1991) 139–194.
  • [16] R.P. Stanley, Enumerative combinatorics. Volume 1. Second edition. Cambridge Studies in Advanced Mathematics 49,
  • [17] A. Varchenko, Bilinear form of real configuration of hyperplanes, Adv. Math. 97 (1993) 110–144.
  • [18] M. L. Wachs, Poset topology: tools and applications. (2006).