Oriented Interval Greedoids
Discrete Comput Geom
Franco Saliola · Hugh Thomas
0 H. Thomas Department of Mathematics and Statistics, University of New Brunswick , Fredericton, NB, E3B 5A3 , Canada
1 F. Saliola ( ) Département de Mathématiques, Université du Québec à Montréal , Montréal, QC, H3C 3P8 , Canada
We propose a definition of an oriented interval greedoid that simultaneously generalizes the notion of an oriented matroid and the construction on antimatroids introduced by L.J. Billera, S.K. Hsiao, and J.S. Provan in Enumeration in convex geometries and associated polytopal subdivisions of spheres (Discrete Comput. Geom. 39(13):123137, 2008). As for oriented matroids, associated to each oriented interval greedoid is a spherical simplicial complex whose face enumeration depends only on the underlying interval greedoid.
Interval greedoid; Oriented matriod; Antimatroid; Convex geometry; Sphericity theorem; CW sphere

1 Introduction
Consider a hyperplane arrangement in Rn, with all of the hyperplanes containing the
origin. Intersecting this arrangement with a sphere centered at the origin, one obtains
a regular cell decomposition of the sphere. Taking the barycentric subdivision, one
obtains a spherical simplicial complex.
Oriented matroids are a generalization of real hyperplane arrangements; the
Sphericity Theorem of Folkman and Lawrence [21] for oriented matroids says that
any oriented matroid induces a certain regular cell decomposition of the sphere (and
thus also a spherical simplicial complex) just as hyperplane arrangements do. (Terms
not defined in the introduction will be defined later in the paper.)
Billera, Hsiao, and Provan showed in [4] that there is also a certain spherical
simplicial complex associated to a convex geometry (or antimatroid). These simplicial
complexes are not a special case of the spheres arising from oriented matroids, but
they are similar in some respects (see Sect. 6.3 in particular).
The goal of this paper is to provide a general theory which includes both of these
as special cases. Following a suggestion in [4] (attributed to Anders Björner), our
approach is via the notion of interval greedoid. The precise definition appears in the
next section, but for now, it suffices to know that interval greedoids are a common
generalization of matroids and antimatroids.
In this paper, we define the notion of an oriented interval greedoid. This is an
additional structure on top of the interval greedoid structure. For a given interval
greedoid, there may be no such additional structure possible, or one, or more than
one.
For an interval greedoid which is a matroid, an oriented interval greedoid structure
amounts to (the collection of covectors defining) an oriented matroid. In contrast,
if the underlying interval greedoid is an antimatroid, it always admits exactly one
oriented interval greedoid structure.
Our main result is an analogue of the Sphericity Theorem for oriented interval
greedoids, providing a CWsphere and (by barycentric subdivision) a spherical
simplicial complex associated to any oriented interval greedoid. Our proof is based on
the proof of the Sphericity Theorem given in [13]. The spherical simplicial complex
associated to the unique oriented structure for an antimatroid, coincides with that
constructed by [4].
Along the way, we give versions for oriented interval greedoids of a number of
constructions for oriented matroids, such as restriction and contraction.
2 Interval Greedoids
In this section we recall basic notions and results on interval greedoids. Much of the
material in this section, except for Sect. 2.2.5, is drawn from [19] or [22]. The maps
μ and ξ introduced in Sect. 2.2.5 seem to be new.
2.1 Definition
Let E denote a finite set and F a set of subsets of E. An interval greedoid is a pair
(E, F ) satisfying the following properties for all X, Y, Z ∈ F :
(IG1) If X = ∅, then there exists an x ∈ X such that X − x ∈ F .
(IG2) If X > Y , then there exists an x ∈ X\Y such that Y ∪ x ∈ F .
(IG3) If X ⊆ Y ⊆ Z and e ∈ E\Z with X ∪ e ∈ F and Z ∪ e ∈ F , then Y ∪ e ∈ F .
The set E is called the ground set of the interval greedoid (E, F ). Elements of
F are called the feasible sets of (E, F ). If F is a nonempty collection of subsets
of E satisfying (IG1), then F is said to be an accessible set system. A greedoid is a
pair (E, F ) that satisfies (IG1) and (IG2). In the literature, (IG3) is often called the
interval property.
We review here several important classes of interval greedoids.
2.1.1 Lower Interval Greedoids (Matroids)
A greedoid (E, F ) satisfying the following strengthening of (IG3) is called a lower
interval greedoid.
(LIP) Suppose X, Y ∈ F with X ⊆ Y . If e ∈ E\Y and Y ∪ e ∈ F , then X ∪ e ∈ F .
The reader familiar with the notion of a matroid will recognize that the notions of
lower interval greedoid and matroid coincide.
This notion is an abstraction of linear independence of vectors. For example, let
V = R2, x = (
−3, 1
), y = (
2, 1
) and z = (
4, 1
), and let F be the collection of subsets
of E = {x, y, z} that consist of linearly independent vectors:
F =
∅, {x}, {y}, {z}, {x, y}, {x, z}, {y, z} .
Then (E, F ) is a lower interval greedoid. Such lower interval greedoids are called
vector matriods.
2.1.2 Upper Interval Greedoids (Antimatroids and Convex Geometries)
A greedoid (E, F ) satisfying the following property is called an upper interval
greedoid.
(UIP) Suppose X, Y ∈ F with X ⊆ Y . If e ∈ E\Y and X ∪ e ∈ F , then Y ∪ e ∈ F .
Upper interval greedoids arise as complements of the closed sets of a convex
geometry. A convex geometry is a pair (E, τ ), where E is a finite set and τ : 2E → 2E
is an increasing, monotone and idempotent function, satisfying:
(AE) If x, y ∈/ τ (X), x = y, and y ∈ τ (X ∪ x), then x ∈/ τ (X ∪ y).
The subsets A ⊆ E satisfying τ (A) = A are called closed sets of the convex
geometry. The extreme points ext(A) of A ⊆ E are the points x ∈ A satisfying x ∈/ τ (A\x).
If (E, τ ) is a convex geometry, then the complements of the closed sets of E are the
feasible sets of an upper interval greedoid on the ground set E. Conversely, every
upper interval greedoid arises from a convex geometry; see [22, Theorem III.1.3] or
[19, Proposition 8.7.3].
The canonical example is a finite subset E ⊆ Rn with τ (A) = conv(A) ∩ E, where
conv(A) is the convex hull of the points in A. As an example, let x, y, z be three
colinear points in the plane, with y between x and z, and consider the convex geometry
with closure operator τ (X) = conv(X) ∩ {x, y, z}. The closed sets are the subsets
∅, {x}, {y}, {z}, {x, y}, {y, z}, {x, y, z}.
Then (E, F ) is an upper interval greedoid , where E = {x, y, z} and F is
F = {x, y, z}, {y, z}, {x, z}, {x, y}, {z}, {x}, ∅ .
Remark 2.1 Upper interval greedoids have been studied under several different
names including antimatroid, APSstructures, discs, and shelling structures. Some
care is required in reading the literature, as some authors have used the term
antimatroid for a convex geometry. By antimatroid, we will always mean an upper interval
greedoid.
2.1.3 Interval Greedoids from Semimodular Lattices
If x and y are two elements of a poset, then we write x y to indicate that x < y and
that there does not exist an element z in the poset satisfying x < z < y.
Let L be a finite lattice. L is said to be (lower) semimodular if it has the following
property for all x = y ∈ L: if x z and y z for some z ∈ L, then x ∧ y x and
x ∧ y y. An element e ∈ L such that e = 1ˆ is called meetirreducible if e = x ∧ y
implies x = e or y = e.
Proposition 2.2 [19, Theorem 8.8.7] Suppose L is a finite lower semimodular lattice.
Let E be the set of meetirreducible elements of L, and let
F = {e1, e2, . . . , ek } ⊆ E : 1ˆ
e1
(e1 ∧ e2)
· · ·
(e1 ∧ e2 ∧ · · · ∧ ek ) .
Then (E, F ) is an interval greedoid.
2.2 The Lattice of Flats
2.2.1 Feasible Orderings
Let (E, F ) denote an interval greedoid. Let X ∈ F . An ordering x1 < x2 < · · · < xr
of the elements of X = {x1, x2, . . . , xr } is denoted by X = {x1 < x2 < · · · < xr }.
Such an ordering is said to be a feasible ordering of X if {x1, . . . , xi } ∈ F for all
1 ≤ i ≤ r = X. Repeated application of (IG1) shows that every X ∈ F has a feasible
ordering.
Proposition 2.3 ([22, Chap. V, Lemma 5.1], [19, Proposition 8.2.5]) Let (E, F ) be
an interval greedoid. Let X, Y ∈ F and X > Y . Suppose X = {x1 < · · · < xr } is
a feasible ordering. Then there is a subset {xi1 < · · · < xik } of X\Y of size X − Y 
such that Y ∪ {xi1 , . . . , xij } ∈ F for all 1 ≤ j ≤ k.
2.2.2 Contractions
Let (E, F ) denote an interval greedoid and let X ∈ F . Let F /X denote the
collection of subsets that can be added to X preserving feasibility:
F /X = {Y ⊆ E\X : X ∪ Y ∈ F }.
The sets in F /X are the feasible sets of an interval greedoid on the ground set
Y ∈F /X Y ; this interval greedoid is called the contraction of (E, F ) by X.
Properties of contractions will be further developed in later sections, but for now we record
the following result, which is crucial to much of what follows.
Proposition 2.4 Suppose (E, F ) is an interval greedoid and let A ⊆ E. Let U and
V be maximal with respect to inclusion among the feasible sets contained in A. Then
F /U = F /V .
Proof Let U and V be two maximal feasible sets contained in A. Then U  = V 
(otherwise we can enlarge the smaller one using (IG2)). Suppose W ∈ F /U with
W = ∅. Then U ∪ W ∈ F . Let U = {u1 < · · · < ur } be a feasible ordering of U .
Repeated application of (IG2) to U and U ∪ W gives a feasible ordering {u1 < · · · <
ur < w1 < · · · < ws } of U ∪ W . Proposition 2.3 applied to U ∪ W and V gives an
ordered subset {z1 < · · · < zt } of U ∪ W such that V ∪ {z1, . . . , zi } ∈ F for each 1 ≤
i ≤ t , where t = U ∪ W  − V  = W . If z1 ∈ U , then V ∪ {z1} ∈ F and V ∪ {z1} ⊆ A,
contradicting the maximality of V . Therefore, z1 ∈ W and the ordering of the zi
implies zi ∈ W for all 1 ≤ i ≤ t . Since t = W , we have W = {z1, . . . , zt }. Thus,
V ∪ W ∈ F , or equivalently, W ∈ F /V . Reversing the roles of U and V gives the
reverse containment F /V ⊆ F /U . Thus, F /U = F /V .
2.2.3 Continuations
Let (E, F ) denote an interval greedoid and let X ∈ F . The set of continuations Γ (X)
of X is the set of elements that can be added to X preserving feasibility:
Γ (X) = {x ∈ E\X : X ∪ x ∈ F }.
For vector matroids, Γ (X) consists of those vectors from E not in the span of X. For
an antimatroid, Γ (X) = ext(E\X).
If X, Y ∈ F and F /X = F /Y , then Γ (X) = Γ (Y ). The converse does not hold
for arbitrary greedoids, but it does hold for interval greedoids.
Proposition 2.5 [22, Chap. V, Theorem 5.10] Suppose (E, F ) is an interval
greedoid. Then for all X, Y ∈ F , we have Γ (X) = Γ (Y ) if and only if F /X = F /Y .
2.2.4 Flats
Let (E, F ) be an interval greedoid. Define an equivalence relation on F by setting
X ∼ Y if and only if F /X = F /Y . In light of Proposition 2.5, X ∼ Y if and only if
Γ (X) = Γ (Y ). We write [X] for the equivalence class of X:
[X] = {Y ∈ F : F /X = F /Y } = Y ∈ F : Γ (X) = Γ (Y ) .
These equivalence classes are called the flats of (E, F ).
The set Φ of flats of (E, F ) is a poset with partial order induced by reverse
inclusion:
[X] ≤ [Y ] iff there exists Z ∈ F /Y such that Y ∪ Z ∼ X.
In particular, if Y ⊆ X, then [X] ≤ [Y ]. (Note that some authors choose to use
inclusion rather than reverse inclusion to induce the partial order on Φ.)
In a vector matroid, [X] ≤ [Y ] iff the span of Y is contained in the span of X. The
poset of flats for the convex geometry on three colinear points is illustrated in Fig. 1.
For an arbitrary antimatroid, [X] = {X} for all X ∈ F , thus the poset of flats Φ for
an antimatroid is isomorphic to F ordered by reverse inclusion. If L is a
semimodular lattice and (E, F ) the interval greedoid described in Proposition 2.2, then the
corresponding poset of flats is naturally isomorphic to L.
The following result implies that Φ is a lower semimodular poset. In fact, Φ is a
semimodular lattice; see Proposition 2.9.
Proposition 2.6 [18, Lemma 8.8.6] Suppose (E, F ) is an interval greedoid. Let X ∈
F and suppose X ∪ x ∈ F and X ∪ y ∈ F . If [X ∪ x] = [X ∪ y], then X ∪ {x, y} ∈ F .
2.2.5 Maps μ and ξ
Let (E, F ) be an interval greedoid and Φ its poset of flats. Define two maps μ :
2E → Φ and ξ : Φ → 2E as follows.
(μ) Define μ : 2E → Φ on arbitrary subsets A ⊆ E by μ(A) = [X], where X is
maximal with respect to inclusion among the feasible sets contained in A.
(ξ ) Define ξ : Φ → 2E for X ∈ F by ξ([X]) = X ∼X X .
It follows from Proposition 2.4 that μ is welldefined. These maps are very important
to what follows, and will be used to describe the meets and joins in Φ.
Proposition 2.7 Suppose (E, F ) is an interval greedoid, Φ its lattice of flats and μ
and ξ the maps defined above.
(
1
) (μ ◦ ξ )(A) = A for all A ∈ Φ. So ξ is injective.
(
2
) μ : (2E , ⊆) → (Φ, ≤) is orderreversing.
(
3
) ξ : (Φ, ≤) → (2E , ⊆) is orderreversing.
(
4
) For all A, B ∈ Φ we have A ≤ B if and only if ξ(B) ⊆ ξ(A).
(
5
) For all Y ∈ F and A ∈ Φ, if Y ⊆ ξ(A), then A ≤ [Y ].
Proof (
1
) Suppose that X is not maximal with respect to inclusion among the feasible
sets contained in ξ([X]). Then there exists x ∈ ξ([X]) − X such that X ∪ x ∈ F .
Therefore, x ∈ F /X and x ∈ X for some X ∼ X with X = X. But X ∼ X if
and only if F /X = F /X , so x ∈ F /X . This is a contradiction since x ∈/ F /X if
x ∈ X . Thus, X is maximal, and so μ(ξ ([X])) = [X].
(
2
) Suppose A ⊆ B . Let X be maximal with respect to inclusion among the
feasible sets contained in A. Then there exists Y such that X ⊆ Y ⊆ B and Y is maximal
among the feasible sets contained in B . Therefore, [Y ] ≤ [X]. Hence, μ(B) ≤ μ(A).
(
3
) Suppose [X] ≤ [Y ]. If e ∈ ξ ([Y ]), then e ∈ Y for some Y ∼ Y . So [X] ≤
[Y ] = [Y ]. Thus, there is a Z ∈ F /Y such that Y ∪ Z ∼ X. Therefore, e ∈ Y ⊆
(Y ∪ Z) ⊆ ξ ([X]).
(
4
) This follows from (
1
), (
2
) and (
3
).
(
5
) Suppose Y ∈ F and Y ⊆ ξ ([X]). Then there exists Z containing Y that is
maximal among the feasible sets contained in ξ ([X]). Then [Y ] ≥ [Z] since Y ⊆ Z
and [Z] = [X] by Proposition 2.4.
Remark 2.8 For a matroid without loops, (ξ ◦ μ)(A) is the smallest closed set
containing A. There is a generalization of the notion of closed sets for any interval greedoid
[18, §8.4B], and in this generality all one can say is that (ξ ◦ μ)(A) is contained in
the smallest closed set containing A. That this containment can be strict follows by
considering the convex geometry on three colinear points.
For a matroid without loops, ξ (Φ) consists exactly of the closed sets of the
matroid. For an antimatroid, ξ (Φ) consists precisely of the feasible sets; for the
interval greedoid arising from a semimodular lattice, ξ ([X]) consists of the set of
meetirreducibles f such that f ≥ φ ([X]).
2.2.6 Lattice of Flats
We have seen that Φ is a lower semimodular poset (Proposition 2.6). It is also graded:
the corank of any element A ∈ Φ is the size of any feasible set in A. It turns out
that Φ is also a lattice [18, Theorem 8.8.7]. The next result reformulates the lattice
operations in terms of the maps μ and ξ .
Proposition 2.9 If (E, F ) is an interval greedoid, then Φ is a lower semimodular
lattice whose lattice operations are given by
A ∨ B = μ ξ (A) ∩ ξ (B)
and
A ∧ B = μ ξ (A) ∪ ξ (B)
for all A, B ∈ Φ . That is, A ∨ B = [X], where X is maximal among the feasible sets
contained in ξ (A) ∩ ξ (B), and A ∧ B = [Y ], where Y is maximal among the feasible
sets contained in ξ (A) ∪ ξ (B).
If (E, τ ) is a convex geometry and (E, F ) the corresponding antimatroid, then
[X] ∨ [Y ] =
E\τ (E\X) ∪ (E\Y )
(2.1)
for all X, Y ∈ F .
3 Oriented Interval Greedoids
Throughout this section (E, F ) will denote an interval greedoid.
3.1 Signed Flats
A signed flat of an interval greedoid (E, F ) is a pair (A, α) consisting of a flat A and
a map α : Γ (A) → {+, −}.
Define a partial order on signed flats as follows. If (A, α) and (B, β) are signed
flats of (E, F ), let (A, α) ≤ (B, β) if A ≤ B (as flats in Φ ) and if α and β agree on
Γ (A) ∩ Γ (B). Reflexivity and antisymmetry are straightforward to verify;
transitivity follows from (IG3).
Define the product (A, α) ◦ (B, β) of two signed flats (A, α) and (B, β) by
otherwise.
This product is welldefined because Γ (A ∨ B) ⊆ Γ (A) ∪ Γ (B) by Proposition 3.3
below.
Example 3.1 (Antimatroids) Suppose (E, F ) is an antimatroid. The continuations
of a feasible set X are the extreme points of the complement E\X in the convex
geometry. Therefore, a signed flat ([X], α) of the antimatroid is an assignment of +
or − to each extreme point of E\X. Figure 2 depicts a closed set C of a convex
geometry; the extreme points of C are labeled by + or −, the nonextreme points in
C are labeled by 1, and the points in the exterior of C are labeled by 0.
The product of two signed flats ([X], α) and ([Y ], β) has a geometric
interpretation. If X = E\X and Y = E\Y , then form a new closed set Z by taking the closure
of X ∪ Y ; that is, Z = τ (X ∪ Y ). Note that the extreme points of Z are contained
in ext(X ) ∪ ext(Y ). The sign for each z ∈ ext(Z ) is α(z) if z ∈ ext(X ), and β(z)
otherwise.
Example 3.2 (Matroids) Suppose (E, F ) is a matroid with no loops. If A is a flat of
the matroid, then ξ(A) = E\Γ (A) is a closed set of the matroid. Therefore, a signed
flat (A, α) is an assignment of a sign + or − to each element of the complement
of the closed set A. If we extend this by assigning 0 to each element of A, then α
induces a covector in the sense of oriented matroids (see Sect. 3.4.1).
Among other things, the following establishes that the product of signed flats is
welldefined.
Proposition 3.3 Let (E, F ) be an interval greedoid, A, B ∈ Φ and X ∈ F .
(
1
) If B ≤ [X] and x ∈ Γ (X), then either x ∈ Γ (B) or B ≤ [X ∪ x].
(
2
) If A ≤ B, then Γ (B) ⊆ Γ (A) ∪ ξ(A).
(
3
) Γ (A ∨ B) ⊆ Γ (A) ∪ Γ (B).
(
4
) Γ (A ∨ B) ∪ ξ(A ∨ B) ⊆ (Γ (A) ∪ ξ(A)) ∩ (Γ (B) ∪ ξ(B)).
Proof (
1
) Pick Y ∈ F such that B = [Y ]. Suppose [Y ] ≤ [X] and let x ∈ Γ (X). Then
there exists Z ∈ F /X such that X ∪ Z ∼ Y . Applying axiom (IG2) repeatedly to X
and X ∪ Z yields a sequence X ⊂ (X ∪ z1) ⊂ (X ∪ {z1, z2}) ⊂ · · · ⊂ (X ∪ Z) of
feasible sets. Put X0 = X and let Xi = Xi−1 ∪ zi for 1 ≤ i ≤ r = Z.
X
X ∪ {z1}
X ∪ {x}
X ∪ {z1, z2}
X ∪ {z1, x}
· · ·
· · ·
X ∪ Z ∼ Y
X ∪ Z ∪ {x}.
Since x ∈ Γ (X), X ∪ x ∈ F . If [X ∪ x] = [X ∪ z1], then [Y ] = [X ∪ Z] ≤ [X ∪
z1] = [X ∪ x] since X ⊆ X ∪ Z. If [X ∪ x] = [X ∪ z1], then X ∪ {x, z1} ∈ F by
Proposition 2.6 since x, z1 ∈ Γ (X). Thus, x, z2 ∈ Γ (X ∪ z1). If [X ∪ {z1, x}] = [X ∪
{z1, z2}], then [Y ] ≤ [X ∪ x] using a similar argument as in the previous case. If
[X ∪ {z1, x}] = [X ∪ {z1, z2}], then X ∪ {z1, z2, x} ∈ F by Proposition 2.6. Continuing
in this manner we get either that [Y ] ≤ [X ∪ x] or (X ∪ Z) ∪ x ∈ F . That is, either
B = [Y ] ≤ [X ∪ x], or x ∈ Γ (Y ) = Γ (B). This proves the statement.
(
2
) Pick X, Y ∈ F such that A = [Y ] and B = [X]. If [Y ] ≤ [X] and x ∈ Γ (Y ),
then x ∈ Γ (Y ) or [Y ] ≤ [X ∪ x] by (
1
). In the latter case, x ∈ ξ(X ∪ x) ⊆ ξ(Y ) since
ξ is orderreversing. Thus, x ∈ Γ (Y ) or x ∈ ξ(Y ).
(
3
) Pick X ∈ F such that A ∨ B = [X]. Let x ∈ Γ (X). Then X ∪ x ∈ F and
[X ∪ x] < [X]. Since [X] = A ∨ B, it follows that [X ∪ x] is not above both A and B.
If A ≤ [X ∪ x], then the above applied to A ∨ B and A gives that x ∈ Γ (A) since
A ≤ [X ∪ x]. Similarly, if B ≤ [X ∪ x], then x ∈ Γ (B). Hence, x ∈ Γ (A) ∪ Γ (B).
(
4
) Since ξ is orderreversing, it follows that ξ(A ∨ B) ⊆ ξ(A) ∩ ξ(B). (
4
) now
follows from (
2
).
The following result summarizes properties of the product and partial order.
Proposition 3.4 Let (A, α) and (B, β) denote two signed flats over an interval
greedoid (E, F ). Then
(
1
) (A, α) ≤ (B, β) if and only if (A, α) ◦ (B, β) = (B, β).
(
2
) (A, α) ≤ (A, α) ◦ (B, β).
(
3
) If A ≤ B, then (B, β) ◦ (A, α) = (B, β).
(
4
) The product ◦ is associative.
(
5
) (A, α) ◦ (B, β) ◦ (A, α) = (A, α) ◦ (B, β).
(6) (A, α) ◦ (A, α) = (A, α).
Proof (
1
) Suppose (A, α) ◦ (B, β) = (B, β). Since (A, α) ◦ (B, β) = (A ∨ B, α ◦ β),
it follows that B = A ∨ B and that β = α ◦ β. Therefore, A ≤ B and β(x) = (α ◦
β)(x) = α(x) for all x ∈ Γ (A) ∩ Γ (B). Thus, (A, α) ≤ (B, β).
Conversely, suppose (A, α) ≤ (B, β). Then A ≤ B and α(x) = β(x) for all x ∈
Γ (A) ∩ Γ (B). Therefore, A ∨ B = B. It remains to show that (α ◦ β)(x) = β(x) for
all x ∈ Γ (A ∨ B) = Γ (B). Let x ∈ Γ (B). If x ∈ Γ (A), then (α ◦ β)(x) = α(x) and
α(x) = β(x) since α and β agree on Γ (A) ∩ Γ (B). If x ∈/ Γ (A), then (α ◦ β)(x) =
β(x). Therefore, β and α ◦ β agree on Γ (A ∨ B).
(
2
) First note that A ≤ A ∨ B by the definition of ∨. We need only show that α
and α ◦ β agree on Γ (A) ∩ Γ (A ∨ B), which follows from the definition of α ◦ β.
Therefore, (A, α) ≤ (A ∨ B, α ◦ β) = (A, α) ◦ (B, β).
(
3
) If A ≤ B, then A ∨ B = B, so Γ (A ∨ B) = Γ (B). So the domains of β ◦ α and
β are the same. And from the definition of ◦, if x ∈ Γ (B), then (β ◦ α)(x) = β(x).
(
4
), (
5
) and (6) are straightforward to verify using similar arguments.
3.2 Covectors
Let (A, α) denote a signed flat. Then α : Γ (A) → {+, −} can be extended to a map
α : E → {0, +, −, 1} as follows:
⎧ α(e), if e ∈ Γ (A),
α(e) = ⎨⎪⎪ 0, if e ∈ ξ(A),
This map α is called the covector of the signed flat (A, α).
Note that a signed flat (A, α) can be recovered from its covector α. Indeed, the set
of indices e ∈ E such that α(e) = 0 is precisely the set ξ(A), from which A can be
recovered (Proposition 2.7). Therefore, there exists a map from the set of covectors
of (E, F ) to the lattice of flats Φ, called the support of the covector,
supp(α) = μ x ∈ E : α(x) = 0 .
The product on signed flats can be formulated for covectors as follows. Let α and
β be the covectors of the signed flats (A, α), (B, β), respectively. Partially order the
symbols 0, +, −, 1 so that 0 and 1 are the (unique) minimal and maximal elements,
respectively, and + and − are incomparable. Define
(α β)(e) =
and
(α ◦ β)(e) =
(α β)(e), if e ∈ Γ (A ∨ B) ∪ ξ(A ∨ B),
1,
Proposition 3.5 Suppose α and β are the covectors of the signed flats (A, α) and
(B, β), respectively. Then the covector γ of (A, α) ◦ (B, β) is α ◦ β.
Proof By definition, the covector γ of the signed flat (A ∨ B, α ◦ β) is given by
γ (e) = (α ◦ β)(e) if e ∈ Γ (A ∨ B); γ (e) = 0 if e ∈ ξ(A ∨ B); and γ (e) = 1 otherwise.
Suppose e ∈/ Γ (A ∨ B) ∪ ξ(A ∨ B). By the definition of the product of covectors,
(α ◦ β)(e) = 1. Hence, (α ◦ β)(e) = γ (e).
Suppose e ∈ ξ(A ∨ B). Then e ∈ ξ(A) and e ∈ ξ(B) since ξ is orderreversing.
This implies that α(e) = β(e) = 0, hence (α ◦ β)(e) = 0. Therefore, (α ◦ β)(e) =
γ (e).
Suppose e ∈ Γ (A ∨ B). By Proposition 3.3(
3
), e ∈ Γ (A) ∪ Γ (B). If e ∈ Γ (A),
then β(e) > α(e), so (α ◦ β)(e) = α(e) = α(e). If e ∈/ Γ (A), then e ∈ Γ (B) ∩ ξ(A)
and so β(e) > 0 = α(e). Hence, (α ◦ β)(e) = β(e) = β(e). Therefore, (α ◦ β)(e) =
(α ◦ β)(e) for all e ∈ Γ (A ∨ B).
Example 3.6 (Antimatroids) If ([X], α) is a signed flat of an upper interval greedoid,
then its covector is obtained by assigning α(x) to the extreme points x ∈ ext(E\X),
1 to the nonextreme points contained in E\X, and 0 to the points in the exterior of
E\X. Figure 3 depicts the product of the two covectors in Fig. 2.
Let α and β be covectors of (E, F ). The separation set of α and β is
S(α, β) = e ∈ E : α(e) = −β(e) ∈ {+, −} .
(Note that S(α, β) ⊆ Γ (supp(α)) ∩ Γ (supp(β)).) The next result develops the
relationship between the product, the partial order and the separation set of covectors.
Lemma 3.7 Let α and β be covectors of an interval greedoid (E, F ).
(
1
) α ≤ β if and only if α ◦ β = β.
(
2
) α ≤ β if and only if α(e) ≤ β(e) for all e ∈ E.
(
3
) α ≤ β if and only if S(α, β) = ∅ and supp(α) ≤ supp(β).
(
4
) If α(e) = 1 or β(e) = 1, then (α ◦ β)(e) = 1 = (β ◦ α)(e).
Proof Let A = supp(α) and B = supp(β). By definition, α ≤ β if and only if A ≤ B
and α and β agree on Γ (A) ∩ Γ (B).
(
1
) This follows from Proposition 3.4 and Proposition 3.5.
(
2
) Suppose α ≤ β and let e ∈ E. If e ∈/ ξ(B) ∪ Γ (B), then β(e) = 1, so α(e) ≤
β(e). If e ∈ ξ(A), then α(e) = 0, so α(e) ≤ β(e). So suppose e ∈ ξ(B) ∪ Γ (B) and
e ∈/ ξ(A). Then e ∈ Γ (A) ∩ Γ (B), by Proposition 3.3. Then α(e) ≤ β(e) because α
and β agree on Γ (A) ∩ Γ (B).
Conversely, suppose α(e) ≤ β(e) for all e ∈ E. Since {e : β(e) = 0} ⊆ {e :
α(e) = 0}, we have supp(α) ≤ supp(β). If e ∈ Γ (A) ∩ Γ (B), then α(e), β(e) ∈
{+, −}, which implies α(e) = β(e) because α(e) ≤ β(e). Thus, α ≤ β.
(
3
) If α ≤ β, then A ≤ B, and S(α, β) = ∅ because S(α, β) ⊆ Γ (A) ∩ Γ (B).
Conversely, if A ≤ B and S(α, β) = ∅, then α(e) = β(e) for all e ∈ Γ (A) ∩ Γ (B),
so α ≤ β.
(
4
) If (α ◦ β)(e) = 1, then e ∈ Γ (A ∨ B) ∪ ξ(A ∨ B) ⊆ (Γ (A) ∪ ξ(A)) ∩ (Γ (B) ∪
ξ(B)) by Proposition 3.3(
4
). Hence, α(e) = 1 and β(e) = 1.
Remark 3.8 The converse of (
4
) is false. Counterexamples amongst the covectors
of the antimatroid on three colinear points are depicted in Fig. 4. They also illustrate
that the following containments can be proper.
Γ (A ∨ B) ∪ ξ(A ∨ B) ⊆ Γ (A) ∪ ξ(A) ∩ Γ (B) ∪ ξ(B) ,
ξ(A ∨ B ) ⊆ ξ(A ) ∩ ξ(B ).
3.3 Oriented Interval Greedoids
For any covector α, let −α be the covector obtained from α by replacing + with −
and − with +.
Definition 3.9 An oriented interval greedoid is a triple (E, F , G), where (E, F ) is
an interval greedoid and G is a set of covectors of (E, F ) satisfying the following
axioms.
(OG1) The map supp : G → Φ is surjective.
(OG2) If α ∈ G, then −α ∈ G.
(OG3) If α, β ∈ G, then α ◦ β ∈ G.
(OG4) If α, β ∈ G, x ∈ S(α, β) and (α ◦ β)(x) = 1, then there exists γ ∈ G such that
γ (x) = 0 and for all y ∈/ S(α, β), if (α ◦ β)(y) = 1, then γ (y) = (α ◦ β)(y) =
(β ◦ α)(y)
Remark 3.10 Note that the set G with the operation ◦ forms a semigroup: by (OG3),
the set G is closed under ◦, and by Proposition 3.4 the operation is associative. This
semigroup also satisfies the identities x2 = x and xyx = xy for all x, y ∈ G
(Proposition 3.4). Semigroups satisfying these identities are called left regular bands, which
have found several recent applications to computing spectra of random walks [2, 5,
11, 14, 15, 27] and to the study of the descent algebras of finite Coxeter groups [1, 7,
14, 23].
As we will see in Sect. 3.4.1, these conditions are modeled on the covector axioms
for oriented matroids. In the next section we will present various examples of oriented
interval greedoids. We record here the following observation.
Lemma 3.11 Suppose α and β are covectors of an oriented interval greedoid. If
(α ◦ β)(y) = (β ◦ α)(y), then α(y) = −β(y) ∈ {+, −} (that is, y ∈ S(α, β)).
Proof Let C = supp(α ◦ β) = supp(β ◦ α). Then (α ◦ β)(y) = 1 iff y ∈/ Γ (C) ∪ ξ(C)
iff (β ◦ α)(y) = 1. Similarly, (α ◦ β)(y) = 0 iff y ∈ ξ(C) iff (β ◦ α)(y) = 0. Thus
α(y), β(y) ⊂ {+, −}. The result follows.
3.4 Examples
3.4.1 Oriented Matroids
This section presents examples of oriented interval greedoids.
Let E be a finite set. An oriented matroid is a collection O of maps from E to
{0, +, −} that satisfies the following axioms.
(OM1) O contains the map z(e) = 0 for all e ∈ E.
(OM2) If α ∈ O, then −α ∈ O.
(OM3) If α, β ∈ O, then α ◦ β ∈ O, where
(OM4) Suppose α, β ∈ O and let S(α, β) = {e ∈ E : α(e) = −β(e) = 0}. For every
e ∈ S(α, β) there exists γ ∈ O with γ (e) = 0 and γ (f ) = (α ◦ β)(f ) = (β ◦
α)(f ) for all f ∈/ S(α, β).
If O is an oriented matroid, then the set of zeros of the elements of O form the
closed sets of a matroid (E, F ). The matroid (E, F ) is the underlying matroid of the
oriented matroid and O is said to be an oriented matroid on (E, F ).
Theorem 3.12 Suppose (E, F ) is a matroid without loops. Then O is an oriented
matroid with underlying matroid (E, F ) if and only if (E, F , O) is an oriented
interval greedoid.
Proof Let (E, F ) be a matroid and let (E, F , G) be an oriented interval greedoid.
Since ξ (A) = E\Γ (A) for any flat A of a matroid without loops, a covector of (E, F )
takes values in {0, +, −}. Therefore, G is a collection of maps from E to {0, +, −},
and G satisfies (OM1)–(OM4) since it satisfies (OG1)–(OG4). So G is an oriented
matroid.
Conversely, suppose that O is an oriented matroid with underlying matroid
(E, F ). If α ∈ O, then the set ζ (α) of zeros of α is a closed set of the matroid,
and there is a unique flat A satisfying ξ (A) = ζ (α). Therefore, α gives a signed flat
(A, αΓ (A)), and the covector of this signed flat is α. So O is a set of covectors of
the interval greedoid (E, F ). It is straightforward to check that the axioms for an
oriented interval greedoid are satisfied by O.
3.4.2 Antimatroids
Next we show that the set of all covectors of an antimatroid forms an oriented interval
greedoid. This collection of covectors, viewed as a poset, is the central object of
study in the work of Billera, Hsiao, and Provan [4]. We also show that this is the only
oriented interval greedoid structure on an antimatroid.
Example 3.13 If α and β are the covectors of Fig. 2, then the separation set S(α, β)
consists of the vertex x circled in Fig. 5. The covector γ depicted in Fig. 5 satisfies
(OG4) for α, β and x ∈ S(α, β).
Theorem 3.14 Suppose (E, F ) is an upper interval greedoid . Let G denote the set
of all covectors of (E, F ). Then (E, F , G) is an oriented interval greedoid.
Proof We will show that (OG1)–(OG4) hold.
Proposition 3.15 Let (E, F ) be an antimatroid. Then the only oriented structure on
(E, F ) is that constructed in Theorem 3.14.
Proof Let (E, F , G) be an oriented interval greedoid. Let α be an arbitrary covector.
We wish to show that α ∈ G.
Let Γ (supp(α)) = X = {x1, . . . , xr }. Let Y = E \ (X ∪ supp(α)). For any x ∈ X,
Yx = supp(α) ∪ (X \ {x}) is feasible. Also, Γ (Yx ) ∩ Γ (supp(α)) = {x}. By (OIG1),
we can find a covector βx ∈ G with supp(βx ) = Yx . By (OIG2), we can choose βx so
that βx agrees with α on x. Now βx1 ◦ · · · ◦ βxr = α is in G.
3.4.3 Complexified Hyperplane Arrangements
An (essential) real hyperplane arrangement is a finite set of hyperplanes {Θ1, Θ2,
. . . , Θn} in Rd satisfying Θi = {0}. Let E = {1, 2, . . . , n} and for each e ∈ E fix
a linear form e : Rd → R such that Θe = ker( e). Extending scalars, we can also
think of e as defining a linear map from Cd to C. Define He to be the kernel of this
map. It is a hyperplane in Cd . The collection A = {H1, . . . , Hn} forms a complexified
hyperplane arrangement. Also define He = {z ∈ Cd : ( e(z)) = 0}.
(Note that not all complex hyperplane arrangements are complexified
arrangements; that is to say, not all complex hyperplane arrangements arise from a real
hyperplane arrangement in the way we have just described.)
For any z = x + iy ∈ C, let
σ (x + iy) =
and for every z ∈ Cd , let
⎧ 1, if y = 0,
⎪⎨⎪⎪⎪⎪ +, if y = 0, x > 0,
⎪⎪ −, if y = 0, x < 0,
⎪
⎩⎪⎪ 0, if y = 0, x = 0,
⎧⎪ +, if y > 0,
σ (x + iy) = ⎪⎨ −, if y < 0,
⎪⎪⎩ 0, if y = 0,
αz(h) =
σ ( i (z)), if h = Hi ,
σ ( i (z)), if h = Hi .
Note that (α(Hi ), α(Hi )) ∈ {(0, 0), (+, 0), (−, 0), (1, +), (
1, −
)} for all 1 ≤ e ≤ n.
Example 3.16 There is a unique complexified hyperplane arrangement in C, namely
A = {H0 = {0}}. In this case
{αz : z ∈ C} = (0, 0), (+, 0), (−, 0), (1, +), (
1, −
)
is the set of covectors of the interval greedoid on the ground set E = {H0, H0 } with
feasible sets F = {∅, {H0 }, {H0, H0 }}. Figure 6 illustrates the partial order on these
covectors.
Let A = {H1, . . . , Hn} be a complexified hyperplane arrangement in Cd . Let L be
the lattice of all intersections of subspaces from the set
EA =
H1, . . . , Hn, H1 , . . . , Hn ,
ordered by inclusion. Then L is a lower semimodular lattice and EA is the set of
meetirreducible elements of L [18]. By Proposition 2.2, (EA, FA) is an interval
greedoid, where
FA = {h1, h2, . . . , hk} ⊆ EA :
Cd
h1
· · ·
(h1 ∩ h2 ∩ · · · ∩ hk) .
(1, +)
(+, 0)
(
1, −
)
(−, 0)
(0, 0)
Lemma 3.17 Let A be a complexified hyperplane arrangement, and let (EA, FA)
be the interval greedoid as defined above. Then, for X ∈ FA, we have
ξ(X) =
h ∈ E :
h ⊆ h ,
h ∈X
Γ (X) =
He ∈ E :
h ⊆ He
∪
He ∈ E :
h ⊆ He and
h ⊆ He .
h∈X
h∈X
h∈X
Proof We prove the first identity holds for any interval greedoid (E, F ) constructed
from a semimodular lattice L as in Sect. 2.1.3. The poset of flats Φ of (E, F ) is
naturally isomorphic to L: the mapping φ (X) = e∈X e from F to L descends to a
poset isomorphism from Φ to L (see [19, Theorem 8.8.7] for details). If X is a
feasible set, then ξ([X]) consists of the set of meetirreducibles f such that f ≥ φ ([X]).
Indeed, if f is in a feasible set Y ∼ X, then φ ([X]) = φ ([Y ]) ≤ f , which proves
one containment. For the other direction, let f ≥ φ ([X]) and let Z be a feasible set
with φ ([Z]) = f . Since f is meetirreducible, f ∈ Z. Since f ≥ φ ([X]), we know
[Z] ≥ [X], which implies that f ∈ Z ⊂ ξ([X]), as desired.
We now prove the second identity. Let M = h∈X h ⊂ Cd . Thinking of Cd as a
2ddimensional real vector space, we can decompose it into real and complex parts
as Cd = (Cd ) ⊕ (Cd ), where each of the summands is a ddimensional real vector
space, and multiplication by i provides an isomorphism from (Cd ) to (Cd ). Note
that Hi and Hi can also be expressed as a direct sum of a real and an imaginary
part. (This relies on the fact that our arrangement is a complexified real arrangement,
rather than being an arbitrary complex arrangement.) Note further that in either case,
the imaginary part corresponds to a subspace of the real part under the map from
(Cd ) to (Cd ). It follows that M , also, can be written as M = (M) ⊕ (M), with
(M) a subspace of (M).
Observe first that if He ≥ M , then, since He is real codimension one in Cd , we
have M M ∩ He , so He ∈ Γ (X). Also, in this case, we have M ∩ He M ∩ He,
because Θe ≥ (M), and thus Θe ≥ (M) either. It follows that in this case He ∈
Γ (X) and He ∈/ Γ (X).
Finally, if He ≥ M , we observe that He is codimension one in He , and thus that
either He ≥ M or M M ∩ He. This completes the proof of the lemma.
Lemma 3.18 Let A = {H1, . . . , Hn} be a complexified hyperplane arrangement
in Cd . Then αz is a covector over the interval greedoid (EA, FA), for every z ∈ Cd .
Proof Recall that a map α : E → {0, +, −, 1} is a covector of an interval greedoid
(E, F ) if and only if there exists X ∈ F such that α(e) = 0 if and only if e ∈ ξ(X),
α(e) = ± if and only if e ∈ Γ (X), and α(e) = 1 otherwise.
Let α = αz be defined as above. Observe that α(h) = 0 if and only if z ∈ h. Let
A = h ∈ E : α(h) = 0 = {Hi : z ∈ Hi } ∪ Hi : z ∈ Hi
⊆ E
and let X be maximal among the elements of FA contained in A.
We show that α(h) = 0 if and only if h ∈ ξ(X) by showing that ξ(X) = A. Suppose
h ∈ ξ(X). Then h ⊇ h ∈X h . Since X ⊆ A, it follows that α(h ) = 0 for all h ∈ X.
Thus, z ∈ h for all h ∈ X. It follows that z ∈ h. Thus, h ∈ A.
Conversely, suppose h ∈ A. Then α(h) = 0. If h = Hi , then {h} ∈ FA, so we
can augment {h} from X until we get a set Y of cardinality X. Since X is maximal
among the feasible sets contained in A and X = Y , Y is maximal as well. Thus,
X ∼ Y , so Y ⊆ ξ(X). In particular, h ∈ ξ(X). On the other hand, if h = Hi , then
i (z) = 0, so Hi ∈ A also. Since {Hi , Hi } ∈ FA, the same argument shows that
h ∈ ξ(X). Thus, A ⊆ ξ(X).
Next we show that α(h) ∈ {+, −} if and only if α(h) ∈ Γ (X). Let h ∈ Γ (X).
Since ξ(X) and Γ (X) are disjoint, it follows from the above that α(h) = 0. So it
suffices to show that α(h) ∈ {0, +, −}. By construction, this is true for h = Hi since
α(Hi ) = σ ( i (z)) ∈ {0, +, −}. If h = Hi , then, by the above description of Γ (X),
we have Hi ∈ ξ(X). So, σ ( i (z)) = 0, which implies α(h) = σ ( i (z)) ∈ {0, +, −}.
Conversely, suppose α(h) ∈ {+, −}. Since α(h) = 0, we have h ∈/ ξ(X), or
equivalently, x∈X x ⊆ h. So if h = Hi , then h ∈ Γ (X). If h = Hi , then we need to show
that Hi ⊇ x∈X x, or equivalently, Hi ∈ ξ(X). Well, σ ( i (z)) = α(Hi ) ∈ {+, −},
so σ ( i (z)) = 0. This implies Hi ∈ ξ(X). Hence, h = Hi ∈ Γ (X).
Finally, it follows from the above that α(h) = 1 if and only if h ∈/ Γ (X) ∪ ξ(X).
Therefore, α is a covector of (EA, FA).
Remark 3.19 The product of two covectors α and β can also be computed as
(α ◦ β)(Hi ), (α ◦ β) Hi
=
(β(Hi ), β(Hi )), if (β(Hi ), β(Hi )) > (α(Hi ), α(Hi )),
(α(Hi ), α(Hi )), otherwise,
where the comparison (β(Hi ), β(Hi )) > (α(Hi ), α(Hi )) is performed in the poset
illustrated in Fig. 6.
Theorem 3.20 If A = {H1, . . . , Hn} is a complexified hyperplane arrangement
in Cd , then G = {αz : z ∈ Cd } is an oriented interval greedoid over (EA, FA).
Proof We show that G satisfies (OG1)–(OG4).
(OG1) Suppose [X] is a flat of (EA, FA) for some X ∈ FA. Let z be a generic
point of x∈X x. The support of αz is the flat [Y ] such that Y is maximal among
feasible sets contained in {h : αz(h) = 0}. Since z is generic, this set is equal to ξ(X).
It follows that Y is equivalent to X, so [X] = [Y ] = supp(αz).
(OG2) Suppose αz ∈ G. Then −αz = α−z because σi ( j (−z)) = σi (− j (z)). So
−αz ∈ G.
(OG3) Let αx , αy ∈ G. For sufficiently small t > 0, we have σ (u + t v) =
σ (u) σ (v) and σ (u + t v) = σ (u) σ (v). It follows that αx+ty = αx ◦ αy for a
sufficiently small t > 0.
(OG4) Let α, β ∈ G and h ∈ S(α, β) such that (α ◦ β)(h) = 1. Pick x, y ∈ Cd such
that α = αx and β = αy . We can assume for all 1 ≤ i ≤ n that the line t x + (1 − t )y,
for 0 < t < 1, does not intersect Hi if x and y are not both contained in Hi (otherwise
perturb x and y slightly).
Since h ∈ S(αx , αy ), we have αx (h) = −αy (h) ∈ {+, −}. Hence, ( h(x)) and
( h(y)) or ( h(x)) and ( h(y)) have opposite signs, where h is the form
associated to h (that is, h = ker( h) or h = ker( h) ). So there exists 0 < t < 1 such that
the real part (or imaginary part) of h(t x + (1 − t )y) is zero. Let γ = αtx+(1−t)y .
Then γ (h) = 0.
Let e ∈/ S(αx , αy ) and (αx ◦ αy )(e) = 1. Suppose first that e = Hi for some i.
Then ( i (x)) = 0 = ( i (y)), for otherwise (αx ◦ αy )(e) = 1. This implies that
(t i (x) + (1 − t ) i (y)) = 0, so γ (e) = σ ( i (t x + (1 − t )y)) is the sign of
i t x + (1 − t )y
= t
i (x) + (1 − t )
i (y) .
Since both of the coefficients t and (1 − t ) are positive and since ( i (x)) and
( i (y)) are not of opposite signs, it follows that γ (e) is the sign of ( i (x)) if it is
nonzero and the sign of ( i (y)) otherwise. This is precisely (αx ◦ αy )(e). Similarly,
if e = Hi ∈/ S(αx , αy ), then γ (e) = (αx ◦ αy )(e).
4 Restriction and Contraction of Oriented Interval Greedoids
4.1 Contraction
This section introduces an operation on oriented interval greedoids that produces an
oriented interval greedoid on the contraction of the underlying interval greedoid. We
begin by studying the relationship between an interval greedoid and its contractions.
4.1.1 Contraction of Interval Greedoids
Let (E, F ) denote an interval greedoid and Φ its lattice of flats. Recall that for
X ∈ F , the contraction of (E, F ) by X is the interval greedoid with feasible sets
F /X = {Y ⊆ E\X : Y ∪ X ∈ F }
and ground set Y ∈F /X Y . We let Φ/X, Γ /X and ξ /X denote the corresponding
notions in the contraction. For Y ∈ F /X, we let (Φ/X)(Y ) denote the flat in the
contraction that contains Y .
Proposition 4.1 Suppose (E, F ) is an interval greedoid and X ∈ F . Then
(
1
) Φ/X ∼= [0ˆ, [X]] ⊆ Φ.
(
2
) If Y ∈ F /X, then (Γ /X)(Y ) = Γ (X ∪ Y ).
(
3
) If Y ∈ F /X, then (ξ /X)(Y ) ⊆ ξ(X ∪ Y ) ∩
Z∈F /X Z.
Proof (
1
) Define a map Φ/X → [0ˆ, [X]] by mapping the flat containing Y (in the
contraction F /X) to the flat [Y ∪ X] of (E, F ). The fact that this map is
welldefined follows from the identity: (F /X)/Y = F /(X ∪ Y ) for X ∈ F and Y ∈ F /X.
This identity also implies that the map is injective. It remains to show that the map
is surjective. Let [Z] ≤ [X]. Then ξ(X) ⊆ ξ(Z). Hence, there exists Z containing
X with Z maximal among the feasible sets contained in ξ(Z). Therefore, Z \X ∈
F /X, and Z \X maps to the flat containing (Z \X) ∪ X = Z , which is [Z] by
Proposition 2.4.
(
2
) Suppose x ∈ (Γ /X)(Y ). Then Y ∪ x ∈ F /X. So (X ∪ Y ) ∪ x ∈ F . That is,
x ∈ Γ (X ∪ Y ). Conversely, suppose x ∈ Γ (X ∪ Y ). Then X ∪ (Y ∪ x) ∈ F , and so
(Y ∪ x) ∈ F /X. That is, x ∈ (Γ /X)(Y ).
(
3
) Let x ∈ (ξ /X)(Y ). Then x ∈ W for some W ∈ F /X that is equivalent (in
F /X) to Y . So x ∈ Z∈F /X Z. And since the map defined in (
1
) is welldefined, we
have [X ∪ Y ] = [X ∪ W ]. Hence, x ∈ W ⊆ ξ(X ∪ W ) = ξ(X ∪ Y ).
We remark that the containment in the previous result can be proper.
4.1.2 Contractions of Oriented Interval Greedoids
Let (E, F , G) be an oriented interval greedoid and Φ = supp(G) the lattice of flats of
(E, F ). Recall that G with the operation ◦ forms a semigroup (Proposition 3.4 and
(OG3)). For A ∈ Φ, let
G≤A = α ∈ G : supp(α) ≤ A .
Then G≤A is a subsemigroup of G. We will show that it is isomorphic to an oriented
interval greedoid over the contraction of (E, F ) by X ∈ F , where A = [X].
Let α be a covector of (E, F ) with supp(α) ≤ [X]. By definition of the
partial order, there exists Y ∈ F /X such that supp(α) = [X ∪ Y ]. Therefore, Y is a
feasible set in the contracted interval greedoid and so it makes sense to talk about
its flat (Φ/X)(Y ). By restricting α to the subset (Γ /X)(Y ), we get a signed flat
((Φ/X)(Y ), α(Γ /X)(Y )) of the contracted interval greedoid. We denote the covector
of this signed flat by conX(α). Then,
⎧ 0, if e ∈ (ξ /X)(Y ),
⎪⎪
conX(α)(e) = ⎨ α(e), if e ∈ (Γ /X)(Y ),
It follows from Proposition 4.1 that if conX(α)(e) = 1, then conX(α)(e) = α(e).
Lemma 4.2 Suppose (E, F ) is an interval greedoid and let X ∈ F . Let α and β be
covectors of (E, F ) with supp(α), supp(β) ≤ [X].
(
1
) (supp /X)(conX(α)) = (Φ/X)(Y ) and (supp /X)(conX(β)) = (Φ/X)(Z),
where Y, Z ∈ F /X satisfy [X ∪ Y ] = supp(α) and [X ∪ Z] = supp(β).
(
2
) conX(α) ◦ conX(β) = conX(α ◦ β).
Proof (
1
) Since conX(α) is the covector of the signed flat ((Φ/X)(Y ), α(Γ /X)(Y )),
it follows from the definition of supp /X that (supp /X)(conX(α)) = (Φ/X)(Y ).
(
2
) We first argue that the supports of the two elements are the same. It follows
from the definition of ◦ that the support of conX(α) ◦ conX(β) is the join of their
supports, so it is (Φ/X)(Y ) ∨ (Φ/X)(Z) by (
1
). Under the isomorphism Φ/X ∼=
[0ˆ, [X]], this corresponds to [X ∪ Y ] ∨ [X ∪ Z], which we can express as [X ∪ W ] for
some W ∈ F /X. Hence, (Φ/X)(Y ) ∨ (Φ/X)(Z) = (Φ/X)(W ). Note that [X ∪ W ]
is also the support of α ◦ β, so (
1
) implies that (supp /X)(conX(α ◦ β)) = (Φ/X)(W ).
Since conX(α) ◦ conX(β) and conX(α ◦ β) are covectors of support (Φ/X)(W ),
to show that they are equal it suffices to show that they agree on (Γ /X)(W ). Let
e ∈ (Γ /X)(W ). Then,
conX(α) ◦ conX(β) (e) =
conX(β)(e), if conX(β)(e) > conX(α)(e),
conX(α)(e), otherwise.
Since (conX(α)◦conX(β))(e) = 1, it follows that neither conX(α)(e) nor conX(β)(e)
is 1. Hence, conX(α)(e) = α(e) and conX(β)(e) = β(e) (see the sentence following
(4.1)). Therefore,
conX(α) ◦ conX(β) (e) =
This is precisely (α ◦ β)(e), which is conX(α ◦ β)(e) by (4.1).
Proposition 4.3 Let (E, F , G) denote an oriented interval greedoid and let X ∈ F .
Then
G/X = conX(α) : α ∈ G and supp(α) ≤ [X]
defines an oriented interval greedoid over the contraction of (E, F ) by X.
Proof (OG1). Let A ∈ Φ/X. Then A = (Φ/X)(Y ) for some Y ∈ F /X, and so [Y ∪
X] ∈ Φ. Since G satisfies (OG1), there exists α ∈ G with supp(α) = [Y ∪ X] ≤ [X].
Then conX(α) ∈ G/X and (supp /X)(conX(α)) = (Φ/X)(Y ) = A by Lemma 4.2.
(OG2) Suppose ν ∈ G/X. Then there exists some β ∈ G such that supp(β) ≤ [X]
and conX(β) = ν. Then −β ∈ G by (OG2), and so −ν = conX(−β) ∈ G/X.
(OG3) Suppose conX(α) and conX(β) are in G/X. Then α ◦ β ∈ G, by (OG3),
and supp(α ◦ β) = supp(α) ∨ supp(β) ≤ [X]. Therefore, conX(α ◦ β) ∈ G/X. By
Lemma 4.2, conX(α ◦ β) = conX(α) ◦ conX(β), so conX(α) ◦ conX(β) ∈ G/X.
(OG4) Suppose conX(α), conX(β) ∈ G/X, and let x ∈ (S/X)(conX(α), conX(β))
such that (conX(α) ◦ conX(β))(x) = 1.
Since conX(α)(x) = − conX(β)(x) ∈ {+, −}, it follows from (4.1) that α(x) =
−β(x) ∈ {+, −}. Hence, x ∈ S(α, β). Since conX(α ◦ β) = conX(α) ◦ conX(β), it
follows that conX(α ◦ β)(x) = 1, which implies that (α ◦ β)(x) = 1. Therefore, (OG4)
applies to α, β and x to guarantee the existence of γ ∈ G satisfying γ (x) = 0 and for
all y ∈/ S(α, β), if (α ◦ β)(y) = 1, then γ (y) = (α ◦ β)(y) = (β ◦ α)(y). We claim
that conX(γ ) satisfies the conditions of (OG4) for G/X.
We first show that supp(γ ) ≤ supp(α ◦ β). Indeed, if (α ◦ β)(y) = 0, then α(y) = 0
and β(y) = 0, so y ∈/ S(α, β). We conclude from (OG4) that γ (y) = (α ◦ β)(y) = 0.
Next we argue that conX(γ )(x) = 0. Since supp(γ ) ≤ supp(α ◦ β), it
follows that (supp /X)(conX(γ )) ≤ (supp /X)(conX(α ◦ β)). Then Proposition 3.3
and the assumption that conX(α ◦ β)(x) = 1 implies that conX(γ )(x) = 1. If
conX(γ )(x) ∈ {+, −}, then γ (x) ∈ {+, −} contradicting the fact that γ (x) = 0.
Therefore, conX(γ )(x) = 0.
Now let y ∈ Z∈F /X Z with y ∈/ (S/X)(conX(α), conX(β)). We claim that y ∈/
S(α, β). If y ∈ S(α, β), then α(y) = −β(y) ∈ {+, −}, and so conX(α)(y) = α(y) =
−β(y) = − conX(β)(y) ∈ {+, −} by (4.1), a contradiction.
Now suppose that conX(α ◦ β)(y) = 1. As above, Proposition 3.3 implies that
conX(γ )(y) = 1. Then the sentence following (4.1) implies that conX(α ◦ β)(y) =
(α ◦ β)(y) and that conX(γ )(y) = γ (y). Hence, (α ◦ β)(y) = 1, so γ (y) = (α ◦
β)(y) = (β ◦ α)(y) by (OG4). Therefore, conX(γ )(y) = (conX(α) ◦ conX(β))(y) =
(conX(β) ◦ conX(α))(y).
The following result identifies G/X with a subsemigroup of G.
Proposition 4.4 Let (E, F , G) denote an oriented interval greedoid and let X ∈ F .
Then there is a semigroup isomorphism
G≤[X] =∼ G/X
given by mapping α ∈ G with supp(α) ≤ [X] to conX(α).
Proof Lemma 4.2 shows this is a semigroup morphism. The morphism is surjective
by definition of G/X. It remains to show that the morphism is injective. Suppose
conX(α) = conX(β). Then supp(α) = supp(β), which can be written as [X ∪ Y ].
Now (Γ /X)(Y ) = Γ (X ∪ Y ), so α and β agree on Γ (X ∪ Y ), and they each are zero
on exactly ξ([X ∪ Y ]), so α and β agree, as desired.
4.2 Restriction
We introduce a restriction operation for an oriented interval greedoid (E, F , G)
that produces an oriented interval greedoid on a restriction of the interval greedoid
(E, F ). We begin by recalling restriction for interval greedoids.
4.2.1 Restriction of an Interval Greedoid
Let (E, F ) denote an interval greedoid and Φ its lattice of flats. If W ⊆ E is an
arbitrary subset, then the restriction of (E, F ) to W is the interval greedoid (W, F W ),
where
F W = {X ∈ F : X ⊆ W }.
To distinguish between objects defined for (E, F ) and (W, F W ), we take the
following convention. If Ξ is an object defined for (E, F ) (for example, its lattice of
flats Φ, the set of continuations Γ ), then Ξ W will denote the corresponding object
defined for (W, F W ) (for example, ΦW , Γ W ).
There is a map Φ → ΦW that maps a flat C ∈ Φ onto the flat μW (W ∩ ξ(C)).
We denote the image of C by CW . Note that if Y ∈ F W , then Y ∈ F and the image
of [Y ] ∈ Φ under Φ → ΦW is the flat in ΦW that contains Y , which by our above
convention is denoted by [Y ]W .
Lemma 4.5 Suppose (E, F ) is an interval greedoid and let W ⊆ E.
(
1
) If Y ∈ F W , then Γ W (Y ) = W ∩ Γ (Y ).
(
2
) If Y ∈ F W and ξ(Y ) ⊆ W , then ξ W (Y ) = ξ(Y ).
(
3
) If A ∈ Φ, then Γ W (AW ) ⊆ W ∩ Γ (A).
(
3
) If A ∈ Φ, then ξ W (AW ) ⊆ W ∩ ξ(A).
Proof (
1
) If Y ∈ F W , then ΓW (Y ) = {y ∈ W \Y : Y ∪ y ∈ F } = W ∩ {y ∈ E\Y :
Y ∪ y ∈ F } = W ∩ Γ (Y ).
(
2
) Suppose Y ∈ F W and ξ(Y ) ⊆ W . The latter assumption implies that all
feasible sets that are equivalent to Y in (E, F ) are contained in W . So they are contained
in F W . Moreover, they are also equivalent in F W since they are all maximal among
the feasible sets contained in ξ(Y ) (see Proposition 2.4).
(
3
) Let Z ∈ AW and let x ∈ Γ W (AW ) = Γ W (Z). By (
1
), x ∈ W ∩ Γ (Z). Since
AW = μW (W ∩ ξ(A)), there exists Y ∈ F containing Z that is maximal among the
feasible sets contained in ξ(A). Thus, [Z] ≥ [Y ] = A, and by Proposition 3.3,
Γ W (Z) = Γ (Z) ∩ W
⊆ Γ (A) ∩ W
∪ ξ(A) ∩ W .
If x ∈ W ∩ ξ(A), then Z ∪ x ∈ W ∩ ξ(A), contradicting that Z is maximal among the
feasible sets contained in W ∩ ξ(A). Therefore, x ∈ W ∩ Γ (A).
(
4
) By definition AW = μW (W ∩ ξ(A)), so the sets contained in AW are the
sets that are maximal among the feasible sets contained in W ∩ ξ(A). Let D be a
maximal feasible set in W ∩ ξ(A), and let C be a set in W that is equivalent to D in
the restriction. We want to show that C is contained in ξ(A).
Since C and D are equivalent in the restriction, they have the same continuations
inside W . Let x be a continuation of C with x ∈/ W . Then D can be augmented
from C ∪ x, and clearly D cannot be augmented from C, so it can be augmented
by x. Thus Γ (C) contains Γ (D), and the converse is also true. So C and D have the
same continuations in the original interval greedoid, and therefore are equivalent. In
particular, C is in ξ(A) as well.
Remark 4.6 The inclusions in (
3
) and (
4
) can be proper, as can be seen in the
following example. Let E = {a, b, c} and F = {∅, {a}, {a, b}, {a, c}}. Then (E, F ) is an
interval greedoid. If W = {b, c}, then F W = {∅}, so
Γ W {a} W = Γ W (∅) = ∅
ξ W {a, b} W = ξ W (∅) = ∅
Γ {a}
ξ {a, b}
∩ W = {b, c},
∩ W = {b, c}.
Proposition 4.7 The map Φ → ΦW defined by C → CW = μW (W ∩ ξ(C)) for all
C ∈ Φ is orderpreserving, surjective and preserves joins: (A ∨ B)W = AW ∨ BW
for all A, B ∈ Φ.
Proof The mapping is orderpreserving since ξ and μW are orderreversing. The
map is surjective since if [Y ]W ∈ ΦW , then it follows that Y ∈ F and that Y is
maximal among the feasible subsets contained in W ∩ ξ([Y ]). So, [Y ] → [Y ]W under
this mapping.
Since A, B ≤ A ∨ B, and since the map is orderpreserving, AW ∨ BW ≤
(A ∨ B)W . Since AW ∨ BW ∈ ΦW , there exists Z ∈ F W such that AW ∨ BW =
[Z]W . Then Z is maximal among the feasible sets contained in ξ W (AW ) ∩
ξ W (BW ) by definition of ∨ (Proposition 2.9). Since AW is the collection of sets
that are maximal among the feasible sets contained in W ∩ ξ(A), it follows that
ξ W (AW ) ⊆ W ∩ ξ(A). Therefore, ξ W (AW ) ∩ ξ W (BW ) ⊆ W ∩ ξ(A) ∩ ξ(B).
So there exists Y containing Z that is maximal among the feasible sets contained
in W ∩ ξ(A) ∩ ξ(B). We now argue that Y is maximal among the feasible sets
contained in W ∩ ξ(A ∨ B). There exists U containing Y that is maximal among
the feasible sets contained in ξ(A) ∩ ξ(B). Thus, [Y ] ≥ [U ] = A ∨ B. This
implies Y ⊆ ξ(Y ) ⊆ ξ(A ∨ B) since ξ is orderreversing. So Y ⊆ W ∩ ξ(A ∨ B).
Since ξ(A ∨ B) ⊆ ξ(A) ∩ ξ(B), it follows that Y is maximal among the feasible
sets contained in W ∩ ξ(A ∨ B). Therefore, (A ∨ B)W = [Y ]W . Since Y ⊇ Z,
we have [Y ]W ≤ [Z]W . Thus, (A ∨ B)W ≤ AW ∨ BW . Therefore, (A ∨ B)W =
AW ∨ BW .
Since F W ⊆ F , there is also a map in the reverse direction ΦW → Φ, defined
by A → [Y ] for any Y ∈ A. Proposition 2.4 implies the map is welldefined, and
the identity (F W )/Y = (F /Y )W = {X ⊆ W \Y : X ∪ Y ∈ F } implies the map is
injective. It is orderpreserving and its image is contained in the interval [[X], 1ˆ],
where X is maximal among the feasible sets contained in W .
Unlike for matroids, for an arbitrary interval greedoid, the lattice of flats ΦW is
not, in general, an interval of Φ. However, if W ⊇ ξ(X), where X is maximal among
the feasible subsets contained in W , then ΦW =∼ [[X], 1ˆ] ⊆ Φ. (This is obtained by
considering the compositions of the maps ΦW → Φ and Φ → ΦW defined above.)
4.2.2 Restricting Covectors
Let W ⊆ E and let α be a covector of (E, F ). Let A = supp(α). It follows from
Lemma 4.5 that Γ W (AW ) ⊆ Γ (A), so (AW , αΓ W (AW ))) is a signed flat of
(W, F W ). Let resW (α) denote the covector of this signed flat:
⎧ 0, if w ∈ ξ W (AW ),
⎪⎪
resW (α)(w) = ⎨ α(w), if w ∈ Γ W (AW ),
for all w ∈ W . Observe that by construction supp W (resW (α)) = supp(α)W . Also
note that by Lemma 4.5, if resW (α)(w) = α(w), then resW (α)(w) = 1.
Example 4.8 (Antimatroid from three colinear points) Let (E, F , G) be the oriented
interval greedoid arising from the convex geometry of three colinear points, x, y, z
with y between x, z. Let W = {x, y}. The covectors of GW are: (±, 1); (0, ±); (0, 0).
For α ∈ G, if α(z) = 0, then resW (α) equals the restriction of α to W . However, if,
for example, α = (+, +, 0), then resW (α) = (+, 1).
As we have just seen in an example, resW (α) cannot necessarily be obtained by
restricting α to W . The following proposition sheds more light on this.
Proposition 4.9 Suppose (E, F ) is an interval greedoid and let W ⊆ E.
(
1
) If α is a covector of (E, F ) and A = supp(α), then resW (α) = αW if and only if
ξ W (AW ) = W ∩ ξ(A) and Γ W (AW ) = W ∩ Γ (A).
(
2
) If α and β are covectors of (E, F ), then resW (α ◦ β) = resW (α) ◦ resW (β).
Proof (
1
) This is obvious from the definitions.
(
2
) Let A = supp(α) and B = supp(β). By Sect. 4.2.1, AW ∨ BW = (A ∨ B)W .
Thus, resW (α) ◦ resW (β) and resW (α ◦ β) have the same support. It is therefore clear
that they coincide.
4.2.3 Restriction of an Oriented Interval Greedoid
The following results shows that the covectors obtained by restricting the covectors of
an oriented interval greedoid (E, F , G) satisfy the first three axioms for an oriented
interval greedoid.
Proposition 4.10 Suppose (E, F , G) is an oriented interval greedoid. If W ⊆ E,
then (W, F W , GW ) satisfies (OG1), (OG2) and (OG3), where
GW = resW (α) : α ∈ G .
Proof By construction, we have that GW is a collection of covectors of (W, F W )
and that supp W (resW (α)) = supp(α)W ∈ ΦW .
(OG1) Let A ∈ ΦW and let Y ∈ A. Then [Y ] ∈ Φ and so there exists α ∈ G with
supp(α) = [Y ]. So resW (α) ∈ GW and supp W (resW (α)) = supp(α)W = AW .
(OG2) If resW (α) ∈ GW , then −α ∈ G. Hence, − resW (α) = resW (−α) ∈ GW .
(OG3) Suppose resW (α), resW (β) ∈ GW and let A = supp(α) and B = supp(β).
Then α ◦ β ∈ G, and so resW (α ◦ β) ∈ GW . By Proposition 4.9, resW (α ◦ β) =
resW (α) ◦ resW (β), so resW (α) ◦ resW (β) ∈ GW .
Although (OG4) may not hold for an arbitrary restriction, it does hold for certain
restrictions, so we get an oriented interval greedoid.
Theorem 4.11 Suppose (E, F , G) is an oriented interval greedoid and let W ⊆ E. If
resW (α) = αW for all α ∈ G, then (W, F W , GW ) is an oriented interval greedoid.
Proof (OG1)–(OG3) hold by Proposition 4.10. The assumption that resW (α) = αW
for all α ∈ G means that (OG4) for G implies (OG4) for (W, F W , GW ).
Sections 4.3 and 4.4 describe restriction to two particular types of subset of E.
4.3 Restriction to Γ (∅)
In this section we treat restriction to Γ (∅).
Proposition 4.12 Suppose (E, F , G) is an oriented interval greedoid. Then the
restriction (Γ (∅), F Γ (∅), GΓ (∅)) is an oriented matroid, and
GΓ (∅) = {αΓ (∅) : α ∈ G}.
Proof Since (Γ (∅), F Γ (∅)) is a matroid, it will follow from Theorem 3.12 that
(Γ (∅), F Γ (∅), GΓ (∅)) is an oriented matroid once we show that it is an oriented
interval greedoid. By Proposition 4.9 and Theorem 4.11, we need only show that
Γ Γ (∅)(AΓ (∅)) = Γ (∅) ∩ Γ (A) and ξ Γ (∅)(AΓ (∅)) = Γ (∅) ∩ ξ (A).
Suppose x ∈ Γ (∅) ∩ Γ (A). Let Y ∈ AΓ (∅). Then Y ⊆ Γ (∅) ∩ ξ (A), so there
exists Z ⊇ Y such that Z is maximal among the sets in F contained in ξ (A). By
Proposition 2.4, Z ∈ A, so Γ (A) = Γ (Z). Hence, Z ∪ x ∈ F since x ∈ Γ (A).
Also, {x} ∈ F since x ∈ Γ (∅). Therefore, (IG3) applied to ∅ ⊆ Y ⊆ Z implies Y ∪
x ∈ F . Since Y ∪ x ⊆ Γ (∅), we have Y ∪ x ∈ F Γ (∅). So x ∈ Γ Γ (∅)(AΓ (∅)). This
establishes one inclusion. The reverse inclusion follows from Lemma 4.5.
It remains to show that ξ Γ (∅)(AΓ (∅)) = Γ (∅) ∩ ξ (A). Suppose x ∈ Γ (∅) ∩
ξ (A). Then {x} ∈ F since x ∈ Γ (∅). Therefore, there exists Y containing x such that
Y is maximal among the feasible sets contained in Γ (∅) ∩ ξ (A). Then Y ∈ AΓ (∅).
Therefore, Y ⊆ ξ Γ (∅)(AΓ (∅)), and so x ∈ ξ Γ (∅)(AΓ (∅)). This, combined with
Lemma 4.5, establishes the equality ξ Γ (∅)(AΓ (∅)) = Γ (∅) ∩ ξ (A).
To simplify notation, we write ξ (X) for ξ ([X]) for any feasible set X ∈ F .
We show that restriction to ξ (X) for X ∈ F produces an oriented interval greedoid
(ξ (X), F ξ(X), Gξ(X)) and that there is a semigroup isomorphism
∼
Gξ(X) −=→ G≥α = {β ∈ G : β ≥ α} = {α ◦ β : β ∈ G},
where α is any covector with supp(α) = [X].
Lemma 4.13 Suppose (E, F ) is an interval greedoid and let X ∈ F and A ∈ Φ .
(
1
) Aξ(X) = A ∨ [X].
(
2
) ξ ξ(X)(Aξ(X)) = ξ (A ∨ [X]) = ξ (A ∨ [X]) ∩ ξ (X).
(
3
) Γ ξ(X)(Aξ(X)) = Γ (A ∨ [X]) ∩ ξ (X).
Proof (
1
) Aξ(X) is μξ(X)(ξ (X) ∩ ξ (A)), the flat that consists of the sets that are
maximal among the feasible sets contained in ξ (X) ∩ ξ (A). By Proposition 2.9, this
is A ∨ [X].
(
2
) This follows from (
1
) since all feasible sets in A ∨ [X] are contained in ξ (X).
(
3
) Write A ∨ [X] = [Y ] for some Y ∈ F . Then [X] ≤ [Y ], so Y ∈ F ξ(X) since
ξ (Y ) ⊆ ξ (X). Thus, Γ ξ(X)(Aξ(X)) = Γ ξ(X)(Y ) = {y ∈ ξ (X)\Y : Y ∪ y ∈ F } =
ξ (X) ∩ Γ (Y ) = ξ (X) ∩ Γ (A ∨ [X]).
Lemma 4.14 Suppose (E, F , G) is an oriented interval greedoid and let X ∈ F .
Gξ(X) = resξ(X)(α) : α ∈ G and supp(α) ≥ [X]
= αξ(X) : α ∈ G and supp(α) ≥ [X] .
Proof We begin by proving the first equality. We show that if β ∈ G, then there
exists α ∈ G with supp(α) ≥ [X] and resξ(X)(α) = resξ(X)(β). Let β ∈ G and let
B = supp(β). Since G satisfies (OG1), there exists γ ∈ G such that supp(γ ) = [X].
Since supp(γ ◦ β) = [X] ∨ B and Bξ(X) = B ∨ [X] = (B ∨ [X])ξ(X) by Lemma 4.13,
Therefore, resξ(X)(γ ◦ β) = resξ(X)(β) if and only if (γ ◦ β)(x) = β(x) for x ∈
Γ ξ(X)(Bξ(X)). So suppose x ∈ Γ ξ(X)(Bξ(X)). By Lemma 4.13, x ∈ ξ(X) ∩ Γ (B ∨
[X]), and by Proposition 3.3, x ∈ ξ(X) ∩ (Γ (B) ∪ Γ (X)). This implies x ∈ Γ (B)
because ξ(X) ∩ Γ (X) = ∅. Therefore, (γ ◦ β)(x) = β(x), and so resξ(X)(γ ◦ β)(x) =
β(x) = resξ(X)(β)(x).
We now prove the second equality. Let resξ(X)(α) such that α ∈ G and A =
supp(α) ≥ [X]. By Lemma 4.13 we have Aξ(X) = A ∨ [X] = A, ξ ξ(X)(Aξ(X)) =
ξ(X) ∩ ξ(A) and Γ ξ(X)(Aξ(X)) = ξ(X) ∩ Γ (A). Therefore, by Proposition 4.9,
resξ(X)(α) = αξ(X) for all α ∈ G such that supp(α) ≥ [X].
Theorem 4.15 Let (E, F , G) denote an oriented interval greedoid and let X ∈ F .
Then (ξ(X), F ξ(X), Gξ(X)) is an oriented interval greedoid.
Proof By Proposition 4.10 we need only show that Gξ(X) satisfies (OG4). Let
resξ(X)(α) and resξ(X)(β) be covectors in Gξ(X). By Lemma 4.14, we can assume
supp(α) ≥ [X], supp(β) ≥ [X], resξ(X)(α) = αξ(X) and resξ(X)(β) = βξ(X).
Let x ∈ S(resξ(X)(α), resξ(X)(β)) with resξ(X)(α ◦ β)(x) = 1. Then x ∈ S(α, β)
and (α ◦ β)(x) = 1. By (OG4) applied to G, there exists γ ∈ G such that γ (x) = 0
and for all y ∈/ S(α, β), if (α ◦ β)(y) = 1, then γ (y) = (α ◦ β)(y) = (β ◦ α)(y).
We show that resξ(X)(γ ) satisfies the conditions of (OG4). Let A = supp(α), B =
supp(β) and C = supp(γ ). Observe that C ∨ [X] ≤ A ∨ B: indeed, if (α ◦ β)(e) = 0,
then α(e) = 0, so e ∈/ S(α, β), which implies that γ (e) = (α ◦ β)(e) = 0.
We first argue that resξ(X)(γ )(x) is 0. By construction, it is either γ (x) or 1.
Suppose it is 1. Then x ∈/ ξ ξ(X)(Cξ(X)) ∪ Γ ξ(X)(Cξ(X)) = ξ(C ∨ [X]) ∪ Γ (C ∨ [X]).
By Proposition 3.3, since C ∨ [X] ≤ A ∨ B, x ∈/ Γ (A ∨ B) ∪ ξ(A ∨ B). This implies
(α ◦ β)(x) = 1, which contradicts (α ◦ β)(x) = 1. Thus, resξ(X)(γ )(x) = γ (x) = 0.
Let y ∈/ S(resξ(X)(α), resξ(X)(β)). Then y ∈/ S(α, β). Suppose resξ(X)(α ◦
β)(y) = 1. Then (α ◦ β)(y) = 1. This implies, as above, that resξ(X)(γ )(y) = 1. Thus,
resξ(X)(γ )(y) = γ (y) = (α ◦ β)(y). By Lemma 4.14, resξ(X)(α ◦ β) = (α ◦ β)ξ(X),
so resξ(X)(γ )(y) = resξ(X)(α ◦ β)(y) = (resξ(X)(α) ◦ resξ(X)(β))(y).
The following result identifies the semigroup Gξ(X) with a subsemigroup of G.
Proposition 4.16 Let (E, F , G) denote an oriented interval greedoid and let X ∈ F .
Then resξ(X)(β) → α ◦ β defines a semigroup isomorphism
∼
=
Gξ(X) −→ G≥α = {β ∈ G : β ≥ α},
where α is any covector with supp(α) = [X].
Proof Define a map g : G≥α → Gξ(X) by β → resξ(X)(β). Then g is a semigroup
morphism by Proposition 4.9. Define a map f : Gξ(X) → G≥α by f (resξ(X)(β)) =
α ◦ β.
We argue that f is welldefined. Suppose β, γ ∈ G with B = supp(β) and C =
supp(γ ), and suppose resξ(X)(β) = resξ(X)(γ ). Then the support of resξ(X)(β) =
resξ(X)(γ ) is B ∨ [X] = C ∨ [X]. This implies that supp(α ◦ β) = supp(α ◦ γ )
because supp(α) = [X]. Therefore, to show α ◦ β = α ◦ γ it suffices to show that they
agree on Γ ([X] ∨ B). Let x ∈ Γ ([X] ∨ B). Then x ∈ Γ (X) ∪ ξ(X) by
Proposition 3.3. If x ∈ Γ (X), then (α ◦ β)(x) = α(x) = (α ◦ γ )(x). So suppose x ∈ ξ(X).
Then (α ◦ β)(x) = β(x) and (α ◦ γ )(x) = γ (x). Moreover, x ∈ Γ ξ(X)(Bξ(X)) and
x ∈ Γ ξ(X)(Cξ(X)) by Lemma 4.13. So resξ(X)(β)(x) = β(x) and resξ(X)(γ )(x) =
γ (x). Since resξ(X)(β) = resξ(X)(γ ), we have (α ◦ β) = (α ◦ γ ).
Now f is a semigroup morphism since α ◦ β ◦ α = α ◦ β for all covectors α and
β (see Proposition 3.4). To complete the proof observe that f ◦ g and g ◦ f are the
identity morphisms of Gξ(X) and G≥α , respectively.
5 Structure of Oriented Interval Greedoids
5.1 G is a Graded Poset
The next result generalizes [13, Lemma 4.1.12] from oriented matroids to oriented
interval greedoids.
Lemma 5.1 Let (E, F , G) denote an oriented interval greedoid. Suppose α, β ∈ G
with supp(α) ≤ supp(β) and α ≤ β. Then there exists δ ∈ G such that δ β and for
all x ∈/ S(α, β), if β(x) = 1, then δ(x) = β(x).
Proof Let A = supp(α) and B = supp(β). Suppose the result is not true. Of all
α, β ∈ G that violate the result choose a pair with S(α, β) minimal. If S(α, β) = ∅,
then α ≤ β by Lemma 3.7, contradicting the assumption that α ≤ β. Therefore,
S(α, β) = ∅. Let y ∈ S(α, β). If (α ◦ β)(y) = 1, then (β ◦ α)(y) = 1 and so β(y) = 1
because A ≤ B implies (β ◦ α) = β. This contradicts the fact that y ∈ S(α, β).
Therefore, (α ◦ β)(y) = 1. (OG4) implies there exists γ ∈ G with γ (y) = 0 and for all
x ∈/ S(α, β), if β(x) = (β ◦ α)(x) = 1, then γ (x) = (α ◦ β)(x) = (β ◦ α)(x) = β(x).
We argue that S(γ , β) S(α, β). Suppose e ∈/ S(α, β). Then either β(e) = 1 or
γ (e) = β(e). In both cases e ∈/ S(γ , β). Since y ∈ S(α, β) and y ∈/ S(γ , β) (because
γ (y) = 0), the inclusion is proper.
Let C = supp(γ ). We argue that C < B by showing that β(e) = 0 implies
γ (e) = 0 and that B = C. If β(e) = 0, then e ∈/ S(α, β), so β(e) = 1 (not possible) or
γ (e) = β(e) = 0. Since γ (y) = 0 and β(y) ∈ {+, −}, we have B = C.
We argue that S(γ , β) = ∅. Suppose S(γ , β) = ∅. Then γ < β by Lemma 3.7. Let
δ ∈ G denote a coatom in the interval [γ , β] of the poset G and let D = supp(δ). We
will argue that δ satisfies the result, contradicting our assumption that no such δ exists.
First note that δ β by the choice of δ. It remains to show that for all x ∈/ S(α, β), if
β(x) = 1, then δ(x) = β(x). Let x ∈/ S(α, β). If β(x) = 1, then γ (x) = β(x). Since
γ (x) ≤ δ(x) ≤ β(x) and γ (x) = β(x), it follows that δ(x) = β(x). And if β(x) = 1,
then γ (x) = 0, so δ(x) ≥ γ (x) > 0.
We argued above that C < B and S(γ , β) = ∅. Hence, γ ≤ β by Lemma 3.7.
Since ∅ = S(γ , β) S(α, β), the minimality of S(α, β) implies there exists δ ∈ G
such that δ β and for all x ∈/ S(γ , β), if β(x) = 1, then δ(x) = β(x). But since
S(γ , β) S(α, β), if x ∈/ S(α, β), then x ∈/ S(γ , β). Thus, for all x ∈/ S(α, β), if
β(x) = 1, then δ(x) = β(x). This contradicts the assumption that no such δ exists for
the pair α and β. We have arrived at a contradiction; so the result is true.
Example 5.2 Figure 7 illustrates Lemma 5.1 for the antimatroid corresponding to the
convex geometry on three colinear points.
The partial order on the set of all covectors restricts to a partial order on G. This
next result shows that G is a graded poset and describes its rank function.
Proposition 5.3 Let G be an oriented interval greedoid over (E, F ). Then supp :
G → Φ is a coverpreserving poset surjection of G onto Φ satisfying
supp(α ◦ β) = supp(α) ∨ supp(β).
In particular, G is graded of rank equal to the rank of Φ. The rank of α ∈ G is the
rank of supp(α) ∈ Φ.
Proof The identity follows immediately because if A = supp(α) and B = supp(β),
then supp(α ◦ β) = A ∨ B, by definition of the product. The fact that supp is a
surjection of posets follows from its definition and axiom (OG1). It remains to show that
supp is coverpreserving.
Suppose α β. Let A = supp(α) and B = supp(β). Suppose there exists C ∈ Φ
such that A < C < B. Let G = Gξ(A). Let α and β be the elements of G
corresponding to α, β.
Since supp : G → [A, 1ˆ] is surjective, there exists ∈ G with supp( ) = C. Let
γ = α ◦ . Then α < γ and supp(γ ) = A ∨ C = C. If γ ≤ β , then γ = β ,
contradicting that C < B. Hence, γ ≤ β . By Lemma 5.1 there exists δ ∈ G such
that δ β and for all x ∈/ S(γ , β ), if β (x) = 1, then δ (x) = β (x). Let D =
supp(δ ). Since δ ∈ G , A < D. Thus α < α ◦ δ < α ◦ β = β (see Proposition 3.4),
contradicting that α β.
5.2 Oriented Interval Greedoids of Rank 1
Let (E, F , G) be an oriented interval greedoid and let Φ be its lattice of flats. The
previous result shows that G is a graded lattice and that its rank is equal to that of Φ.
We define the rank of (E, F , G) to be the rank of G (equivalently, the rank of Φ).
We first make a useful observation about arbitrary oriented interval greedoids.
Lemma 5.4 Suppose (E, F , G) is an oriented interval greedoid. Let 0ˆ be the
minimal element of Φ. Then there is a unique element of G with support 0ˆ.
Proof By (OG1), there exists ε ∈ G with supp(ε) = 0ˆ . Then Γ (supp(ε)) = ∅, so
ε(e) ∈ {0, 1} for all e ∈ E. Thus, ε is determined by F , and consequently is the
unique element of G with support 0ˆ .
We will consistently denote the unique element of G with support 0ˆ by ε.
The next result describes the oriented interval greedoids of rank 1.
Proposition 5.5 Suppose (E, F , G) is an oriented interval greedoid of rank 1. Then
G contains exactly three elements, and its Hasse diagram is
−β
ε
Proof Since G has rank 1, Φ contains exactly two elements, its minimal and maximal
elements 0ˆ and 1ˆ = [∅], respectively.
By the previous lemma, ε is the unique element of G with support 0ˆ. We now show
that there exist exactly two elements in G of support 1ˆ. By (OG1), there exists β ∈ G
such that supp(β) = 1ˆ. Then −β ∈ G by (OG2). Since Γ (∅) = ∅, β = −β. So G
contains at least two elements of support 1ˆ.
Let α ∈ G, α = β and supp(α) = 1ˆ. Let y ∈ Γ (∅), y ∈/ S(α, β). Then let δ be the
vector guaranteed by Lemma 5.1. Then δ(y) = β(y). But δ = ε, so this is impossible.
It follows that S(α, β) = Γ (∅); in other words, α = −β.
5.3 Oriented Interval Greedoids of Rank 2
Proposition 5.6 Suppose (E, F , G) is an oriented interval greedoid of rank 2. Then
G is isomorphic to the semigroup of covectors of an oriented matroid of rank 2, or
the Hasse diagrams of G and Φ are, respectively, the following two posets.
−β
γ
Proof There are two cases to consider.
Case 5.6.1 Suppose Φ contains at least two coatoms. By Proposition 4.12, the
restriction GΓ (∅) is an oriented matroid. The map C → [Y ] for any Y ∈ C embeds
ΦΓ (∅) into the interval [[X], 1ˆ] of Φ, where X is the maximal among the feasible
sets contained in Γ (∅) (see Sect. 4.2.1). Since every coatom of Φ is of the form
[{x}] for some x ∈ Γ (∅), there is a bijection between the coatoms of Φ and those of
ΦΓ (∅). Therefore, [X] = 0ˆ, so ΦΓ (∅) =∼ Φ. So GΓ (∅) is a rank 2 oriented matroid.
We argue that the map γ → γ Γ (∅) is an isomorphism G =∼ GΓ (∅). By
Proposition 4.12 we need only show that this is an injection.
Let γ , γ ∈ G and suppose γ Γ (∅) = γ Γ (∅). Then supp(γ )Γ (∅) = supp(γ )Γ (∅).
Since ΦΓ (∅) =∼ Φ, it follows that supp(γ ) = supp(γ ). This implies that γ (x) is 0
or 1 if and only if γ (x) is 0 or 1, respectively. Let C = supp(γ ) = supp(γ ).
If C = 0ˆ, then γ = γ since there is a unique element of G with support 0ˆ. If
C = 1ˆ = [∅], then γ Γ (∅) = γ Γ (∅) implies that γ = γ since they agree on Γ (∅).
Let C 0ˆ . Suppose γ = γ . Arguing as in the end of Proposition 5.5, we conclude
Γ (C) = S(γ , γ ). Since γ and γ agree on Γ (∅) and disagree on Γ (C), it follows
γ Γ (∅) = γ Γ (∅) takes values in {0, 1}. Thus, CΓ (∅) = 0ˆ, which implies C = 0ˆ,
contradicting that C 0ˆ. Thus, γ = γ .
Case 5.6.2 Suppose Φ contains exactly one coatom. Then [X] is this coatom, so
X = {x} for some x ∈ E. By Lemma 5.4, ε is the unique element of G with support 0ˆ.
By (OG1), there exists γ ∈ G such that supp(γ ) = [X]. By arguing as in
Proposition 5.5, we conclude that γ = −γ and that if ν ∈ G with supp(ν) = [X], then ν = γ
or ν = −γ . Hence, there are exactly two elements in G of support [X].
By (OG1), there exists β ∈ G such that supp(β) = [∅]. By (OG2) −β ∈ G. As
above, we have β = −β. Γ (∅) ∩ Γ (X) = ∅, since if y ∈ Γ (∅) ∩ Γ (X), then y ∈/
ξ(X), so [y] = [x], contradicting our assumption that Φ has only one coatom. Thus
γ , −γ < β, −β.
Let ν ∈ G such that supp(ν) = 1ˆ. By arguing as before (using Lemma 5.1), it
follows that ν = β or ν = −β.
5.4 Intervals of Length Two
Let Gˆ denote the poset obtained from G by adjoining a maximal element 1ˆ. We prove
that all intervals of length two in Gˆ contain exactly four elements.
Proposition 5.7 Suppose G is an oriented interval greedoid. Then all intervals in Gˆ
of length two contain exactly four elements.
Proof Let α, β, γ ∈ Gˆ such that α γ β. The case where β = 1ˆ was proved in
Proposition 5.5, so suppose β ∈ G. Let A = supp(α), B = supp(β) and C = supp(γ ).
By Proposition 5.3, A C B in Φ.
Let supp(α) = [X] for some X ∈ F . By Proposition 4.16 and Proposition 3.4,
Gξ(X) =∼ G≥α (as posets), so {δ ∈ G : α δ β} ∼= {δ ∈ Gξ(X) : αξ(X) δ βξ(X)}.
Thus, by passing to Gξ(X) we can suppose that A = 0ˆ.
Let supp(β) = [Y ] for some Y ∈ F . By Proposition 4.4, G/Y ∼= G≤[Y ] = {ν ∈ G :
supp(ν) ≤ [Y ]}. Since Φ/Y ∼= [0ˆ, [Y ]] ⊆ Φ (Proposition 4.1), by passing to G/Y , we
can suppose that B = 1ˆ, and therefore that Φ is a lattice of rank 2.
Proposition 5.6 classified the oriented interval greedoids of rank 2 as being either
an oriented matroid of rank 2 or having the Hasse diagram shown in the statement of
Proposition 5.6. For the latter situation a quick inspection of the given poset
establishes the result. And for the former situation, it is wellknown that this result holds
for oriented matroids [13, Theorem 4.1.14].
5.5 The Underlying Oriented Matroid
Let (E, F , G) be an oriented interval greedoid. The top element in the poset of flats
Φ is [∅], and by Proposition 3.3 it follows that Γ (∅) ⊆ Γ (A) ∪ ξ(A) for any flat
A ∈ Φ. This implies that α(x) ∈ {0, +, −} for any α ∈ G and any x ∈ Γ (∅).
Moreover, Γ (∅) is the largest subset of E with this property: if α is maximal in G, then
supp(α) = [∅] and α(x) = 1 if and only if x ∈/ Γ (∅). This observation implies that
the restriction to Γ (∅) produces an oriented interval greedoid whose covectors take
values in {0, +, −}. Thus, GΓ (∅) is an oriented matroid. Alternatively, one can note
that the restriction (Γ (∅), F Γ (∅)) is a matroid and appeal to Theorem 3.12.
Definition 5.8 Let G denote an oriented interval greedoid over (E, F ). The
underlying oriented matroid of G is G = GΓ (∅).
The lattice of flats Φ of G is a geometric lattice because G is an oriented matroid.
Moreover, it is isomorphic to the sublattice of Φ generated by all the coatoms.
5.6 The Tope Graph
A tope of an oriented interval greedoid (E, F , G) is a covector that is maximal in G
with respect to the partial order on covectors. Alternatively, topes are covectors whose
support is 1ˆ = [∅]. A subtope of G is a covector in G that is covered by some tope.
From Proposition 5.7 it follows that every subtope is covered by exactly two different
topes. Two topes are said to be adjacent if there exists a subtope that is covered by
both topes.
The tope graph T(G), or just T, of G is the graph with one vertex for each tope of
G and an edge between two vertices if the corresponding topes are adjacent.
Lemma 5.9 Suppose G is an oriented interval greedoid. Then the tope graph of G is
isomorphic to the tope graph of the underlying oriented matroid G of G.
Proof First we will show that topes of G are in onetoone correspondence with
the topes of G. Suppose α is a tope in G. Then supp(α) = [∅] = {∅}, and so
∅
supp Γ (∅)(resΓ (∅)(α)) = { } = 1ˆ ∈ ΦΓ (∅). Thus, resΓ (∅)(α) is a tope of G.
Conversely, suppose resΓ (∅)(α) is a tope of G. Then supp Γ (∅)(resΓ (∅)(α)) =
[∅]Γ (∅) = {∅}. Let A = supp(α). Then AΓ (∅) = {∅}, so ∅ is maximal among
the feasible sets contained in Γ (∅) ∩ ξ(A). This implies that Γ (∅) ∩ ξ(A) = ∅ (if
x ∈ Γ (∅) ∩ ξ(A), then {x} ∈ F because x ∈ Γ (∅), contradicting that maximality
of ∅). If A = [∅], then A ≤ [{y}] for some y ∈ Γ (∅). Hence, y ∈ Γ (∅) ∩ ξ(A),
contradicting that Γ (∅) ∩ ξ(A) = ∅. Thus, A = [∅]. So α is a tope of G.
Let α and β be topes in G and suppose resΓ (∅)(α) = resΓ (∅)(β). We show
that α = β by showing that they agree on Γ (supp(α)) = Γ (supp(β)) = Γ (∅).
Since supp(α) = supp(β) = [∅], we have Γ (∅) = Γ Γ (∅)(∅) = Γ Γ (∅)([∅]Γ (∅)).
Hence, resΓ (∅)(α)(w) = α(w) and resΓ (∅)(β)(w) = β(w) for all w ∈ Γ (∅). It
follows that α(w) = β(w) for all w ∈ Γ (∅). This establishes the onetoone
correspondence.
Suppose α, β ∈ G are two adjacent topes and let γ ∈ G with γ α, β. Then
supp(γ ) supp(α) = supp(β) = [∅]. Since resΓ (∅) is a semigroup morphism, it
follows that resΓ (∅)(γ ) ≤ resΓ (∅)(α). We cannot have equality since this would
imply that both are topes of GΓ (∅), contradicting that γ is not a tope. We have
supp(γ ) = [{y}] for some y ∈ Γ (∅) since all coatoms of Φ are of this form. Hence,
resΓ (∅)(γ ) = [{y}]Γ (∅) [∅]Γ (∅). Since supp Γ (∅) is coverpreserving, it follows
that resΓ (∅)(γ ) resΓ (∅)(α). Similarly, resΓ (∅)(γ ) resΓ (∅)(β). So resΓ (∅)(α)
and resΓ (∅)(β) are adjacent topes.
Let resΓ (∅)(α), resΓ (∅)(β) ∈ GΓ (∅) be adjacent topes and let resΓ (∅)(γ ) ∈
GΓ (∅) with resΓ (∅)(γ ) ∈ G with resΓ (∅)(γ ) resΓ (∅)(α), resΓ (∅)(β). Since
resΓ (∅)(γ ◦ α) = resΓ (∅)(γ ) ◦ resΓ (∅)(α) = resΓ (∅)(α) and since γ ◦ α and α
are both topes, we have α = γ ◦ α. So γ < α. To show that γ α, it suffices
to show that supp(γ ) [∅]. Let C = supp(γ ). If C is not covered by [∅], then
C ≤ [{x, y}] for some x, y ∈ Γ (∅), x = y. Thus, {x, y} ⊆ Γ (∅) ∩ ξ(C). Let
Y ⊇ {x, y} be maximal among the feasible sets contained in ξ(C) ∩ Γ (∅). By
definition, supp Γ (∅)(resΓ (∅)(γ )) = CΓ (∅) is the flat containing Y . Since Y  > 2, it
follows that supp Γ (∅)(resΓ (∅)(γ )) is not a coatom of ΦΓ (∅), contradicting that it
is. Hence, γ α. Similarly, γ β. Therefore, α and β are adjacent topes.
6 CWSpheres from Oriented Interval Greedoids
6.1 CWSpheres
The Sphericity Theorem is an important result for oriented matroids which asserts
that there is a certain regular CWsphere associated to any oriented matroid, whose
cells correspond to the covectors of the oriented matroid. It is originally due to
Folkman and Lawrence [21]; see also [13, Theorem 4.3.3]. In this section and the next,
we will prove the corresponding result for oriented interval greedoids.
We recall some topological definitions, following [13, Sect. 4.7].
A ball is a topological space homeomorphic to the usual closed d dimensional
ball, for some nonnegative integer d .
A regular cell complex Δ is a finite set of balls in a Hausdorff topological space
Δ = σ ∈Δ σ with the properties that:
• The interiors of the balls σ ∈ Δ partition Δ.
• For each σ ∈ Δ, the boundary of σ is the union of some elements τ ∈ Δ.
This definition of a regular cell complex is (nontrivially) equivalent to the usual
definition of a regular CWcomplex. (See [13, Sect. 4.7].)
A cell complex Δ is called a regular CWsphere if its geometric realization Δ is
homeomorphic to a sphere.
The face poset of a cell complex is the poset structure on the cells of Δ, ordered
by containment. The augmented face poset of a cell complex is the face poset with a
maximal element 1ˆ adjoined.
We can now state our main theorem for this section more precisely.
Theorem 6.1 For (E, F , G) an oriented interval greedoid, Gˆ is isomorphic to the
augmented face poset of a regular CWsphere.
In the oriented matroid setting, a trivial case of the Sphericity Theorem is when the
oriented matroid arises from a hyperplane arrangement (with, say, all the hyperplanes
containing the origin): taking a small sphere about the origin, and thinking of it as
sliced into pieces by the hyperplanes, one immediately obtains a regular CWsphere
as in the statement of the theorem. Similarly, complexified hyperplane arrangements
(see Sect. 3.4.3) yield examples where it is clear that the above theorem holds, and
which can serve as useful illustrations.
Example 6.2 A more interesting example is the following. We construct a greedoid
on E = {a, b, c, e}. Let F = {a, b, c, ab, ac, bc}. Let the feasible sets be as follows:
∅
F = { } ∪ F ∪
X ∪ {e}  X ∈ F .
Consider a twodimensional sphere, draw three great circles on it, each passing
through the north and south poles, and mark the points where each great circle crosses
the equator. The result is a regular CWsphere with 6 twocells, 12 onecells, and 8
zerocells, and it is clear how the cells correspond to a collection of covectors (which
do indeed form an oriented interval greedoid structure for (E, F )).
This example can be generalized to start with F being the nonempty feasible sets
of an arbitrary matroid; then any oriented structure for that matroid gives rise to an
oriented interval greedoid structure for the corresponding greedoid.
The order complex of a bounded poset P is the simplicial complex consisting of
chains in P \ { 0ˆ, 1ˆ}. Taking a barycentric subdivision of the CWsphere in the previous
theorem, we obtain the following.
Corollary 6.3 The order complex of Gˆ is a simplicial sphere.
Proof of Theorem 6.1 As in the proof of the Sphericity Theorem in [13], the main
technical tool required in the proof is the notion of recursive coatom ordering.
Let P be a graded, bounded poset. A linear ordering of its coatoms, q1, . . . , qr , is
called a recursive coatom ordering if it satisfies:
(i) [0ˆ, qi ] admits a recursive coatom ordering in which the coatoms of [0ˆ, qi ] which
lie below some qj with j < i, come first;
(ii) any element lying below qi and also below some qj for j < i, necessarily lies
below a coatom of [0ˆ, qi ] which lies below some qk with k < i.
The unique linear order on the (single) coatom of a bounded poset of rank 1 is
considered to be a recursive coatom ordering, so any linear order on the coatoms of a
bounded poset of rank 2 is a recursive coatom ordering. Recursive coatom orderings
are dual to recursive atom orderings, which go back to [16]. The concept has been
extended to nongraded posets [17], but we shall not need that here.
The fundamental technical result is the following lemma, whose proof we defer to
Sect. 6.2.
Lemma 6.4 Gˆ admits a recursive coatom ordering.
A poset is called thin if all intervals of length 2 have cardinality four. By
Proposition 5.7, we know that Gˆ is thin. The following theorem completes our proof.
Theorem 6.5 ([9], [13, Theorem 4.7.24]) P is isomorphic to the face poset of a
shellable regular cell decomposition of the sphere iff P is thin and admits a recursive
coatom ordering.
(We shall not discuss the significance of the “shellable” in the above theorem; the
interested reader is directed to [13].)
We now turn to the proof of the corollary.
Proof of Corollary 6.3 The order complex of the augmented face poset of a regular
cell complex Δ is homeomorphic to Δ [13, Proposition 4.7.8]. (In fact, the order
complex should be thought of as the barycentric subdivision of the regular cell
complex.) The corollary follows.
6.2 A Recursive Coatom Ordering for Gˆ
This section is devoted to the proof of Lemma 6.4, which asserts that Gˆ has a recursive
coatom ordering.
If one chooses a particular tope α of G then there is a natural poset structure on
the topes with respect to which α is the minimum element and −α is the maximum
element, and the Hasse diagram is (a suitable orientation of) the tope graph. This
poset is called T (G, α). (Since the topes of G are identified with the topes of G, this
follows from the analogous statements for oriented matroids; see [13, Sect. 4.2].)
Let α be a tope of G. Consider a maximal chain β in T (G, α), say α = β0 < · · · <
βr = −α. Choose γi to be a common facet of βi−1 and βi . Let Gi = supp(γi ). The
Gi are distinct and include all the coatoms of Φ. Thus, β induces a linear order on
the coatoms of Φ. However (unlike the situation for oriented matroids) this does not
immediately yield a linear order on the coatoms of [ε, α], because there may be more
than one coatom with the same support.
For 1 ≤ i ≤ r , let Gi be the oriented matroid obtained by contracting G to Gi .
Consider the tope poset T (Gi , γi ).
Let Δ be the set of facets of α. Let Δi be the set of facets of α whose support is Gi . (This set could be empty.)
A linear extension of T (Gi , γi ) will be called adapted to α if it contains in order:
(
1
) first, the topes of Gi that lie on the same side as γi of some Gj for j < i,
(
2
) then, the topes that are facets of α,
(
3
) finally, the remaining topes of Gi .
We will need the following lemma:
Lemma 6.6 T (Gi , γi ) admits a linear extension adapted to α.
Proof It is certainly possible to define a linear extension of T (Gi , γi ) which begins
with the elements (
1
) above, since they form a lower order ideal in T (Gi , γi ). In order
to be able to construct a linear extension such that the next elements are those from (
2
)
above, we need to show that any tope below a tope from (
2
) not in (
2
), is contained
in (
1
). If δ is a tope of Δi which is a facet of α, and is a tope lying below δ which
is not a facet of α, it must be separated from α by some Gj with j < i, which shows
that is in (
1
). Thus the linear extension, whose beginning was already described,
can be continued with the set of facets of α, followed by the remaining topes of Gi .
A linear order on Δ will be said to be compatible with β if
(
1
) the elements of Δ are arranged first of all in increasing order by support (so Δ1
comes first, then Δ2, etc.),
(
2
) the elements of Δi are arranged according to a linear order on T (Gi , γi ) which
is adapted to α.
Now we will prove the following:
Proposition 6.7
(
1
) For a tope α in G, and a maximal chain β in T (G, α), any order on the coatoms
of [ε, α] compatible with β is a recursive coatom order.
(
2
) For a tope α in G, any linear extension of T (G, α) is a recursive coatom ordering
for Gˆ.
Proof The proof will be by induction on the rank of G. The base case, when the rank
of G is 1, is trivial. We will assume that (
1
) and (
2
) hold for oriented interval greedoids
of rank less than n; we will prove (
1
) for oriented interval greedoids of rank n, and
then make use of (
1
) to prove (
2
) for oriented interval greedoids of rank n.
Proof of (
1
) Pick a coatom order for [ε, α] which is compatible with β. As part
of this, we are given γi a common facet of βi−1 and βi . Let Gi be the support of γi .
Let Δi be the coatoms of α with support Gi . As part of our coatom order for [ε, α],
we are given a linear order on Δi which is the restriction of a linear extension of
T (Gi , γi ) adapted to α. Fix such a linear extension.
Let δ ∈ Δi be a coatom of [ε, α]. We must define a coatom order for [ε, δ].
Using our chosen linear extension of T (Gi , γi ), we can apply (
2
) to Gˆi , obtaining a
recursive coatom order for [ε, δ]. We must show that this order satisfies the necessary
conditions.
Now, δ is a coatom of two different posets, [ε, α] and Gˆi . Let X be the set of
coatoms of [ε, α] which precede δ with respect to the coatom order on [ε, α], and
let Y be the set of coatoms of Gˆi which precede δ with respect to the fixed linear
extension of T (Gi , γi ). Let Xˇ be the coatoms of [ε, δ] lying below an element of X,
and let Yˇ be the coatoms of [ε, δ] lying below an element of Y . We will now show
that Xˇ and Yˇ coincide.
Let be a coatom of [ε, δ]. By Proposition 4.16, G≥ is itself an oriented greedoid,
so we may assume that = ε, or, in other words, that G is rank 2. By Proposition 5.6,
we know that G is either isomorphic to a rank 2 oriented matroid, or else it is of
the special form described in that proposition. In either case, it is straightforward to
check that ∈ Xˇ iff ∈ Yˇ .
Since we know property (i) of recursive coatom orders holds for our fixed linear
extension of T (Gi , γi ), property (i) also follows for our coatom ordering on [ε, α].
Next, we check property (ii). Let ∈ [ε, δ], which lies under some element ζ ∈ X.
We must show that it also lies below some element of Xˇ .
Again, by restricting, we may assume that = ε. The fact that lies under an
element of X implies, in particular, that X is nonempty, and thus that δ is not the
first coatom in our coatom order on [ε, α]. We will now show that Y is nonempty. If
δ is not the first coatom with support Gi in our recursive coatom order on [ε, α] then
this is clear. So suppose that δ is the first coatom with support Gi in our recursive
coatom order. Since δ is not the first coatom overall, it must be that i > 1. Therefore
γi is not a facet of α, so γi ∈ Y .
Now, since we have assumed that = ε, the fact that Y is nonempty means that
there are elements of Y lying over . Therefore, by property (ii) for the fixed linear
extension of T (Gi , γi ), we know that there are elements of Yˇ lying over . Since
Yˇ = Xˇ , we are done.
Proof of (
2
), assuming (
1
) Pick a linear extension of T (G, α). For each coatom
δ of Gˆ, pick a maximal chain β in T (G, δ) which includes α. Then we claim that
any linear order on the coatoms of [ε, δ], compatible with β, satisfies the necessary
conditions. First of all, it is a recursive coatom order by (
1
).
Second, define Qδ to be the set of coatoms of [ε, δ] which also lie under some ξ
preceding δ in the linear extension of T (G, α). The coatoms of Qδ precede the other
coatoms of δ in any order compatible with β. (In fact, for this, it suffices to know
that an order compatible with β agrees with the order induced by β on the coatoms
of [0ˆ, supp(δ)].) This proves (i).
Thirdly, we check that
[ε, ζ ] = [ε, δ] ∩
[ε, ξ ].
ζ ∈Qδ
ξ preceding δ
The containment of the lefthand side in the righthand side is obvious. For the
opposite inclusion, let ∈ [ε, δ] ∩ [ε, ξ ] for some ξ preceding δ. The topes of G that
contain are exactly the topes of G that contain Γ (∅). By [13, Lemma 4.2.12], this
is an interval I in T (G, α). Since is contained in some ξ preceding δ, we know that
δ is not the minimum element of the interval. Let ρ be covered by δ in I . Since ρ lies
below δ in T (G, α), it precedes δ in the linear extension of T (G, α). Since ρ is in I ,
∈ [0ˆ, ρ]. Finally, since ρ and δ are adjacent topes, they have a common subtope σ
in G. Since, in Gˆ, ρ and δ lie over Γ (∅), σ lies over Γ (∅). Thus supp(σ ) lies over
supp( ).
Let φ be covector of G, such that φΓ (∅) = σ . Now consider ◦ φ. This lies over
, and its support is supp( ) ∨ supp(φ) = supp(φ). Since, in Gˆ, φ and lie below
both δ and ρ, the same is true of ◦ φ, and we are done: we can take ◦ φ as the
common coatom of [ε, δ] and [ε, ρ] lying over .
6.3 Face Enumeration
Here, we prove formulas counting chains in an oriented interval greedoid G. These
results generalize results for oriented matroids [13, Proposition 4.6.2] and for oriented
antimatroids [4].
Let P be a poset. Recall that the Möbius function of P , denoted μP , is the unique
function from pairs (x, y) with x ≤ y in P to Z, such that:
• μP (x, x) = 1.
• For x < y, x≤z≤y μP (z, y) = 0.
Theorem 6.8 Let (E, F , G) be an oriented interval greedoid. Let A1 > · · · >
Ak+1 = 0ˆ be a chain of flats in Φ. Then:
supp−1(A1, . . . , Ak+1) =
k
i=1 B∈[Ai+1,Ai ]
μΦ (B, Ai ) ,
where μΦ is the Möbius function of Φ, and supp−1(A1, . . . , Ak+1) denotes the chains
α1 > · · · > αk+1 in G such that supp(αi ) = Ai .
First, we state and prove the following special case, which generalizes [13,
Theorem 4.6.1].
Proposition 6.9 Let (E, F , G) be an oriented interval greedoid. Then the number of
topes of G is
μΦ (B, 1ˆ) .
Proof One could adapt the proof for oriented matroids to this setting, thus reproving
the result for oriented matroids, but we prefer to assume the result if G is an oriented
matroid; this is [13, Theorem 4.6.1].
Recall that the topes of G are the same as those of G. Applying the proposition
to G, and writing Φ for Φ(G), we need now only show that
B∈Φ
μΦ (B, 1ˆ) =
B∈Φ
Consider the orderpreserving map i : Φ → Φ defined by i([X]) = [X], as
discussed in Sect. 4.2.1. We prove a few more properties of it here.
Lemma 6.10
(
1
) i is a poset isomorphism onto its image.
(
2
) For A, B ∈ Φ, we have i(A ∧ B) = i(A) ∧ i(B).
Proof (
1
) Proposition 4.7 provides a restriction map from Φ to Φ defined by
AΓ (∅) = μΓ (∅)(ξ(A) ∩ Γ (∅)), which is orderpreserving. Since i(A)Γ (∅) = A,
we know i is a poset isomorphism onto its image.
(
2
) Let A, B ∈ Φ. Let C = i(A) ∧ i(B), and let D = i(CΓ (∅)). It is immediate
that D ≥ C. However, we know D ≤ i(A) and D ≤ i(B), so D = C. This implies
that C is in the image of i, so, by (
1
), C = i(A ∧ B).
Thanks to Lemma 6.10 (
1
), we can identify Φ as a subposet of Φ.
Let x ∈ E such that {x} ∈ F . Then, by definition, x ∈ Γ (∅). It follows that every
coatom of Φ is in Φ. Further, since Φ is a geometric lattice, every element of Φ can
be written as a meet (in Φ) of coatoms. Thanks to Lemma 6.10(
2
), it follows that Φ
consists exactly of those elements of Φ that can be written as a meet of coatoms in Φ.
Lemma 6.11
(
1
) If A ∈ Φ \ Φ, then μΦ (A, 1ˆ) = 0.
(
2
) If A ∈ Φ, then μΦ (A, 1ˆ) = μΦ (A, 1ˆ).
Proof (
1
) Since A ∈/ Φ, A cannot be expressed as a meet of coatoms of Φ. It follows
that the meet of the coatoms of [A, 1ˆ] is strictly greater than A. The Crosscut Theorem
(see [10]) now implies μΦ (A, 1ˆ) = 0.
(
2
) We induct on the corank of A. The statement is obvious for A = 1ˆ . For A of
positive corank, we use the formula:
μΦ (A, 1ˆ) = −
μΦ (B, 1ˆ).
Now we observe that, by (
1
), only the terms with B ∈ Φ contribute. By induction,
these terms agree with μΦ (B, 1ˆ), which proves the result.
Proof of Theorem 6.8 The proof goes exactly as in the oriented matroid case, now
that the preparations have been made.  supp−1(Ak) is the number of topes of G/Ak ,
which is Ak≥B μ(B, Ak ), and then the rest of the chain lies in Gξ(Ak ), which
accounts for the remaining terms.
Since Gˆ is an Eulerian poset, it is natural to ask about its cd index. Using the results
of [3], it develops that one can compute the cd index of Gˆ directly from the lattice of
flats Φ of G.
We begin by recalling the necessary notions. Let P a graded poset of rank
n + 1, with minimum element 0ˆ and maximum element 1ˆ. Let S be a subset of
[n] := {1, . . . , n}. The rankselected subposet PS is the subposet of P consisting only
of the elements whose ranks are in S. Let α(S) be the total number of maximal chains
in PS , and let β(S) = T ⊆S (−1)S−T α(T ).
Let Z a, b be the noncommutative polynomial ring in the variables a, b. Define a
monomial uS ∈ Z a, b , for S ⊆ [n] by saying that the ith letter in uS is a if i ∈/ S and
it is b if i ∈ S. The abindex of P is the polynomial Ψ (P ) = S β(S)uS ∈ Z a, b .
When P is an Eulerian poset, Fine noticed that Ψ (P ) can be expressed as a
polynomial with integer coefficients in the noncommuting variables c = a + b and
d = ab + ba [12]. The expression in terms of c and d is called the cd index of P . If
Ψ (P ) can be expressed as a polynomial with integer coefficients in c and 2d , then
the expression is called the c2d index of P .
There is an operation ω : Z a, b → Z c, 2d that acts on monomials by taking
each consecutive pair ab and replacing it by 2d , and replacing all other letters with c.
There is also an involution on Z a, b that reverses monomials. This involution is
denoted by ∗. For P a poset, we denote by P ∗ the dual poset obtained by reversing
the relations of P . It is clear that Ψ (P ∗) = Ψ (P )∗.
The following follows from Theorem 6.8 and the proof of the main result of [3].
Corollary 6.12 Let G be an oriented interval greedoid and Φ its lattice of flats. Then
the c2d index of Gˆ is given by
Ψ (Gˆ) = ω a · Ψ (Φ∗) ∗.
We close with two final remarks. By Proposition 2.9, Φ is a lower semimodular
lattice. It was shown by Stanley [24] that such posets admit an Rlabeling, and thus
there is a combinatorial interpretation of β(S) [8, 25].
Our second remark is that the c2d index of an Eulerian poset P can also be
interpreted as giving the expansion of the Ehrenborg quasisymmetric function of P
[20] with respect to a certain basis, see [6, Corollary 2.2]. From this perspective, the
map ω above corresponds to Stembridge’s θ map [26], which takes quasisymmetric
functions to the peak subalgebra of the quasisymmetric functions. See [6, Sect. 3.3]
for the equivalence between ω and θ . Corollary 6.12 can therefore be rephrased in
the language of quasisymmetric functions. A version of Corollary 6.12, expressed in
those terms, and specialized to the case of oriented antimatroids, is given in [4].
Acknowledgements The authors would like to thank Richard Ehrenborg for pointing out the relevance
of [3], and the referee for helpful comments which improved the exposition. H.T. was partially funded by
an NSERC Discovery Grant.
References
(OG1) Suppose A ∈ Φ is a flat. Let (A, α) be a signed flat of (E, F ) and let α denote the covector of this signed flat. Then α ∈ G and supp(α) = A.
(OG2) Suppose α ∈ G. Let A = supp(α). Then (A, −αΓ (A)) is a signed flat of (E, F ) . The covector of this signed flat is precisely −α . So −α ∈ G.
(OG3) If α , β ∈ G, then α ◦ β is a covector of (E, F ), so α ◦ β ∈ G.
(OG4) Suppose α, β ∈ G and x ∈ S(α, β) satisfies (α ◦ β)(x) = 1 . Let (E, τ ) denote the convex geometry that is complementary to (E, F ) (see Sect. 2.1.2). Let A = supp(α) and B = supp(β). Then A = [X] and B = [Y ] for some X , Y ∈ F . Let X = E\X and Y = E\Y . Then X and Y are closed sets in (E, τ ).
Step 1: We show that x is an extreme point of τ (X ∪ Y ) . Since (α ◦ β)(x) ∈/ {0, 1}, we have x ∈ Γ ( A ∨ B). Since A ∨ B = [E\ τ (X ∪ Y )] by (2.1), we have Γ (A ∨ B) = ext(τ (X ∪ Y )) . Thus, x ∈ ext(τ (X ∪ Y )) .
Step 2: We define γ . Let Z = τ (X ∪ Y ) − x. Since x is an extreme point of τ (X ∪ Y ), it follows that x is not in Z . Hence Z is a closed set not containing x . Let Z = E\Z . Then Z ∈ F and x ∈ Z. Define a map γ : Γ (Z) → {+, −} for y ∈ Γ (Z) as follows: if y ∈ Γ (A ∨ B), then set γ (y) = (α ◦ β)(y); otherwise arbitrarily set γ (y) to be + or − . Let γ be the covector of the signed flat ([Z], γ ).
Step 3: γ has the desired properties. First note that γ ∈ G since γ is a covector of (E, F ) . Next observe that γ (x) = 0 since x ∈ Z ⊆ ξ([Z]) . Let y ∈/ S(α , β).
Suppose that (α ◦ β)(y) = 1 . Then y ∈ Γ (A ∨ B) ∪ ξ(A ∨ B). Since ξ(A ∨ B) = E\τ (X ∪ Y ) ⊆ Z ⊆ ξ([Z]), if y ∈ ξ(A ∨ B), then γ (y) = 0 = (α ◦ β)(y) = (β ◦ α)(y). On the other hand, if y ∈ Γ (A ∨ B) = ext(τ (X ∪ Y )), then y is an extreme point of Z = τ (X ∪ Y ) − x (since y = x) . Equivalently, y ∈ Γ (Z). So , by definition of γ and because y ∈/ S(α, β), γ (y) = (α ◦ β)(y) = (β ◦ α)(y). 1 . Aguiar , M. , Mahajan , S. : Coxeter Groups and Hopf Algebras . Fields Institute Monographs,
vol. 23 . Am. Math. Soc. , Providence ( 2006 ). With a foreword by Nantel Bergeron . MR2225808
(2008d:20072) 2 . Brown , K.S., Diaconis , P. : Random walks and hyperplane arrangements . Ann. Probab. 26 ( 4 ), 1813 
1854 ( 1998 ). MR1675083 (2000k:60138) 3 . Billera , L.J. , Ehrenborg , R. , Readdy , M.: The c2dindex of oriented matroids . J. Comb. Theory , Ser.
A 80( 1 ), 79  105 ( 1997 ). MR1472106 (98h:05051) 4 . Billera , L.J. , Hsiao , S.K. , Provan , J.S.: Enumeration in convex geometries and associated poly
topal subdivisions of spheres . Discrete Comput. Geom . 39 ( 13 ), 123  137 ( 2008 ). MR2383754
(2009b:52046) 5 . Bidigare , P. , Hanlon , P. , Rockmore , D.: A combinatorial description of the spectrum for the Tsetlin
library and its generalization to hyperplane arrangements . Duke Math. J . 99 ( 1 ), 135  174 ( 1999 ).
MR1700744 (2000m:52032) 6 . Billera , L.J. , Hsiao , S.K., van Willigenburg , S. : Peak quasisymmetric functions and Eulerian enumer
ation. Adv. Math. 176 ( 2 ), 248  276 ( 2003 ). MR1982883 (2005a:06003) 7 . Bidigare , T.P. : Hyperplane arrangement face algebras and their associated Markov chains . Thesis
(Ph.D. )University of Michigan. ProQuest LLC , Ann Arbor, MI, 1997 . MR2695489 8 . Björner , A. : Shellable and CohenMacaulay partially ordered sets . Trans. Am. Math. Soc . 260 ( 1 ),
159 183 ( 1980 ). MR570784 (81i:06001) 9 . Björner , A. : Posets, regular CW complexes and Bruhat order . Eur. J. Comb . 5 ( 1 ), 7  16 ( 1984 ).
MR746039 (86e:06002) 10 . Björner , A. : Topological methods . In: Handbook of Combinatorics , pp. 1819  1872 . Amsterdam, El
sevier ( 1995 ). MR1373690 (96m:52012) 11 . Björner , A. : Random walks, arrangements, cell complexes, greedoids, and selforganizing libraries .
In: Building Bridges. Bolyai Soc. Math. Stud. , vol. 19 , pp. 165  203 . Springer, Berlin ( 2008 ).
MR2484640 (2010e:05051) 12 . Bayer , M.M. , Klapper , A.: A new index for polytopes . Discrete Comput. Geom . 6 ( 1 ), 33  47 ( 1991 ).
MR1073071 (91k:52024) 13 . Björner , A. , Las Vergnas , M. , Sturmfels , B. , White , N. , Ziegler , G.M. : Oriented Matroids . Ency
clopedia of Mathematics and Its Applications , vol. 46 . Cambridge Univ. Press, Cambridge ( 1993 ).
MR1226888 (95e:52023) 14 . Brown , K.S.: Semigroups, rings, and Markov chains . J. Theor. Probab . 13 ( 3 ), 871  938 ( 2000 ).
MR1785534 (2001e:60141) 15 . Brown , K.S.: Semigroup and ring theoretical methods in probability . In: Representations of Finite Di
mensional Algebras and Related Topics in Lie Theory and Geometry . Fields Inst. Commun. , vol. 40 ,
pp. 3  26 . Am. Math. Soc. , Providence ( 2004 ). MR2057147 (2005b:60118) 16 . Björner , A. , Wachs , M. : On lexicographically shellable posets . Trans. Am. Math. Soc . 277 ( 1 ), 323 
341 ( 1983 ). MR690055 (84f:06004) 17 . Björner , A. , Wachs , M.L. : Shellable nonpure complexes and posets . I. Trans. Am. Math. Soc . 348 ( 4 ),
1299 1327 ( 1996 ). MR1333388 (96i:06008) 18 . Björner , A. , Ziegler , G.M.: Combinatorial stratification of complex arrangements . J. Am. Math. Soc.
5 ( 1 ), 105  149 ( 1992 ). MR1119198 (92k:52022) 19 . Björner , A. , Ziegler , G.M. : Introduction to greedoids . In: Matroid Applications. Encyclopedia Math.
Appl., vol. 40 , pp. 284  357 . Cambridge Univ. Press, Cambridge ( 1992 ). MR1165545 (94a:05038) 20 . Ehrenborg , R.: On posets and Hopf algebras . Adv. Math . 119 ( 1 ), 1  25 ( 1996 ). MR1383883
(97e:16079) 21 . Folkman , J. , Lawrence , J.: Oriented matroids . J. Comb. Theory, Ser. B 25 ( 2 ), 199  236 ( 1978 ).
MR511992 (81g:05045) 22 . Korte , B. , Lovász , L. , Schrader , R.: Greedoids. Algorithms and Combinatorics , vol. 4 . Springer, Berlin
( 1991 ). MR1183735 (93f:90003) 23 . Saliola , F.V. : On the quiver of the descent algebra . J. Algebra 320 ( 11 ), 3866  3894 ( 2008 ).
MR2464797 (2009i:20078) 24 . Stanley , R.P. : Finite lattices and JordanHölder sets . Algebra Univers . 4 , 361  371 ( 1974 ).
MR0354473 ( 50 #6951) 25 . Stanley , R.P. : Enumerative Combinatorics, vol. 1 . Cambridge Studies in Advanced Mathematics,
vol. 49 . Cambridge Univ. Press, Cambridge ( 1997 ). With a foreword by GianCarlo Rota , Corrected
reprint of the 1986 original. MR1442260 (98a:05001) 26 . Stembridge , J.R. : Enriched P partitions . Trans. Am. Math. Soc . 349 ( 2 ), 763  788 ( 1997 ). MR1389788
(97f:06006) 27 . Steinberg , B. : Möbius functions and semigroup representation theory . J. Comb. Theory, Ser. A 113 ( 5 ),
866 881 ( 2006 ). MR2231092 (2007c:20144)