#### The Number of Triangulations of the Cyclic Polytope C (n , n -4)

Discrete Comput Geom
The Number of Triangulations of the Cyclic Polytope C.n, n¡4/¤
M. Azaola 0
F. Santos 0
0 Departamento de Matema ́ticas, Estad ́ıstica y Computacio ́n, Universidad de Cantabria , Av. de los Castros s/n, 39005 Santander , Spain
We show that the exact number of triangulations of the standard cyclic polytope C .n; n ¡ 4/ is .n C 4/2.n¡4/=2 ¡ n if n is even and ..3n C 11/=2/ 2.n¡5/=2 ¡ n if n is odd. These formulas were previously conjectured by the second author. Our techniques are based on Gale duality and the concept of virtual chamber. They further provide formulas for the number of triangulations which use a specific simplex. We also compute the maximum number of regular triangulations among all the realizations of the oriented matroid of C .n; n ¡ 4/. ¤ This research was partially funded by CajaCantabria and by Grant PB97-0358 of the Spanish Direccio´n General de Investigacio´n Cient´ıfica y Te´cnica
Introduction
By a triangulation of a finite point set A ½ Rd we mean a simplicial complex
geometrically realized in Rd with the vertex set contained in A and which covers the convex
hull of A. If A is the vertex set of a polytope P this definition agrees with the standard
definition of the triangulation of P . The collection of all triangulations of a fixed point
set has attracted attention in recent years for its connections to algebraic geometry [18],
combinatorial topology [
6
], [16] and optimization [8].
The cyclic polytope C .n; d / of dimension d and with n vertices .n > d / is the
convex hull of any n distinct points on the affine moment curve of degree d , defined as
0d .t / :D .t ; t 2; : : : ; t d / 2 Rd , t 2 R. Cyclic polytopes play a central role in geometric
combinatorics for several reasons: They are neighbourly, which implies that they have the
maximum possible number of faces of each dimension among all polytopes of dimension
d with n vertices [19, Theorem 8.23]. They are universal in the sense that for each fixed
n and d there is a number N .n; d / such that any N .n; d / points in general position in
Rd will contain n points which are the vertices of a cyclic polytope C .n; d/ (and no
other polytope has this property) [7, Proposition 9.4.7]. In a context closer to this paper,
the set of triangulations of a cyclic polytope has a bit more structure than the set of
triangulations of an arbitrary polytope. Edelman and Reiner defined two (conjectured to
be isomorphic) poset structures on this set [10], Rambau proved that all triangulations
of C .n; d/ are connected under bistellar operations [13] (while point configurations and
polytopes without this property exist [17]), and the so-called generalized Baues problem
in the case of cyclic polytopes is essentially solved [
1
], [
2
], [15].
Our main result in this paper is a proof of the following closed formula for the number
of triangulations of the cyclic polytope C .n; n ¡ 4/. This formula had previously been
conjectured by the second author. As pointed out by Reiner [16, page 325], this and the
well-known Catalan number for the number of triangulations of a convex polygon are
the only known non-trivial closed formulas counting triangulations of a polytope.
Theorem 1. The number of triangulations of the cyclic polytope C .n; n ¡ 4/ is
² .n C 4/2.n¡4/=2 ¡ n, if n is even (Theorem 2.6) and
² ..3n C 11/=2/ 2.n¡5/=2 ¡ n, if n is odd (Theorem 2.9).
The odd case formula can be written in a way formally closer to the even case as
.®n C ¯/ 2.n¡4/=2 ¡ n, with ® D 3=2p2 D 1:061 and ¯ D 11=2p2 D 3:889. It
is interesting to relate our result to the following numbers of triangulations for other
parameters of the cyclic polytope:
² C .n; 0/ (n copies of the same point) has n triangulations, if multiple points are
dealt with in the natural way.
² C .n; 1/ (n different points along a line) has 2n¡2 triangulations.
² C .n; 2/ (a convex n-gon) has .1=.n ¡ 1//¡2nn¡¡24¢ D 2 ¡22n=n3=2¢ triangulations.
² C .n; n ¡ 4/ (this paper) has 2.n2n=2/ triangulations.
² C .n; n¡3/, C .n; n¡2/ and C .n; n¡1/ have, respectively, n, 2 and 1 triangulations.
The case of C .n; n ¡ 3/ follows from [11]. The others are trivial.
Concerning C .n; n ¡ 5/, its number of triangulations for n up to 12 appears in [
1
] and
[15]. The numbers for C .13; 8/, C .14; 9/, C .15; 10/ and C .16; 11/ are, respectively,
35,789, 159,613, 499,900 and 2,677,865. We thank Jo¨rg Rambau for these numbers,
computed by him with his public software TOPCOM [14]. Dividing the number of
triangulations of C .n; n ¡ 5/ by the number of triangulations of C .n; 2/ gives, for n from 5
to 16, the following intriguing sequence:
1; 0:875; 1; 0:96; 1:20; 1:16; 1:48; 1:33; 1:64; 1:30; 1:49; 0:9987:
The sequence stays surprisingly close to 1, and is neither decreasing nor increasing, even
if we separate odd and even values of n.
¤
Our methods are based on Gale duality and oriented matroid theory. More precisely,
on the concept of virtual chamber introduced in Section 5 of [8]. This same tool was
used in [
4
] to prove that the flip-graph of triangulations of d C 4 points in dimension d
is always connected.
Let A D fa1; : : : ; ang be a finite point configuration in Rd . We homogenize A, which
means that we embed Rd as an affine hyperplane not passing through the origin in RdC1
and consider A as a vector configuration in RdC1. We say that d C 1 is the rank of A and
n ¡ d ¡ 1 its corank.
Definition 2. A triangulation of a vector configuration A is any collection of linear
bases contained in A whose positive spans are (the maximal elements of) a simplicial
fan covering the positive span of A.
For a vector configuration obtained by homogenization of a point configuration, the
definitions of triangulation in the vector and point contexts agree.
Let B be a Gale transform of A. This is any vector configuration B D fb1; : : : ; bng ½
Rn¡d¡1 such that the kernels of the linear maps ei 7! ai and ei 7! bi are orthogonal
complements of one another (fe1; : : : ; eng denotes the standard basis in Rn). Equivalently,
such that PinD1 ai bi D 0 in RdC1 Rn¡d¡1. Observe that there is an implicit bijection
between A and B in this definition, given by the labels.
If A and B are Gale transforms of each other, then a vector .¸1; : : : ; ¸n/ is the
sequence of values of a linear functional on one of them if and only if it is the sequence
of coefficients of a linear dependence in the other. In particular, the oriented matroids of
A and B are dual to each other and, hence, any information of A which depends only
on its underlying oriented matroid such as the set of triangulations of A (see [8]) can be
retrieved from the oriented matroid of B. In the case of interest to us, this translates our
corank 3 problem into a rank 3 one, but we have to characterize triangulations of A in
terms of B. This is done with the concept of virtual chamber, defined below. Observe
that, by Gale duality, each linear basis of A is the complement of a linear basis of B and
vice versa, under the identification between the elements of A and B by their labels.
Definition 3 [8, Section 5]. Let A and B be vector configurations which are dual to
each other. A virtual chamber of B is a collection of linear bases of B whose complements
form a triangulation of A.
In the following result, being in general position for a vector configuration of rank k
means that every k elements form a basis. Our configuration B will have this property
since the property holds for the vertex set of a cyclic polytope and is preserved by Gale
duality.
Lemma 4 [8]. Let C be a collection of linear bases of a vector configuration B in
general position. Then the following two conditions are equivalent:
(i) C is a virtual chamber of B.
(ii) C shares exactly one basis with every triangulation of B.
The implication from (i) to (ii) holds without the general position assumption. This
implication, in a sense, explains the name “virtual chamber”. The chamber fan of the
vector configuration B is the common refinement of all its triangulations. The same
definition for a point configuration gives what is known as the chamber complex of the
configuration. The chambers are the maximal cells in either case. For any given chamber,
the collection of linear bases of B whose positive spans contain that chamber are a
particular example of virtual chamber and two different chambers produce different virtual
chambers. Hence, a chamber can be considered as a special case of virtual chamber.
In fact, chambers of B are the virtual chambers corresponding to the so-called regular
triangulations of A [
5
], [11], of interest to us in Section 4.
Following the analogy with (geometric) chambers, when a basis ¿ is an element of
a virtual chamber C we say that C lies on ¿ . With this, Lemma 4 can be rephrased as
“C is a virtual chamber of B if and only if it lies on exactly one maximal simplex of
every triangulation of B”. As a consequence, for every triangulation T of B, one has
that the number of virtual chambers of B equals P¿2T ]fvirtual chambers lying on ¿ g.
¤
The structure of this paper is as follows. Section 1 outlines our method and translates
the problem of counting triangulations of a corank 3 point configuration A to that of
counting ideals in certain posets arising from the Gale transform of A. Section 2 makes
use of the specially nice structure of the Gale transform of C .n; n¡4/ to prove Theorem 1.
In Section 3 we compute, for even n, the number of triangulations of C .n; n¡4/ which use
any specific full-dimensional simplex (Theorem 3.3). Finally, in Section 4, we compute
the exact maximum number of regular triangulations of point configurations having the
same oriented matroid as C .n; n ¡ 4/ (Theorem 4.3). This number is a polynomial of
degree four (as follows from Theorem 5.7 of [
5
]) and contrasts with the exponential
number of all triangulations. The polynomial versus exponential behaviour of regular
versus non-regular triangulations of C .n; n ¡ 4/ had already been pointed out in Section
5 of [8], although the exponential lower bound for the number of triangulations stated
there is Ä.2n=4/, instead of the 2.n2n=2/ proved here.
1. Counting Virtual Chambers in Rank 3
Virtual Chambers in Rank 3
Let B be a rank 3 vector configuration in general position, such as the Gale transform
of the vertex set of C .n; n ¡ 4/. The oriented matroid of a vector confiuration does not
change under a positive scaling of an element. Hence, without loss of generality we may
assume that all the elements of B lie on the unit 2-sphere S2. The positive span of a
subset S ½ B is then substituted by its spherical convex hull (i.e. the intersection of the
positive span with the unit sphere) and we denote it conv.S/. This motivates that we call
triangles the bases of B. We will often refer to independent sets (i.e. subsets of bases)
as simplices. We call the simplices with one and two elements, points (or vertices) and
edges, respectively. In this setting, a triangulation of B is a geometric triangulation of
conv.B/ ½ S2 by (spherical) triangles of the configuration.
The relative interior of a subset S of B is the sphere S2 intersected with the relatively
open cone of strictly positive linear combinations of elements of S. We say that two
simplices of a configuration overlap if their relative interiors have non-empty intersection.
If two edges l1 and l2 of B overlap, then their relative interiors meet in a single point.
We say that l1 and l2 cross each other.
A simplex ¾ of B is said to be empty if conv.¾ / \ B D ¾ . It is clear that if an empty
edge l overlaps an empty triangle ¿ , then either l crosses two edges of ¿ or l \ ¿ is a
vertex of both l and ¿ and l crosses the opposite edge of ¿ . Note also that since B is in
general position, all its edges are empty.
Definition 1.1. Let B be a rank 3 configuration. An empty triangle ¿ of B is said to be
admissible with respect to one of its vertices p if no edge of B which overlaps ¿ has p
as a vertex.
Remark 1.2. If ¿ has an edge l which is not crossed by any other edge of the
configuration, then ¿ is admissible with respect to the vertex opposite to l. The converse is not
true.
An edge l of B defines two sides, which are the two hemispheres in which S2 is
divided by the unique great circle which contains l. A virtual chamber C is said to lie
on a certain side of an edge l if there is a triangle of C contained in that (closed) side of
l. By part (i) of the following lemma, a virtual chamber cannot lie on both sides of an
edge. However, it can happen that it lies in none of the two sides of it.
Lemma 1.3. Let C be a virtual chamber of a rank 3 configuration B in general position.
Then:
(i) Any two triangles of C overlap.
(ii) If S ½ B contains a triangle of C, then any triangulation of S contains an element
of C.
(iii) If an edge l overlaps an empty triangle of C, then C lies on some side of l.
Proof. Parts (i) and (iii) are Lemma 1.4 and Proposition 3.2 in [
4
], respectively. Part
(ii) is the specialization of Lemma 2.7 in [
3
] to triangulations.
Virtual Chambers in Rank 3 as Poset Ideals
Given a poset (partially ordered set) .P; </ and two subsets I; F ½ P, we say that I is
an ideal if for every pair of elements x ; y 2 P with x < y and y 2 I , it holds that x 2 I .
Let ¿ :D f p; q; r g be a triangle of a rank 3 configuration B, admissible with respect to
p. We denote by Ä.¿ / the set of edges which overlap ¿ and we define the following binary
relation in Ä.¿ /: l1 Áp l2 if l1 6D l2 and l2 does not overlap the triangle l1 [ f pg. Observe
that if l1 Áp l2, then the rays from p through q and from p through r meet conv.l1/
before conv.l2/. The converse is not true, but this property implies that the relation Áp
does not have cycles. Hence, its transitive closure is a partial order, that we denote <p.
If l1 <p l2 we say that l1 is closer to p than l2. A chain l1 <p ¢ ¢ ¢ <p lk of edges in
.Ä.¿ /; <p/ is called a strong chain if every consecutive pair of edges share a vertex.
Let C be a virtual chamber lying on ¿ . By part (iii) of Lemma 1.3, C lies on a side
of every edge of Ä.¿ /. Let I .C/ be the set of edges in Ä.¿ / which have p and C on
opposite sides. Equivalently, I .C/ :D fl 2 Ä.¿ /: l [ f pg 62 Cg.
Proposition 1.4. For every virtual chamber C lying on ¿ , I .C/ is an ideal of .Ä.¿ /; <p/.
Moreover, the correspondence C 7! I .C/ is a bijection between virtual chambers which
lie on ¿ and ideals of .Ä.¿ /; <p/.
Proof. Let us see that I .C/ is an ideal. Let l1; l2 2 Ä.¿ / with l1 <p l2 and let l2 2 I .C/.
First suppose that l1 Áp l2. We consider a triangulation T of ¿ [ l2 which uses the
triangle l2 [ f pg. The remaining triangles of T are contained in the side of l2 opposite to
p. Since C lies on the side of l2 opposite to p, C lies on some of the remaining triangles.
The condition l1 Áp l2 implies that none of the remaining triangles overlaps l1 [ f pg,
which implies l1 2 I .C/. Hence, C 7! I .C/ is a well-defined map from virtual chambers
lying on ¿ to ideals in .Ä.¿ /; <p/.
Conversely, for each ideal I of Ä.¿ /, let C.I / be the following collection of triangles
of B: (a) Triangles whose convex hull contains ¿ . (b) Triangles containing p in their
convex hulls and with exactly one edge in Ä.¿ /nI . (c) Triangles not containing p in
their convex hulls and with exactly one edge in Ä.¿ / \ I . We claim that C.I / is a virtual
chamber, i.e. that it contains one triangle from each triangulation T of B.
If no edge of a triangulation T overlaps ¿ , then the triangle of T containing ¿ in its
convex hull is the only triangle of T in C.I /. Otherwise, since ¿ is empty and admissible
with respect to p, the collection of edges of T which overlap ¿ forms a a strong chain
l1 <p ¢ ¢ ¢ <p lk . Observe that there is a unique triangle ¾ of T containing p in its convex
hull and overlapping ¿ , that ¾ must have a unique edge in Ä.¿ / and that such an edge
must be l1. If l1; 62 I , then ¾ is the only triangle of T in C.I /. Finally, if l1 2 I , then let i
be the biggest index such that li 2 I . Clearly, the triangle of T incident to li on the side
opposite to p is the only triangle of T in C.I /.
The fact that the correspondences C 7! I .C/ and I 7! C.I / are inverse of each other
is straightforward.
Remark 1.5. As a conclusion, there is a bijection between ideals in the poset .Ä.¿ /;
<p/ and triangulations of A using the simplex An¿ . We do not work out the details, but
this bijection extends to a bijection between polyhedral subdivisions of A that use An¿
and pairs of ideals I1 ½ I2 such that I2nI1 is an antichain.
Remark 1.6. Proposition 1.4 translates the problem of counting virtual chambers in
an admissible triangle to counting ideals in a certain poset. This will suffice for counting
triangulations of corank 3 cyclic polytopes since, as we will see, their Gale transforms
can be triangulated with admissible triangles.
However, in fact, the technique can be modified to deal with non-admissible
triangles as well. Let ¿ be a non-admissible triangle in a rank 3 configuration B. We
can assume that ¿ is empty, otherwise we triangulate its convex hull by empty
triangles. Let p be a vertex of ¿ and let l D fq; r g be the opposite edge. We consider
the ordered sequence (from q to r ) of edges f p; pi g, i D 2; : : : ; k which overlap
¿ . Set p1 :D q and pkC1 :D r . The triangles ¿i :D f p; pi ; piC1g, i D 1; : : : ; k ¡
1, are admissible with respect to p, they intersect properly and, since ¿ is empty,
they cover ¿ . In rank 3, any set of triangles which intersect properly can be
completed to a triangulation, which implies that each virtual chamber of B lying on ¿
lies on exactly one of the triangles ¿i . Moreover, the bijection between virtual
chambers of B lying on ¿i and ideals of .Ä.¿i /; <p/ restricts to a bijection between
virtual chambers lying both on ¿i and ¿ and ideals of .Ä.¿i /; <p/ not containing l.
Summing up, we can count virtual chambers in ¿ by adding the numbers of ideals
of .Ä.¿i /; <p/ which do not contain the edge l. An analogous generalization can be
done for the case of subdivisions and pairs of ideals mentioned in the previous
remark.
2. Counting Virtual Chambers of C.n; n ¡ 4/¤
Combinatorial Structure of C .n; n ¡ 4/¤
A sign sequence of length n is any element of f¡1; 0; C1gn. The support of a sign
sequence is its set of non-zero coordinates. Recall that in oriented matroid theory the
circuits of a vector configuration A are the sign sequences with minimal support produced
by the coefficients of linear dependences in A, and the cocircuits of A are the sign
sequences with minimal support produced by the values of non-zero linear functionals
on A. Either circuits or cocircuits suffice to characterize the oriented matroid of A and
two oriented matroids are dual to each other if and only if the circuits of one are the
cocircuits of the other.
Let C .n; n ¡4/ D fa1; : : : ; ang denote the vertex set of a cyclic polytope of dimension
n ¡ 4 with n vertices. Let p1; : : : ; pn be points in a non-great circle ° in S2, taken in
order along the circle. Let C .n; n ¡ 4/¤ :D fb1; : : : ; bng, where bi :D .¡1/i pi , for
i D 1; : : : ; n. Part (ii) of the following statement appears in [19, Exercise 6.13], and part
(i) is essentially the Gale evenness criterion for cyclic polytopes.
Lemma 2.1.
(i) For each quadruple fbi1 ; bi2 ; bi3 ; bi4 g with i1 < i2 < i3 < i4 the signs sign.bij / D
.¡1/ij C j , j 2 f1; 2; 3; 4g, give one of the two (opposite) circuits with support in
that quadruple.
(ii) The oriented matroids of C .n; n ¡ 4/ and C .n; n ¡ 4/¤ are dual to each other.
Observe that, in general, C .n; n ¡4/¤ is not going to be a Gale transform of C .n; n ¡4/
according to our definition, in which the points of C .n; n ¡4/ are taken along the moment
curve. However, it has the same oriented matroid as a Gale transform, and hence the same
collection of triangulations and virtual chambers.
To simplify the notation, from now on we refer to each point bi 2 C .n; n ¡ 4/¤ by its
label i 2 f1; : : : ; ng. C .n; n ¡ 4/¤ is contained in two opposite circles in S2, one with
the set of “odd points” f1; 3; 5; : : :g and the other with the “even points” f2; 4; 6; : : :g.
The points in each circle define a spherical polygon, so that the sphere is divided into
two polygons (odd and even) plus a topological band between them.
A triangle ¿ of C .n; n ¡ 4/¤ will be said to lie on one of the polygons if its convex
hull is contained in that polygon and will be said to lie on the band if its convex hull is
contained in the band.
Lemma 2.2.
(i) An edge in the boundary of one of the polygons is crossed by no other edge of
C .n; n ¡ 4/¤. In particular, it overlaps no empty triangle.
(ii) Every triangulation of C .n; n ¡ 4/¤ by empty triangles uses all the boundary
edges of the two polygons. Hence, all its triangles lie either on one of the polygons
or on the band.
(iii) A triangle lies on one of the polygons if and only if its three vertices are in that
polygon.
(iv) For a triangle ¿ which has two vertices in one polygon and the third in the other
polygon the following conditions are equivalent:
(a) ¿ is admissible with respect to this third vertex.
(b) ¿ is empty.
(c) ¿ lies on the band.
(d) ¿ has one edge in the boundary of one of the polygons and ¿ is not one of
the following triangles: fi; i C 1; i C 2g with i 2 f1; : : : ; n ¡ 2g or, if n is
even, fn; 1; 2g or fn ¡ 1; n; 1g.
Proof. That an edge l in the boundary of the polygons does not cross any other edge
can be proved geometrically; or it can be derived from Lemma 2.1 that l cannot be
the positive part of a circuit of C .n; n ¡ 4/¤. The second part of part (i) holds because
any edge overlapping an empty triangle must cross at least one of the edges of the
triangle.
Part (ii) is a consequence of part (i): let l be a boundary edge of one of the polygons
and let x be a point in the relative interior of l. If T is a triangulation by empty simplices,
x must lie on the relative interior of a unique simplex of T . This simplex cannot be a
triangle or an edge other than l itself, by part (i).
Part (iii) is trivial. In part (iv), the equivalence of (d) with any of (a), (b) or (c) is easy
to establish: let l be the edge of ¿ having its two vertices in the same polygon. For ¿ to
be either empty, admissible or to lie on the band it is clearly necessary that l be an edge
of that polygon. The converse is true unless ¿ is one of the triangles excluded in part (d),
whose convex hull contains the whole polygon.
Let C be a virtual chamber and let T be a triangulation of C .n; n ¡ 4/¤ all of whose
triangles lie either on one of the polygons or on the band (we will see later that such
triangulations exist). We say that C lies on the odd polygon, on the even polygon or
on the band if it lies on a triangle of T on the odd polygon, the even polygon or the
band, respectively. According to part (ii) of Lemma 1.3 applied to the two polygons this
definition is independent of T . In what follows we count the number of virtual chambers
on each polygon and on the band.
Virtual Chambers of a Polygon
Let M be the configuration consisting of the vertices of one of the two polygons of
C .n; n ¡ 4/¤.
Lemma 2.3. There is a natural bijection between virtual chambers of C .n; n ¡ 4/¤
lying on M and virtual chambers of M (as a configuration by itself).
Proof. By part (i) of Lemma 2.2, every triangle of M containing an edge in the boundary
of M is admissible with respect to the opposite vertex, both in M and in C .n; n ¡ 4/¤.
Clearly, M can be triangulated with triangles of that type: take any point p in M and
triangulate M by coning p to every boundary edge of M not containing p.
Moreover, only edges in M overlap conv.M / and, hence, for those triangles, the poset
Ä.¿ / is the same in M and in C .n; n ¡ 4/¤. By Proposition 1.4 the number of virtual
chambers of M and of C .n; n ¡ 4/¤ lying on each of those triangles is the same.
Lemma 2.3 allows us to forget C .n; n ¡ 4/¤ for a while and speak rather of a polygon
P whose vertices are labelled from 1 to m. We compute the number of virtual chambers
in P by adding the virtual chambers which lie on the triangles ¿i :D f1; i ¡ 1; i g
for i D 3; : : : ; m since these triangles are admissible with respect to 1 and define a
triangulation of P . In Ä.¿i / we consider the ordering <1 of “being closer to 1”. Our task
is to count ideals of .Ä.¿i /; <1/.
Note that
Ä.¿i / D ff j; kg: 2 · j · i ¡ 1; i · k · mgnffi ¡ 1; i gg;
with the partial order f j; kg ·1 f j 0; k0g if and only if j · j 0 and k ¸ k0. We extend Ä.¿i /
to a larger poset Ä\.¿i / :D Ä.¿i / [ ffi ¡ 1; i gg by setting fi ¡ 1; i g as the maximum of
Ä\.¿i /. The ideals of Ä\.¿i / are those of Ä.¿i / plus Ä\.¿i / itself. The Hasse diagram of
Ä\.¿i / is shown in Fig. 1(a).
Proposition 2.4.
(i) The poset Ä.¿i / has ¡ mi¡¡21¢ ¡ 1 ideals.
(ii) A polygon with m vertices has 2m¡1 ¡ m virtual chambers.
Proof. Ideals in Ä\.¿i / are in bijection with maximal left-to-right monotone paths
through the lattice points in a .i ¡ 2/ £ .m ¡ i C 1/ rectangle as shown in Fig. 1.
The number of such paths is
This proves part (i). For part (ii), add the virtual chambers in each triangle ¿i , i.e. the
number of ideals in each poset Ä.¿i /.
The number 2m¡1 ¡ m equals the number of maximal straightline thrackles with
vertices in a convex m-gon, computed in [9]. There is actually a bijection between these
thrackles and virtual chambers in the polygon, in which edges of the thrackle correspond
to “flippable edges” of the virtual chamber.
Virtual Chambers on the Band: the Even Case
We now assume n to be even and let m D n=2. All indices will be regarded modulo n.
Let Teven be the following set of triangles, all lying on the band according to part (iv) of
Lemma 2.2:
Teven D ff2i C 1; 2i C 2; 2i C 4g; f2i C 1; 2i C 3; 2i C 4g: 0 · i · m ¡ 1g:
These triangles form a triangulation of the band, by which we mean that adding
triangulations of the odd and even polygons to them we get a triangulation of C .n; n¡4/¤.
Figure 2 displays Teven. In this figure and the subsequent ones, the following flattened,
twisted, planar representation of the band is used: the odd and even points are placed
on two parallel lines in the plane, with 2i ¡ 1 and 2i facing each other. The sequence
of points is meant to be repeated infinitely, or the left and right ends of the figure be
identified. For a given odd and a given even vertex there are different ways to draw a
straight line segment joining them. We choose to take the one of greatest positive slope,
considering a vertical line as having infinite positive slope. With this choice, two edges
cross in the representation if and only if they cross in C .n; n ¡ 4/¤.
Since Teven is a “triangulation of the band”, in order to count the virtual chambers in
the band we just add the virtual chambers in the triangles of Teven. Due to the symmetry
of Teven, this gives the same result as multiplying the number of virtual chambers in
f1; 2; 4g by 2m.
2m-2
2m
2m-3
2m-1
2
1
4
3
6
5
Fig. 2. The set of triangles Teven in our twisted representation of the band.
As stated in Lemma 2.2, the triangle ¿ D f1; 2; 4g is admissible with respect to the
vertex 1. Therefore, our task is to compute the number of ideals of .Ä.¿ /; <1/. Figure 3
represents the triangle ¿ (with m D 5) and the edges in Ä.¿ /. These edges are those
crossing either f1; 2g or f1; 4g which, by Lemma 2.1, are respectively ff2i; 2 j C 1g: 2 ·
i · j · m ¡ 1g and ff2i; 2 j C 1g: 3 · i · j · m ¡ 1g. Since the second set is contained
in the first, we have
Ä.¿ / D ff2i; 2 j C 1g: 2 · i · j · m ¡ 1g:
The poset structure in Ä.¿ / is that .2i; 2 j C 1/ ·1 .2i 0; 2 j 0 C 1/ if and only if i ¸ i 0
and j ¸ j 0. Hence, the Hasse diagram of .Ä.¿ /; <1/ is as depicted in Fig. 4(a).
Proposition 2.5.
(i) The number of virtual chambers lying on the triangle f1; 2; 4g of C .2m; 2m ¡4/¤
is 2m¡2.
(ii) The number of virtual chambers of C .2m; 2m ¡ 4/¤ lying on the band is m2m¡1.
(iii) The total number of virtual chambers of C .2m; 2m ¡ 4/¤ is .m C 2/2m¡1 ¡ 2m.
Proof. Let ¿ D f1; 2; 4g. Ideals of .Ä.¿ /; <1/ are in bijection with maximal
left-toright monotone paths in the “dual diagram” shown in Fig. 4(c). These, in turn, are in
bijection with maximal monotone paths in the complete binary tree of depth m ¡ 2. This
proves part (i) and the symmetry remarks stated above prove part (ii).
2m-4
2m-2
2m-5
2m-3
2m-1
For part (iii) we have to add the m2m¡1 virtual chambers on the band to the 2m¡1 ¡ m
on each of the two polygons, which gives the stated number.
Theorem 2.6. The cyclic polytope C .2m; 2m¡4/ has .mC2/2m¡1¡2m triangulations.
Virtual Chambers on the Band: the Odd Case
The configuration C .2m ¡ 1; 2m ¡ 5/¤ (up to oriented matroid equivalence) can be
obtained from C .2m; 2m ¡ 4/¤ by deleting the element 2m. We intend to apply the same
technique to C .2m ¡ 1; 2m ¡ 5/¤ as in the even case. We start by choosing a triangulation
of the band (see Fig. 5):
Todd D ff2i C 1; 2i C 2; 2i C 4g; f2i C 1; 2i C 3; 2i C 4g: 0 · i · m ¡ 3g
[ ff2m ¡ 3; 2m ¡ 2; 2g; f2m ¡ 3; 2m ¡ 1; 2g; f2m ¡ 1; 1; 2gg:
The triangles of Todd are all admissible (this is automatic, by Lemma 2.2(iv)). Hence,
we can compute the number of virtual chambers in each triangle of Todd using Proposition
1.4. The only new difficulty with respect to the even case is that Todd is not preserved
under any non-trivial oriented matroid symmetry of C .2m ¡1; 2m ¡5/¤. In fact, C .2m ¡
1; 2m ¡ 5/¤ has only one non-trivial symmetry: the reversal of indices.
Lemma 2.7. The number of virtual chambers of C .2m ¡ 1; 2m ¡ 5/¤ in each triangle
of Todd is:
(i) 2m¡2 ¡ Pik¡D10 ¡mk¡3¢ for f2i C 1; 2i C 2; 2i C 4g, with i 2 f0; : : : ; m ¡ 3g.
(ii) 2m¡2 ¡ PikD0 ¡mk¡3¢ for f2i C 1; 2i C 3; 2i C 4g, with i 2 f0; : : : ; m ¡ 3g.
(iii) 2m¡2 for f2m ¡ 1; 1; 2g.
(iv) 2m¡3 for f2m ¡ 3; 2m ¡ 2; 2g and for f2m ¡ 3; 2m ¡ 1; 2g.
Proof. C .2m ¡ 1; 2m ¡ 5/¤ can be obtained from C .2m; 2m ¡ 4/¤ by deleting any
element. Triangles of parts (i)–(iii) belong to Teven, so we can assume them to be f1; 2; 4g
as long as we choose the appropriate vertex to be removed (instead of 2m). That is, we
{4,2m-1}
{2m-2i,2m-1}
{4,5}
{2m-2i,2m-2i+1}
i-1
.
.
.
k
.
.
.
0
Proposition 2.8. The number of virtual chambers of C .2m ¡ 1; 2m ¡ 5/¤ lying on
triangles of Todd is .3m ¡ 2/2m¡3.
Proof.
By Lemma 2.7 the number N of virtual chambers in Todd is
PimD¡03 ³2m¡2 C 2m¡2 ¡
Pi
kD0
³¡m¡3¢
k
C ¡ mk¡¡13¢´´ C 2m¡2 C 2 ¢ 2m¡3
D .4m ¡ 4/2m¡3 ¡ mX¡3 Xi µm ¡ 2¶
:
If we call
Ai D
Xi µm ¡ 2¶
kD0
k
we have 2m¡2 D Ai C Am¡i¡3 which implies .m ¡ 2/2m¡2 D 2.PimD¡03 Ai /. Hence
N D .4m ¡ 4/2m¡3 ¡ .m ¡ 2/2m¡3 D .3m ¡ 2/2m¡3:
Theorem 2.9. The number of triangulations of the cyclic polytope C .2m ¡ 1; 2m ¡ 5/
is .3m C 4/2m¡3 ¡ .2m ¡ 1/.
Proof. To the number obtained in Proposition 2.8 we have to add the numbers of
virtual chambers of the two polygons defined by C .2m ¡ 1; 2m ¡ 5/¤, which have m
and m ¡ 1 vertices, respectively. By Proposition 2.4, these numbers are 2m¡1 ¡ m and
2m¡2 ¡ .m ¡ 1/, respectively.
3. Triangulations of C.2m; 2m ¡ 4/ Which Use a Fixed Simplex
In this section we count the number of virtual chambers in any particular triangle of
C .2m; 2m ¡ 4/¤, although the technique will still use some particular properties of
this configuration rather than the general method outlined in Remark 1.6. One can do
analogous calculations in the case of C .2m ¡ 1; 2m ¡ 5/¤ but there are many more cases
to be studied due to much less symmetry, so we prefer to restrict our study to the even
case.
We first prove two additional results:
Lemma 3.1. Let M be a convex m-gon. Let S be a subconfiguration consisting of k
consecutive vertices of M . Then the number of virtual chambers of M lying on S (i.e.
lying on triangles of any triangulation of S) equals
Xk¡2 µm ¡ 1¶
lD0
l
¡ .k ¡ 1/:
Proof. Consider the vertices of M labelled from 1 to m, and without loss of generality
suppose that S D f1; : : : kg. Then the following is a triangulation of S:
ff1; l ¡ 1; lg: l 2 f3; : : : ; kgg:
By Proposition 2.4, the triangle f1; l ¡ 1; lg contains ¡ ml¡¡21¢ ¡ 1 virtual chambers of M .
Adding this number for l from 3 to k gives
Xk µm ¡ 1¶
lD3
l ¡ 2
¡ .k ¡ 2/ D
¡ .k ¡ 2/;
Xk¡2 µm ¡ 1¶
lD1
l
which coincides with the number stated.
Lemma 3.2. Let ¿ be any empty triangle in C .2m; 2m ¡ 4/¤ not contained in one of
the polygons. Then there are exactly 2m¡2 virtual chambers of C .2m; 2m ¡ 4/¤ in ¿ .
Proof. The subgroup of combinatorial symmetries of C .2m; 2m ¡ 4/¤ generated by
i 7! i C 2 and i 7! 2m ¡ i C 1 (which has 2m elements), applied to ¿ , produces 2m
empty triangles not contained in either of the polygons, i.e. lying on the band by Lemma
2.2. It is easy to check that these 2m triangles form a triangulation of the band, very
similar to the triangulation Teven depicted in Fig. 2. By symmetry, all the 2m triangles
contain the same number of virtual chambers, i.e. the m2m¡1 virtual chambers in the
band divided by 2m.
Theorem 3.3. Let ¿ be a triangle in C .2m; 2m ¡ 4/¤.
(i) If ¿ is contained in one of the two polygons, let ¿ D fi; j; kg with i < j < k. The
number of virtual chambers of C .2m; 2m ¡ 4/¤ lying on ¿ equals
2m¡1 ¡
Xa µm ¡ 1¶
lD0
l
¡
lD0
l
¡
lD0
Xb µm ¡ 1¶
Xc µm ¡ 1¶
;
where a, b and c are the numbers of points in the polygon and between each
two vertices of ¿ . That is, a D . j ¡ i /=2 ¡ 1, b D .k ¡ j /=2 ¡ 1 and c D
.2m C i ¡ k/=2 ¡ 1.
(ii) If ¿ does not lie on either of the polygons, then there is a combinatorial symmetry of
C .2m; 2m ¡4/¤ which sends ¿ to a triangle ¿ 0 D fi; j; kg with i < j < k and with
i odd and j and k even. Then the number of virtual chambers of C .2m; 2m ¡ 4/¤
lying on ¿ equals
Proof. The case of ¿ lying on a polygon is easy in light of Lemma 3.1: Joining
triangulations of three configurations as the one in Lemma 3.1 with the parameter k taking the
values a C 2, b C 2 and c C 2 to ¿ produces a triangulation of the whole polygon, which
has 2m¡1 ¡ m virtual chambers. Hence, the number of virtual chambers in ¿ equals
2m¡1 ¡ m ¡
Xa µm ¡ 1¶
lD0
Xb µm ¡ 1¶
l
l
¡
lD0
C .a C 1/
C .b C 1/ ¡
Xc µm ¡ 1¶
lD0
C .c C 1/;
as desired since a C b C c D m ¡ 3.
In part (ii), ¿ has two vertices in one polygon and the third vertex in the other polygon.
The combinatorial symmetry of the statement can be taken as one of the two which send
this third vertex to the vertex 1. Without loss of generality we assume in the rest of the
proof that ¿ D fi; j; kg with i odd, j and k even and i < j < k.
A point l is in the interior of the triangle ¿ exactly when the circuit with support in
fi; j; k; lg has the same sign in i , j and k and the opposite sign in l. By Lemma 2.1 this
l
l
:
l
happens if and only if l is even and between j and k. We consider the subconfiguration
S D fi; j; j C 2; j C 4; : : : ; kg consisting of the vertices of ¿ and its interior points. We
can triangulate the part of conv.S/ in the even polygon as in Lemma 3.1, which gives
l
¡
virtual chambers, and the part in the band with the .k ¡ j /=2 empty triangles ffi; l; l C
2g: l 2 f j; j C 2; j C 4; : : : ; k ¡ 2gg, each containing 2m¡2 virtual chambers by
Lemma 3.2.
Hence, the following table gives the number of triangulations of the cyclic polytope
C .2m; 2m ¡ 4/ which use the simplex C .2m; 2m ¡ 4/ n fi; j; kg under the assumption
that i < j < k and depending on the parities of k ¡ j and j ¡ i . The first two rows are
just the formulas in Theorem 3.3, where the value of a, b and c can be found. The last
two rows are the translation of the second row to the case in which k ¡ j is odd.
j ¡ i and k ¡ j even
j ¡ i odd, k ¡ j even
j ¡ i even, k ¡ j odd
j ¡ i odd, k ¡ j odd
2m¡1 ¡ Pa
lD0
µm ¡ 1¶
l
¡ Pb
lD0
µm ¡ 1¶
l
¡ Pc
lD0
µm ¡ 1¶
l
A regular triangulation of a d-dimensional point configuration A is a triangulation of A
which can be obtained as the orthogonal projection of the lower envelope of a .d C
1/dimensional polytope (see [
5
] or [19] for details). The bijection between triangulations
of A and virtual chambers of its Gale transform B sends the regular triangulations to
the geometric chambers of B, i.e. the full-dimensional cones in the chamber fan that we
defined in the Introduction.
The chamber fan of a configuration does not depend only on the oriented matroid.
We intend here to compute the maximum possible number of chambers among all the
coordinatizations of the oriented matroid of C .n; n ¡ 4/¤. That is to say, the maximum
possible number of regular triangulations of a polytope realizing the oriented matroid of
C .n; n ¡ 4/.
For the next lemma, we recall that in oriented matroid theory a vector configuration is
called totally cyclic if its positive span is the whole space. The statement holds equally if
B is not totally cyclic, except that the formula changes by 1 because of the use of Euler’s
formula for a ball instead of a sphere.
Lemma 4.1. Let B be a rank 3 totally cyclic vector configuration in general position.
Let N be the number of (opposite pairs of) circuits of B having two positive and two
negative elements. Then the maximum number of chambers produced by realizations of
the oriented matroid of B is
N C
µn¶
2
¡ n C 2:
Moreover, the maximum is achieved in any realization in which no three edges cross in
a common point. This happens if B is sufficiently generic among the realizations of its
oriented matroid.
Proof. We first prove that there is a realization of the oriented matroid of B in which
no three edges have a common crossing. Indeed, let k be the number of triplets of edges
crossing in a point. If k · 1, then a slight perturbation of the coordinates of one of the
six vertices involved in a triple crossing decreases the number of such crossings, and
does not change the oriented matroid because of our general position assumption. On the
other hand, if k D 0, then sufficiently small perturbations cannot create triple crossings.
This proves the assertion.
Next we prove that if B has no triple crossings, then it has exactly the stated number
of chambers. This, together with the fact that small perturbations cannot decrease the
number of chambers, implies that the stated number is indeed the maximum.
Embedding B in the 2-sphere as we have done in this paper, the chamber complex of B
(i.e. the intersection of the chamber fan with the unit sphere) is a polyhedral subdivision
of the sphere whose numbers of cells of dimensions 2, 1 and 0 we denote f2, f1 and f0.
The number of chambers equals f2. The number f0 equals n plus the number of crossing
points between edges of B, which under the assumption of no triple crossings equals N :
On the other hand, 2 f1 equals the number of incidences between 0-cells and 1-cells
in the cell decomposition. That is to say,
f0 D N C n:
2 f1 D 4N C n.n ¡ 1/;
where the term n.n ¡ 1/ comes from the fact that n ¡ 1 edges are incident at each point
of B. By Euler’s formula for the 2-sphere,
f2 D f1 ¡ f0 C 2 D N ¡ n C
µn¶
2
C 2:
Proposition 4.2. The number of (pairs of) circuits with two elements of each sign in
C .n; n ¡ 4/¤ equals
² 6¡m4 ¢ C 3¡m3 ¢ if n D 2m is even, and
² 6¡m4 ¢ C 2 ¡ m C 1 if n D 2m ¡ 1 is odd.
¡m¢
Proof. Suppose first that n D 2m is even. We know that no edge of C .2m; 2m ¡ 4/¤
crosses the boundary of any of the two polygons defined by C .2m; 2m ¡ 4/¤, so if
two edges cross, then either both are edges of the band or both are edges of one of the
polygons. Let B and P be the numbers of crossings of edges of the band and of the even
polygon, respectively. Then
The number P is the number of crossings between edges of an m-gon, but any four
vertices of an m-gon define a unique crossing, so
N D B C 2 P:
P D
µm¶
4
:
For computing the number B, we first compute the number of edges of C .2m; 2m ¡4/¤
crossing a certain edge fa; bg of the band. We assume a D 1 and, hence, b D 2 j is
even, in order that fa; bg be in the band. Let fa0; b0g be another edge in the band, and
assume that a0 is odd and b0 is even. According to Lemma 2.1, under these assumptions
f1; bg and fa0; b0g cross each other if and only if 1 < a0 < b0 < b or 1 < b <
b0 < a0. Taking into account the parities of a0, b0 and b, the first case gives ¡ j¡1¢
2
possibilities and the second gives ¡m¡ j¢ (for the first number, observe for example that
2
each pair of indices 1 · i 0 < j 0 · j ¡ 1 gives the edge having a0 D 2i 0 C 1 and
b0 D 2 j 0).
Adding up these numbers for j 2 f1; : : : ; mg we conclude that the total number of
crossings between edges of the band, one of which contains the point 1, equals
Xm µ j ¡ 1¶
Xm µm ¡ j ¶
jD1
2
C
jD1
2
D
µm¶
µm¶
Now, by the symmetries of C .2m; 2m ¡ 4/¤ the same is valid for any other vertex:
the number of crossings in the band between edges, one of which contains any specific
vertex, is 2¡m3 ¢. Since each crossing uses four vertices, the number of crossings in the
band equals
B D 4
2m µm¶
2
3
D m
µm¶
3
D m
µm ¡ 1¶
3
C m
µm ¡ 1¶
2
µm¶
D 4 4
C 3 3
µm¶
:
Hence,
N D B C 2 P D 6
µm¶
4
C 3 3
µm¶
:
For the odd case, remember that C .2m¡1; 2m¡5/¤ can be obtained from C .2m; 2m¡
4/¤ by deleting the point 2m. Then the number of crossings in C .2m ¡ 1; 2m ¡ 5/¤ equals
the total number of crossings in C .2m; 2m ¡4/¤ minus the crossings involving the vertex
2m. This number is 2¡m3 ¢ crossings in the band plus ¡m¡1¢ crossings in the even polygon.
3
Hence, in the odd case we have
Theorem 4.3.
The number of regular triangulations of C .n; n ¡ 4/ is at most:
(i) 6¡m4 ¢ C 3¡m3 ¢ C 4¡m2 ¢ ¡ m C 2 if n D 2m for some positive integer m.
(ii) 6¡m4 ¢ C 5¡m2 ¢ ¡ 4m C 5 if n D 2m ¡ 1 for some positive integer m.
Moreover, these formulas give the exact number of regular triangulations in any
sufficiently generic coordinatization of the oriented matroid of C .n; n ¡ 4/.
This is straightforward from Lemma 4.1 and Proposition 4.2, taking into account
which if n D 2m gives
and if n D 2m ¡ 1 gives
4m2 ¡ 4m C 1 ¡ 6m C 3 C 4
2
D 2m2 ¡ 5m C 4 D 4µm2 ¶
¡ 3m C 4:
Remark 4.4. One may ask about the minimal, instead of maximal, number of regular
triangulations in realizations of the oriented matroid of C .n; n ¡ 4/. This would
correspond to computing the number of chambers in the “least generic” realization of the
dual oriented matroid. It is reasonable to think that, if n D 2m is even, the realization
in which each half of the points form a regular m-gon is a good approximation of this
“least-generic” case. The number of chambers in a regular m-gon has been computed in
[12]. The result is m4=24 § 2.m3/, exactly as in the most-generic case. This leads to the
conjecture that the number of regular triangulations in every realization of C .n; n ¡ 4/
is n4=64 § 2.n3/, as in the generic case.
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