THE DIMENSION, DEGREE, AND INITIAL COMPLEX OF THE
SECANT IDEAL OF THE SECOND HYPERSIMPLEX
UMUT VAROLGUNES
Abstract. Sullivant finds a Gröbner basis for the second secant ideal of the
second hypersimplex with respect to a circular term order. We use this basis
to combinatorially analyze the initial complex of the ideal.
We first compute the dimension of its variety. Then, we work on the degree
of the variety, and rephrase the problem as a simple counting problem. Finally,
we show that the initial complex is pure, by combinatorial methods, which is
also a corollary of an algebro-geometric theorem proven by Sturmfels et al.
1. Introduction
The second secant ideal of the toric ideal associated to the second hypersimplex
arises naturally in the factor analysis with two hidden variables. Some computations about the codimension and the degree of this ideal are made in [2]. In
[6], Sullivant finds a Gröbner basis for this ideal.
{2}
The reduced Gröbner basis of In with respect to a circular order has polynomials with leading terms as the odd cycles of the non-crossing graph, more precisely,
{2}
In
(1)
in (In{2} ) = hxV | V ⊆ V (Gn ), GV is an odd cyclei.
In this paper, we use this square-free basis to characterize the initial complex
{2}
of In with respect to a circular order, and therefore to work on the dimension
{2}
and the degree of V (In ). A similar work is done in [1] by De Loera et al. for
the initial complex of the toric ideal In itself. In that case the maximal simplices
correspond to the n − 1 dimensional simplices of a certain triangulation of the
second hypersimplex, which makes the computations easier. We also present a
{2}
combinatorial proof of the pureness of the initial complex of In .
The paper is organized as follows. We begin with a review of the literature about
{2}
the ideal In , and describe the relations between the ideal and its initial complex in
Background. Then, we describe the combinatorial structure of the initial complex
(Section 3).
{2}
In Section 4, we compute the dimension of the variety of In . Later we further
analyze the top dimensional simplices in order find the degree of the variety. Even
though we can not compute its numerical value, we manage to restate the problem
in a more approachable form (Section 5). We finish our paper with a combinatorial
{2}
proof of the pureness of the initial complex of In , which is also implied by the
algebraic geometrical work of Sturmfels et al. (Section 6).
Note that throughout the paper we will be assuming that n ≥ 5, since the small
{2}
cases do not reflect most of the general features of In . Similar results can be
easily found for small cases, but we omit this.
1
2
UMUT VAROLGUNES
This paper was done as a SPUR project in the summer of 2010, under the
mentorship of Hoda Bidkhori.
2. Background
Let C[x] := C[xij | 1 ≤ i < j ≤ n] and C[t] := C[ti | 1 ≤ i ≤ n], and let φ be the
ring homomorphism such that
(1)
φ : C[x] −→ C[t], xij 7−→ ti tj .
The toric ideal kerφ = In is the vanishing ideal of the toric variety of the second
hypersimplex.
We use the term circular embedding of Kn as the drawing of the complete graph
with n vertices such that the vertices are equispaced points on the circle, and the
edges are drawn as straight line segments between those points.
A circular term order ≺ is any block term order for the monomials in C[x] such
that xi1 j1 ≺ xi2 j2 whenever the edge i1 j1 is a longer line segment than i2 j2 .
In [1], they show that
(2)
{xij xkl − xik xjl , xil xjk − xik xjl | 1 ≤ i < j < k < l ≤ n}
form a reduced Gröbner basis for In with respect to any circular term order.
By the properties of the circular term order and that particular basis, the initial
ideal of In turns out to be the edge ideal of a particular graph. An edge ideal is
an ideal I(G) associated with an undirected graph G with vertex set [k], in such a
way that I(G) = hxi xj | ij ∈ E(G)i.
The noncrossing graph Gn has vertex set consisting of all two element subsets of
[n]. A pair ij and kl form an edge of the noncrossing graph if an only if the edges
ij snd kl do not cross (possibly at endpoints) in the circular embedding of Kn . It
is clear that in≺ (In ) is the edge ideal of Gn .
The rth secant variety X {r} of a projective variety X ⊂ Pm−1 is the closure of
the union of all planes in Pm−1 spanned by r points in X.
Using the characterization of secant ideals of edge ideals in [4], and other alge{2}
braic geometrical techniques, Sullivant finds a Gröbner basis for In which has a
nice combinatorial structure.
{2}
The reduced Gröbner basis of In with respect to a circular order has polynomials with leading terms as the odd cycles of the non-crossing graph, more precisely,
(3)
in (In{2} ) = hxV | V ⊆ V (Gn ), GV is an odd cyclei.
Independence complex ∆(I) of an ideal I of k[x1 , . . . xn ] is defined as,
(4)
∆(I) := {X ⊆ {x1 , . . . xn } | k[X] ∩ I = {0}}.
Consequently, initial complex of an ideal I is defined as the independence complex
of its initial ideal.
The initial complex of In is well studied, mainly in [1]. We study the initial
{2}
complex of In in this paper.
An abstract simplicial complex is simply a set X and a subset S of 2X satisfying
the below conditions,
(1) A ∈ S =⇒ ∀B ⊂ A, B ∈ S.
(2) ∀a ∈ X, {a} ∈ S
SPUR PROJECT
3
1
13
45
5
2
14
23
35
12
15
4
24
3
34
25
Figure 1. The circular embedding of K5 and G5
If we are given an abstract simplicial complex ∆ on the set of variables of a
polynomial ring over a field, then its Stanley-Reisner ideal I∆ is the monomial
ideal generated by the square-free monomials corresponding to the non-faces of ∆.
{2}
Therefore the initial ideal of In with respect to a circular order is the StanleyReisner ideal of its initial complex.
{2}
It is possible to interpret the dimension of V (In ) in terms of the initial complex.
{2}
As discussed in [5], the dimension of V (In ) is equal to the the maximum dimension
{2}
of a simplex of the initial complex of In . Note that the variety is projective, and
the dimension of a simplex in the initial complex is its cardinality minus one, hence
this equality follows.
One of the most important invariants of an algebraic variety is its degree. For a
projective variety V the number of points in V ∩ L, where L is a linear subspace in
general position with codimension equal to the dimension of V is the degree of V.
The degree of an ideal is equal to the degree of its initial ideal. On the other hand,
the degree of the Stanley-Reisner ideal of an abstract simplicial complex is equal
{2}
to the number of its top dimensional simplices. Hence, the degree of In is equal
to the number of top dimensional simplices of its initial complex.
We say that an (abstract) simplicial complex is pure if and only if all of its
simplices are contained in at least one top dimensional simplex, or in other words
all of its maximal simplices have the same dimension.
In is the kernel of a homomorphism of polynomial rings over complex numbers,
therefore it is prime. Since any secant ideal of a prime ideal is also prime [3], the
{2}
pureness of the initial complex of In is implied by [5]. We give a combinatorial
proof of this pureness in this paper.
{2}
3. The initial complex of In
In this section we study the combinatorial properties of the initial complex,
especially its top dimensional simplices. Some of the results in this section are also
present in [1], but for the sake completeness we give their proofs too. Also, the
4
UMUT VAROLGUNES
proofs in that paper use some properties of triangulations. We prove everything by
purely combinatorial methods.
{2}
The initial complex of In is clearly isomorphic to the abstract simplicial complex ∆, on the universal set V (Gn ) with the family of subsets as,
(1)
{X ⊂ V (Gn ) | @Y ⊂ X such that the induced subgraph of Y is an odd cycle}
by the obvious map,
(2)
xij 7−→ ij, for ∀1 ≤ i < j ≤ n.
{2}
Therefore we use ∆ to understand the properties of the initial complex of In .
We begin by interpreting the simplices of ∆ as the subsets of the edge set of the
circular embedding of Kn satisfying certain intersection conditions, by using basic
graph theory.
Lemma 1. Let G be a simple graph, and V be a subset of its vertex set. Then,
there is a V 0 ⊆ V such that GV 0 is an odd cycle if and only if GV has an odd cycle.
Proof. It is not hard to see one direction. So let us assume that GV has an odd
cycle. Take the smallest odd cycle of GV . If there is another edge between the
vertices of this cycle, then there must be a smaller odd cycle of GV . Therefore the
vertices of this odd cycle satisfy the condition, which proves the statement
Corollary 1. The simplices of ∆ are the subsets of V (Gn ) such that the corresponding induced subgraphs are bipartite.
Proof. By Lemma 1 the simplices of ∆ are the subsets of V (Gn ) such that the
corresponding induced graphs do not contain any odd cycle. Yet, by the fact that
a simple graph is bipartite if and only if it does not contain any odd cycles, the
corollary follows.
We can interpret this in terms of the circular embedding of Kn . From now on we
will refer to the circular embedding of Kn , by just Kn . Also we will see the vertices
of Gn as the edges of Kn .
Definition 1. A subset of E(Kn ) is intersecting if and only if any two edges in
the subset cross. Moreover if it has at least n elements call it n − intersecting.
In Figure 2, the union of blue and red edges give an intersecting, in particular an
n-intersecting set.
We finish this section with the following corollary.
Corollary 2. The simplices of ∆ are the subsets of E(Kn ) that can be represented
as a union of two disjoint intersecting sets.
Definition 2. A subset of E(Kn ) is a star if and only if it is intersecting and its
elements are the edges of a cycle of Kn , without any repeated vertices. In Figure
2, the blue edges form a star.
Lemma 2. Let S be a star of Kn . Then there exists a positive integer m and 2m+1
vertices i1 < · · · < i2m+1 of Kn such that S = {ik ik+m | k ∈ {1, . . . , 2m + 1}},
where k + m in the second term is considered modulo 2m + 1.
SPUR PROJECT
5
1
8
2
3
7
6
4
5
Figure 2. An example of 8-intersecting sets
Proof. Let S be the cycle j1 j2 . . . jk . Let t ∈ [k], the angle jt−1 jt jt+1 seperates the
set of vertices except jt−1 , jt , jt+1 . Call these parts X and Y. If we show that the
sum of the degrees of the vertices in X and Y are the same (this is the degree in the
induced graph of the star), it follows that their cardinalities are the same, because
all the vertices in the cycle has degree 2. Since S is a star, then any edge that has
no endpoint in the set {jt−1 , jt , jt+1 } adds exactly one to the sums of the degrees
of the vertices of both X and Y . If we consider the remaining 4 edges, we see that
they add 1 to both sets in total, which proves that X and Y have the same number
of elements. This implies that k is odd. So if we set k = 2m + 1, then both X
and Y must have m − 1 elements, which proves the rest of the statement when the
vertices of the cycle are written in increasing order.
Definition 3. Let S be a star of Kn with the vertex set i1 < · · · < i2m+1 , then
the set of edges of Kn with one endpoint ik that stays in between line segments
ik ik+m and ik ik+m+1 , for some k of [2m + 1] is called the shadow of S (note that
the elements of the star are not in the shadow of the star). In Figure 2, the red
edges form the shadow of the star made by the blue edges.
Now we can explicitly characterize n-intersecting sets in terms of stars and their
shadows.
Lemma 3. Any n-intersecting set is a union of a star and its shadow, and this
representation is unique.
Proof. Let X be any n-intersecting set, and G be the subgraph of Kn with the
vertex set [n] and the edge set X. Now since G has n vertices and at least n edges,
it contains a cycle. Since any subset of an intersecting set is clearly intersecting,
the edges of this cycle form a star. It is easy to see that any other edge which is
not in the shadow of this star does not intersect with at least one of the edges of
it. Also, the total number of elements of a star and its shadow is n, which proves
the statement. Uniqueness is clear from the construction.
6
UMUT VAROLGUNES
We conclude the following corollary, which is also a special case of Conway’s
thrackle conjecture.
Corollary 3. The cardinality of an intersecting subset is at most n.
4. The dimension of the variety
In this section, we compute the maximal dimension of a simplex of ∆, as de{2}
scribed in Corollary 2, and hence find the dimension of the variety of In .
The main result of this section is the following theorem, which implies that the
{2}
dimension of In is less than 2n − 1.
Theorem 1. Let A and B be two disjoint intersecting subsets of E(Kn ), then the
total number of elements of A and B is less than 2n.
Before starting the proof, we prove another lemma about graph theory.
Lemma 4. Let G be an directed graph with at least one incoming edge to each of
its vertices, then G has an oriented cycle.
Proof. Take the longest oriented path of G. Assume that this path starts from the
vertex v. There must be an incoming edge to v. The initial vertex of this edge
must be one of the vertices in this path, since this path is the longest path in graph.
This proves the statement.
Proof of Theorem 1. We proceed by contradiction. Assume that there are such A
and B that has at least 2n elements in total. By Corollary 3, A and B should both
have exactly n elemets.
Call their stars SA and SB with the vertex sets i1 < · · · < i2m+1 and j1 < · · · <
j2k+1 respectively. Then we construct a directed bipartite graph K with the pairs
(A, is ) and (B, jl ) as its vertices. There is an oriented edge from (A, is ) to (B, jl ) if
and only if is jl is in A, and similarly for B. (There is no other edge.) Since A and
B are n-intersecting sets, every vertex of K has an incoming edge. In Figure 3, an
example of constructing the graph K is given. The orange and blue edges are the
edges of the stars of A and B respectively, and the red edges are the edges between
the vertices of two stars, except the edges that pass in the stars.
Therefore K has a cycle, by Lemma 4. In terms of Kn , this implies that there is a
cycle a1 b1 a2 b2 . . . ar br of Kn such that {ai bi | i ∈ [r]} ⊂ A and {bi ai−1 | i ∈ [r]} ⊂ B
are intersecting sets.
If r = 1, it means that A and B have at least one edge in common, which is a
contradiction. So assume that r > 1.
Look at the angle a1 b1 a2 , and also note that a1 , b1 , and a2 are different vertices.
Claim 1. All the ai for i > 1 lies on different sides of the line segment b1 a2 with
a1 , or lies on it.
Proof. We will prove this by induction on i. It is clearly true for i = 2. Assume
that it is true for i = k − 1, where k > 2, and we will prove that it is also true for
i=k
Since bk−1 ak crosses b1 a2 , ak lies on the closed arc a1 b1 that does not contain
a2 . Similarly since ak bk crosses a2 b2 , bk lies on the closed arc a2 b1 that does not
contain a1 , which finishes the induction.
SPUR PROJECT
7
1
8
(B, 1)
2
3
7
6
(A, 1)
(B, 2)
(A, 6)
(B, 3)
(A, 8)
(B, 5)
(B, 7)
4
5
Figure 3. An example of graph K
Yet this claim contradicts the fact that ar+1 = a1 . Theorem 1 is proven.
{2}
Corollary 4. The dimension of the variety V (In ) is equal to 2n − 2, for n > 4.
Proof. We only have to find an example of a simplex with 2n − 1 elements. Take
the following set,
(1)
{1j | j ∈ {2, . . . , n}} ∪ {23 ∪ {4j | j ∈ {2, 3, 5 . . . , n}} ∪ {35}
It is straightforward to check that this example works.
We finish this section with a strenghtening of Theorem 1, which will be useful
later.
Proposition 1. Let A and B be two n-intersecting sets with only one element in
common. Then, this common element can not belong to the stars of either of them.
Proof. We proceed by contradiction. Let this common edge be in the star of A.
Clearly one of the endpoints of this edge belongs to the vertex sets of the stars
of both A and B, which means that this vertex paired with B has at least two
incoming edges in the graph G above. If we erase the one coming from the common
edge of A and B, it will still have an incoming edge together with all the other
vertices. Now we can apply the proof of Theorem 1 in order to show that there is
another edge in common.
5. The degree of the variety
{2}
The purpose of this section is to find the degree of the variety of In . The degree
of the variety is equal to the number of top dimensional simplices in its initial
complex, as discussed in the Background section. Even though we are not able
to find a closed form of this number, we find the structure of the top dimensional
simplices, and we reformulate the problem in an easier form.
As discussed before, the top dimensional simplices are the 2n−1-element subsets
of the edge set of Kn which can be represented as a union of two disjoint intersecting
8
UMUT VAROLGUNES
sets. We denote such subsets as maximal bipartite subsets. By the Corollary 3
of the previous section, any maximal bipartite set is a union of one n-intersecting
set and one intersecting set with n − 1 elements. Since we know the shape of nintersecting sets, we try to represent the maximal bipartite sets in terms of the
n-intersecting sets. The main result of this section is the following theorem
Theorem 2. The maximal bipartite sets can uniquely be represented as a union of
two n-intersecting sets.
We first give a combinatorial proof of the pureness of the initial complex of In .
Lemma 5. Every intersecting set with k < n elements can be extended to an
n-intersecting set by adding n − k more edges.
Proof. Look at the graph with vertex set same as Kn and edge set as this intersecting set with k elements. It is enough to prove that we can add one edge such
that the new set is still intersecting. then we can iterate the process until we get
to an n-intersecting set.
If the graph contains a cycle, then this cycle is a star and the only edges we can
add to a star are the edges of its shadow. So there are n − k edges from the shadow
which is not in the set. When we add one of them, we are done.
In case it does not have a cycle, first we can assume that if there are two edges
with the same endpoint, then we can ignore the vertices stay strictly inside them.
Because there can be only one edge incident to such a vertex. If it is not inside the
set, we add it. Otherwise we can just ignore that vertex and edge incident to it,
since every edge that crosses the initial two edges, crosses that edge too.
In the remaining graph there must be a vertex with degree at most 1. It is easy
to see that we can add one edge incident to this vertex, with the other endpoint
right next to the only vertex that is adjacent to it.
Remark 1. This is also implied by [5], according to the fact that In is prime.
Corollary 5. Any maximal bipartite set can be represented as the union of two
n-intersecting sets with only one element in common. We will call this a decomposition of the maximal bipartite set.
Proof. Let the maximal bipartite set be the union of A and B such that A is nintersecting. Now by Lemma 5, we can add an edge to B so that it becomes an
n-intersecting set. Yet this edge should belong to A, since any two n-intersecting
sets can not be disjoint.
Moreover, because of the structure of an n-intersecting set, for any decomposition
of a maximal bipartite set every vertex of Kn has at least one incident edge from
both of the n-intersecting sets.
We now characterize the maximal bipartite sets by showing that the decomposition is unique. This is important because this will imply that the maximal bipartite
subsets correspond to pairs of odd subsets of [n] which satisfy some certain conditions.
First we prove that the common edge in a decomposition of an maximal bipartite
set is uniquely determined. If an edge is the common edge of the two n-intersecting
sets in some decomposition, call it common.
Lemma 6. For any maximal bipartite set S there is only one common edge.
SPUR PROJECT
9
Figure 4. Common edge and the critical edges of its endpoints
Proof. If an element is common, then it crosses all the elements of S. Assume that
there are two elements of S which cross every other edge. Take any decomposition
A ∪ B = S, one of these two edges does not pass in both A and B, without loss of
generality assume that it does not belong to A. Now if we actually add this to A,
it implies that there is an intersecting set with n + 1 elements. Contradiction! Clearly, for any maximal bipartite set, there are at least two edges incident to
any vertex of Kn from that set. We call the two that has every other incident edge
inside the angle made by those edges, the critical edges of that vertex.
The following lemma is really straightforward but important about the maximal
bipartite sets.
Lemma 7. Assume that there are two edges of an maximal bipartite with the same
endpoint v, which belong to the same n-intersecting set for some decomposition, then
all the other edges incident to v that are in between those two edges should belong
to the maximal bipartite set, and moreover should belong to the same n-intersecting
set that the initial two belong to, for that certain decomposition.
Proof. This is clear by the structure of an n-intersecting set.
Recall that the common edge is in the shadow part of both n-intersecting sets
by Proposition 1. Therefore common edge is not a critical edge of either of its
endpoints. Combining this with Lemma 7, we get the following corollary.
Corollary 6. The critical edges of the vertices except the endpoints of the common
edge can not belong to the same n-intersecting set, whereas the ones of the endpoints
of the common edge belong to the same n-intersecting set, for any decomposition of
any maximal bipartite set.
We now come to the crucial part of the proof of the uniqueness.
Lemma 8. We can uniquely determine the sets of the critical edges of all the
vertices, up to permutation, for any decomposition of any maximal bipartite set.
Proof. We denote the endpoints of the common edge by v1 and v2 . The critical
edges of v1 and the critical edges of v2 belong to different n-intersecting sets in the
decomposition. If a critical edge of another vertex belongs to the set that the edges
incident to v1 belong to call them Type 1, and Type 2 otherwise.
We claim that if we take any vertex which is not v1 or v2 , then the critical edge
of that vertex which intersects the common edge in a closer point to v1 than the
other critical edge is Type 1, and the other is Type 2.
10
UMUT VAROLGUNES
v1
v
v2
Figure 5. The proof of Lemma 8
There are 3 different ways for the critical edges of v1 and v2 to intersect with
each other, which are shown in the Figure 4. For the vertices that stay inside one
of the angles that are made by the 2 critical edges of v1 and v2 it is easy to see that
the claim is true by the basic proporties of n-intersecting sets and Lemma 8. This
finishes the proof for the first 2 cases, because every edge of the maximal bipartite
set has at least one endpoint at one such vertex.
Now assume that the claim is wrong, for some vertices in a maximal bipartite set
like the third figure. Take the vertex which does not satisfy the claim and closest
to v1 arcwise. Call that vertex v. The other endpoint of the Type 1 critical edge
of v, should have a Type 2 critical edge, and the other endpoint of that edge is
a vertex which is closer to v1 arcwise than v. Yet this vertex should also have a
Type 1 critical edge, which should intersect the Type 1 critical edge of v. This
contradicts the assumption that v is the closest one that does not satisfy the claim.
This contradiction is illustrated in Figure 5.
Lemma 9. For any edge of the maximal bipartite set except the common edge there
is at least one critical edge which does not cross it.
Proof. Every edge except the common edge has at least one edge in the maximal
bipartite which does not cross it. If this is a critical, edge we are done. Otherwise,
it easy to see that one of the critical edges of an endpoint of that edge does not
cross it.
The following corollary proves Theorem 2.
Corollary 7. The decomposition of an maximal bipartite set is unique up to permutation.
Proof. We first determine the common edge and the sets of the critical edges, up
to permutation. Then the sets that the others belong to are uniquely determined
by Lemma 9.
SPUR PROJECT
11
Finally we give some remarks on how this can be used to enumarate the number
of maximal bipartite sets.
Definition 4. Let f ((m, s), k) denote the number of almost intersecting subsets of
K2s+2m+2−k such that in its decomposition the star parts of the (2s + 2m + 2 − k)intersecting sets have m and s elements.
Proposition 2. f ((m, s), k) is equal to half the number of ways to choose two
subsets A = {ai | 1 ≤ a1 < · · · < a2m+1 ≤ 2s + 2m + 2 − k} and B = {bi | 1 ≤ b1 <
· · · < b2s+1 ≤ 2s + 2m + 2 − k} of {1, . . . , 2s + 2m + 2 − k} satisfying the conditions
below;
• A and B have exactly k elements in common.
• There is only one pair (i, j) such that both ai ai+n bj ai+n+1 and bj bj+m ai bj+m+1
are convex quadrilaterals on the circle.
• If ai = bj for some i, j, then ai ai+n ai+n+1 bj+m bj+m+1 is a convex pentagon
on the circle.
The following corollary is a direct application of Corollary 7 and Proposition 1.
{2}
Corollary 8. The degree of In is equal to the summation below;
X
(1)
C(n, 2s + 2m + 2 − k)f ((m, s), k)
2s+2m+2−k≤n
{2}
Therefore in order to compute the degree of V (In ), it is enough to compute
f ((m, s), k).
As the last result of this section we give a big family of examples of pairs of
subsets that satisfy the conditions of Proposition 2. In fact, it is checked by hand
that all such pairs are in this form for n < 11. The smallest counterexample we
could find is for n = 14, which is given after this proposition.
Proposition 3. Let A = {ai | 1 ≤ a1 < · · · < a2m+1 ≤ 2s + 2m + 2 − k}
and B = {bi | 1 ≤ b1 < · · · < b2s+1 ≤ 2s + 2m + 2 − k} be two subsets of
{1, . . . , 2s + 2m + 2 − k} such that the following conditions hold;
• A and B have k elements in common.
• There is either ∃ p ∈ [2m + 1] such that there are at least s + 1 elements
of B in the arc ap ap+1 that does not contain any other elements of A, or
∃ q ∈ [2s + 1] such that there are at least m + 1 elements of A in the arc
bq bq+1 that does not contain any other elements of B.
• In case that such p in the second bullet exists, then bp+s+1 is not contained
in A, and the same statement for q.
Then they satisfy the conditions of Proposition 2
Remark 2. For n = 14, if we take A = {1, 2, 3, 6, 7, 11, 12} and B = {4, 5, 8, 9, 10, 13, 14}.
They satisfy the conditions of Proposition 2, so they are counted in f ((7, 7), 0), but
they are not in the form described in Proposition 3.
6. The pureness of the initial complex
{2}
As discussed in the background part, the fact that the initial complex of In
is pure can be shown by using algebraic geometry. We now give a combinatorial
proof of this fact.
12
UMUT VAROLGUNES
If a subset of E(Kn ) can be represented as a union of two disjoint intersecting
sets, we call it bipartite. We need to show that any bipartite set is contained in at
least one maximal bipartite set.
We proceed by contradiction. Assume that there are some bipartite sets which
are not contained in any maximal bipartite set. Let B be one of such sets with
maximal cardinality.
Lemma 10. B is a union of two n-intersecting sets.
Proof. Let B be the union of disjoint intersecting sets X and Y . Let X 0 and Y 0
be n-intersecting sets, containing X and Y respectively. If X 0 * B, then we can
add at least one more edge to X which is not contained in Y such that it keeps
being intersecting. This can not happen because of the maximality of B. The same
statement is true for Y 0 too. Therefore B ⊃ X 0 ∪ Y 0 ⊃ X ∪ Y = B, which implies
that B = X 0 ∪ Y 0 .
We take the common edge definition in Section 5 similarly. Any edge of a bipartite set which lies in the intersection of two n-intersecting sets of which union
is the bipartite set is called common. Clearly, an edge is common if and only if it
crosses every other edge in the bipartite set.
We define the neighbour edges of ij ∈ E(Kn ) as the edges of the form (i+1)(j−1)
and (i − 1)(j + 1). Note that every edge has at least one neighbour edge (see Figure
(a)).
A1
1
8
2
3
7
6
A4
A2
4
5
A3
(a) Neighbour edges of 14
(b) Different ways to cross two common edges
Lemma 11. Assume that there are 2 common edges of B which don’t have a
common endpoint. Call these edges e1 and e2 , and denote their endpoints by A1
and A3 , and A2 and A4 respectively. All the other edges cross either A1 A2 and
A3 A4 , or A1 A4 and A3 A2 . If the corresponding vertices of two edges of B are
connected in the induced subgraph of B in non-crossing graph Gn , then they cross
the same pair (see Figure (b)) .
Proof. If we prove that any two adjacent vertices cross the same pair, the lemma
follows. Yet, this follows from the fact that an edge that cross A1 A2 and A3 A4 ,
always cross another edge that cross A1 A4 and A3 A2 .
SPUR PROJECT
13
Because of the way we choose B, when we add another edge to B, it is not
bipartite anymore. This means that the induced graph of these vertices in Gn
should have an odd cycle.
Let e be a common edge of B, and e0 be a neighbour edge of e. Since e0 do not
cross e, e0 can not be contained in B. Therefore, the induced graph of B ∪{e0 } in Gn
should include an odd cycle C with one of its vertices as the vertex corresponding
to e0 . Let the edges correspond to the vertices that are adjacent to e0 in that cycle
be e1 and e2 . It is easy to see that e1 and e2 both have a common endpoint with e,
and because of the definition of non-crossing graph e1 and e2 don’t cross with e0 .
Lemma 12. e, e1 , and e2 do not have the same endpoint (see Figure (c)).
Proof. Assume the contrary. Since e1 and e2 cross, C has at least 5 vertices. Let
e1 be between e and e2 . Then if an edge of B does not cross e1 , then it does not
cross e2 either. Therefore the vertex that is adjacent to the vertex corresponding
to e1 other than e is also adjacent to e2 . This means that B itself has an odd cycle,
which contradicts the fact that B is bipartite.
e1
e0
e1
e0
e2
e
e
e2
(c) Lemma 12
(d) The last case
Now we divide the problem into two parts. Since B is not maximal bipartite,
there are at least two common edges.
First, assume that all the common edges of B are incident to the same vertex.
We take any two that are adjacent. It is clear that one of those edges has at least
one neighbour edge that cross the other common edge. If we take that common
edge and its neighbour edge as our e and e0 above, we see that e, e1 , and e2 have
the same endpoint. Hence this case gives a contradiction.
Otherwise, there are two common edges with no common endpoints (see Figure
(d)). Take one of them as e, one neighbour of e as e0 . The corresponding edges
of the vertices that are incident to e0 in the odd cycle C, e1 and e2 , should be
connected in the induced subgraph of B. Moreover, because of Lemma 12, they
can not have a common endpoint. Therefore they have to cross different pairs of
edges of the quadrilateral with vertices as the endpoints of the two common edges,
as described in Lemma 11. This is a contradiction. Pureness is shown.
14
UMUT VAROLGUNES
References
[1] J. De Loera, B. Sturmfels, and R. Thomas. Gröbner bases and triangulations of the second
hypersimplex. Combinatorica 15 (1995), no. 3, 409–424.
[2] M. Drton, B. Sturmfels, and S. Sullivant. Algebraic factor analysis: tetrads, pentads and
beyond. Probab. Theory Related Fields 138 (2007), no. 3-4, 463–493.
[3] B. Sturmfels and S. Sullivant. Combinatorial secant varieties. Quarterly Journal of Pure and
Applied Mathematics 2 (2006) 285–309, (Special issue: In Honor of Robert MacPherson).
[4] A. Simis and B. Ulrich. On the ideal of an embedded join. J. Algebra 226 (2000) 1–14.
[5] M. Kalkbrener and B. Sturmfels. Initial complexes of prime ideals. Advances in Mathematics.
116 (1995), 365–376.
[6] S. Sullivant. A Groebner basis for the secant ideal of the second hypersimplex. Journal of
Commutative Algebra. 1 (2009), no. 2