A Zonotope and a Product of Two Simplices with Disconnected Flip Graphs

We give an example of a three-dimensional zonotope whose set of tight zonotopal tilings is not connected by flips. Using this, we show that the set of triangulations of Δ4×Δn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta ^4 \times \Delta ^n$$\end{document} is not connected by flips for large n. Our proof makes use of a non-explicit probabilistic construction.


Introduction
We consider the poset P of polyhedral subdivisions of a polytope P or zonotopal tilings of a zonotope Z , ordered by refinement. This poset is called the Baues poset of P or Z . The minimal elements of this poset are, respectively, the triangulations of P or the tight zonotopal tilings of Z . Two minimal elements of P are connected by a flip if there is an element of P whose only proper refinements are these two minimal elements. The flip graph of P or Z is the graph whose vertices are minimal elements of P and whose edges are flips.
We are concerned in this paper with connectivity of the flip graph. For zonotopal tilings, the flip graph is known to be connected for cyclic zonotopes [19] or if the zonotope has dimension two [5]. Our first result is the first example of a zonotope whose Editor in Charge: Kenneth Clarkson Gaku Liu gakuliu@gmail.com flip graph is not connected. This answers a question of Reiner in [9]. Our example is a three-dimensional permutohedron with many copies of each of its generating vectors. The number of copies of each vector is determined by a probabilistic argument to be around 100.
Using a related construction, we also show that the flip graph of triangulations of the product of two simplices is not generally connected. Santos [10] gave the first example of a polytope whose flip graph of triangulations is not connected. However, the case when P is a product of two simplices remained of special interest due to the appearances of these triangulations in various branches of mathematics; see [2,Chap. 6.2] for an overview. Santos [14] proved that the flip graph of 2 × n is connected for all n, and the author [6] proved that the flip graph of 3 × n is connected for all n. However, we show that the flip graph of 4 × n is not connected for n ≈ 4 · 10 4 .
Triangulations of lattice polytopes are closely related to toric varieties in algebraic geometry. Each lattice polytope defines a toric ideal and an associated toric Hilbert scheme. For a totally unimodular polytope, the associated toric Hilbert scheme is connected if and only if the flip graph of the polytope is connected; see [17,Chap. 10] or [7]. Thus, our result implies that the toric Hilbert scheme of 4 × n is not connected for large n. While non-connected toric Hilbert schemes had previously been constructed [13], our proof demonstrates non-connectivity for the first time for a totally unimodular polytope. In addition, the toric ideal associated to m × n is the well-studied determinantal ideal generated by 2 × 2 minors of an (m + 1) × (n + 1) matrix; its zero-locus is the Segre variety.
Connectivity of flip graphs is also related to the generalized Baues problem, formulated by Billera et al. [1], which concerns the topology of the Baues poset P. Specifically, the problem asks if the order complex of P minus its maximal element is homotopy equivalent to a sphere. While the problem has been resolved for most cases of interest, it remains open for zonotopes, and is of particular interest in this case because it is equivalent to the extension space conjecture for oriented matroids, which states that the extension space of a realizable oriented matroid is homotopy equivalent to a sphere. See [11,18] for definitions and known results. In general, the generalized Baues problem and the question of flip graph connectivity do not imply each other. However, there are situations where the answer to one can be used to answer the other. The generalized Baues problem for triangulations was answered in the negative by Santos [15] by constructing a point set in general position with disconnected flip graph. The problem for zonotopal tilings remains open, but would be answered in the negative if a zonotope with generating vectors in general position was found to have disconnected flip graph; see [9,Lem. 3.1].
The paper is organized as follows. Section 2 reviews triangulations and the product of two simplices. Section 3 reviews mixed subdivisions, zonotopal tilings, and the Cayley trick. Section 4 constructs our zonotope and proves that its flip graph is not connected. Section 5 proves that the flip graph of 4 × n is not connected. Section 6 is an appendix proving several propositions used in the paper.

Triangulations and the Product of Two Simplices
We begin with a quick overview of triangulations, flips, and the product of two simplices. We refer to De Loera et al. [2] for a more comprehensive treatment.

Subdivisions and Triangulations
Throughout this section, let A ⊂ R m be a finite set of points. A cell of A is a subset of A. A simplex is a cell which is affinely independent. A face of a cell C is a subset F ⊆ C such that there exists a linear functional φ ∈ (R m ) * such that F is the set of all points which minimize φ on C. For any cell C, let conv(C) denote the convex hull of C.
The subdivision consisting of A and all faces of A is the trivial subdivision. A subdivision all of whose elements are simplices is a triangulation.
If S is a subdivision of A and F is a face of A, then S induces a subdivision For subdivisions S , S , we say that S is a refinement of S if every element of S is a subset of an element of S . Refinement gives a poset structure on the set of all subdivisions of A. The maximal element of this poset is the trivial subdivision and the minimal elements are the triangulations.

Flips
As stated in the introduction, two triangulations are connected by a flip if there is a subdivision whose only proper refinements are these two triangulations. We will now give an equivalent definition of a flip which will be easier to use.
A circuit is a minimal affinely dependent subset of R m . If X = {x 1 , . . . , x k } is a circuit, then the elements of X satisfy an affine dependence equation where λ i ∈ R \ {0} for all i, i λ i = 0, and the equation is unique up to multiplication by a constant. This gives a unique partition X = X + ∪ X − of X given by X + = {x i : λ i > 0} and X − = {x i : λ i < 0}. We will write X = (X + , X − ) to denote a choice of which part we call X + and which we call X − .
A circuit X = (X + , X − ) has exactly two non-trivial subdivisions, which are the following triangulations: Given a subdivision S and a cell C ∈ S , we define the link of C in S as We can now state the definition of a flip, in the form of a proposition.
is a triangulation of A. We say that T has a flip supported on (X + , X − ), and that T is the result of applying this flip to T . The flip graph (on triangulations) of T is the graph whose vertices are triangulations of T and with an edge between two triangulations if one is obtained from the other by a flip.
The following fact is a convenient way to determine whether a flip supported on a certain circuit exists. It is proved in [6].

Proposition 2.3
Let T be a triangulation of A and let X = (X + , X − ) be a circuit in A. Suppose that X − ∈ T . Then T has a flip supported on (X + , X − ) if and only if there is no maximal simplex τ ∈ T with X − ⊆ τ and |X ∩ τ | ≤ |X | − 2.

Regular Subdivisions
Let A ⊆ R m be as before. Let ω : A → R be any function. For a cell C ⊆ A, we define the lift of C to be the set C ω ⊂ R m ×R given by We call a subset F ⊆ A ω a lower face of A ω if either F is empty or there is a linear functional φ ∈ (R m × R) * such that φ(0, 1) > 0 and F is the set of all points which minimize φ on A ω . Then the collection of all C ⊆ A such that C ω is a lower face of A ω is a subdivision of A. We call this the regular subdivision of A with respect to ω, and denote it by F . Both triangulations of a circuit are regular, as stated below.

The Product of Two Simplices
We now consider m−1 × n−1 , the product of two simplices of dimensions m − 1 and n − 1. Following the conventions of the previous section, we will understand m−1 × n−1 to mean the set of vertices of m−1 × n−1 rather than the polytope itself.
Let m−1 := {e 1 , . . . , e m } be the standard basis for R m and n−1 : Let G := K m,n be the complete bipartite graph with vertex set m−1 ∪ n−1 and Then C is a simplex if and only if G(C) is acyclic, and C is a circuit if and only if G(C) is a cycle. If C is a circuit, then G(C) alternates between edges corresponding to positive and negative elements of the circuit.

Zonotopal Tilings
We now define zonotopal tilings in terms of mixed subdivisions. We review the Cayley trick which shows that mixed subdivisions can be thought of as polyhedral subdivisions of another polytope. The information in this section was developed in [4], [14], and [16].

Mixed Subdivisions
Let A 1 , . . . , A n be finite subsets of R m . The Minkowski sum of A 1 , . . . , A n is the set of points A i = A 1 + · · · + A n := {x 1 + · · · + x n : x i ∈ A i for all i}.
In this paper, we want the phrase " A i " to identify a set of points but also retain the information of what A 1 , . . . , A n are. In other words, A i will formally mean an ordered tuple (A 1 , . . . , A n ) but by abuse of notation will also refer to the Minkowski sum.
A mixed cell is fine if all the B i are simplices and lie in independent affine subspaces. A face of a mixed cell B i is a mixed cell F i of B i such that there exists a linear functional φ ∈ (R m ) * such that for all i, either F i = ∅ or F i is the set of all points which minimize φ on B i .
A mixed subdivision is fine if all of its elements are fine.
For mixed subdivisions S , S , we say that S is a refinement of S if every element of S is a mixed cell of an element of S . Refinement gives a poset structure on the set of mixed subdivisions of A i whose minimal elements are the fine mixed subdivisions.

Basic Sums and Zonotopes
Let A 1 , . . . , A n be as in the previous section. Let m−1 = {e 1 , . . . , e m } be the standard basis of R m . We will say that A i is basic if A i ⊆ m−1 for all i. Let A 1 , . . . , A n be as in the previous section, and assume additionally that |A i | = 2 for all i. Then A i is called a zonotope, and its mixed subdivisions are called zonotopal tilings. A fine zonotopal tiling is called a tight zonotopal tiling.

Coherent Mixed Subdivisions
Let A 1 , . . . , A n be finite subsets of R m . For each i = 1, . . . , n, let ω i : A i → R be a function. For a mixed cell We call a mixed cell  [3]) Let A i be basic, and let ω i : A i → R be functions. For any x = (x 1 , . . . , x m ) ∈ R m and A i , let type(x, ω, A i ) be the set of all e j ∈ A i such that

The Cayley Trick
Let A 1 , . . . , A n be as before. Let n−1 = { f 1 , . . . , f n } be the standard basis of R n . We define the Cayley embedding of A i to be the following set in R m × R n : The Cayley trick says the following.

Example 3.5 If
can be written as We say that two fine mixed subdivisions T and T differ by a flip if the triangulations C(T ) and C(T ) differ by a flip. We define the flip graph analogously.

A Zonotope with Disconnected Flip Graph
We are now ready to construct a zonotope and a nontrivial component of its flip graph. Explicitly, this zonotope is the three-dimensional permutohedron, but with each of its generating vectors repeated a large number of times. The idea of the proof will be to generate certain random tilings of this zonotope, and then show that most of the generated tilings will be in different components of the flip graph. We begin by developing the machinery needed to generate the random tilings.

Zonotopal Tilings of the 3-Permutohedron
It will be notationally easier to work with Cayley embeddings of zonotopes rather than zonotopes themselves. Thus we will identify a zonotope with its Cayley embedding. We first set up some notation. For a set S, let k S denote the set of all ordered k-tuples as k n . The main object we will focus on is the Cayley embedding of the 3-dimensional permutohedron. This object was described in Example 3.6; here we will redefine it using slightly different indices. Let {e 1 , . . . , e 4 } be the standard basis for R 4 and let { f α } α∈ 2 4 be the standard basis for R 6 . Then is the (Cayley embedding of the) 3-dimensional permutohedron. (Note we have written 3 instead of C( 3 ) here for convenience.) Our proof will focus on specific circuits of size 6 in the permutohedron. For any  We will now construct eight different tilings of 3 , each indexed by a different element of 3 4 and ω(x) = 0 for all other x ∈ 3 . We define T γ 3 := S ω 3 . It is easy to check (using Theorem 3.3, for example) that T γ 3 is a triangulation. Let us take a closer look at T γ 3 where γ = (i jk). For each γ ∈ 3 4 , the circuit X γ is a face of 3 , and thus T γ 3 contains the triangulation it induces on X γ . From the definition of ω, we have Thus, by Proposition 2.4, T γ 3 induces the following triangulations on the circuits X γ : (4.1)

A Group Action on 3 4
The key property of T γ 3 is that it only has flips on the circuits X (i jl) , X ( jkl) , and X (kil) . The idea will be to tile a larger zonotope with 3-permutohedra and then tile each 3-permutohedron with some T γ 3 so that in the end, no circuit of size six can be flipped. To help with this construction, we will take some time to define a group action on 3 4 .
for all S ∈ [4] 3 . It is easy to check that γ is determined by o γ . Now, we will map each α ∈ 2 4 to a permutation π α : 3 4 → 3 4 . This map is completely determined by the following rules: For any distinct i, j, k, l ∈ [4], we have We can check that for all distinct i, j, k, l ∈ [4] and γ ∈ 3 4 , we have In other words, π (i j) reverses the value of o γ on S ∈ [4] 3 if {i, j} ⊆ S and leaves it the same otherwise. Let G 3 4 be the permutation group of 3 4 generated by all the π α .

Proposition 4.2
The following are true. is an involution and G 3 4 is abelian. It is also easy to check from (4.2) that every element of 3 4 has orbit of size 8, and hence G 3 4 is transitive. From (4.2), we see that H l maps 3 4 (i jk) to itself and every element of 3 4

A Zonotope and a Component of Its Flip Graph
We are now ready to go into the main proof. Let N be a positive integer to be determined later. For each distinct i, j ∈ [4] and −N ≤ r ≤ N , we create a variable f r i j , and we make the identification of variables We think of the f r i j as copies of the variable be the 3-dimensional permutohedron with 2N + 1 copies of each generating vector. As a zonotope, is the unit 3-permutohedron scaled by 2N + 1. We now prove our first main result.

Theorem 4.3 For large enough N , the flip graph of is not connected.
We will use the following lemma which identifies a component of the flip graph of based on the triangulations of certain circuits of .
and (X rst

Lemma 4.4 Let C be a collection of triangulations of the form T rst i jk such that for all
Let S C be the set of all triangulations of which contain every element of C as a subset. Then S C is closed under flips.
The proof of this lemma follows immediately from the following two facts.
On the other hand, the only circuits of are of the form X r s t i j k or Of these circuits, the only circuit whose negative part is contained in X rst i jk is X rst i jk itself. Since X = X rst i jk by assumption, we must therefore have X − X rst i jk , and hence T rst i jk ⊆ T .
Then T does not have a flip supported on X rst i jk .
Proof Consider the cell C is a face of , as seen by the linear functional φ such that φ(e a , 0) = 0 for all a, otherwise .
Since C is a simplex, we must have C ∈ T . Thus, there is some maximal simplex τ ∈ T such that C ⊆ τ . Since τ is maximal, it must contain one element from each of the sets Suppose (e i , f r i j ) ∈ τ . Then since C ⊆ τ , we have On the other hand, since T This is a contradiction, because the opposite parts of a circuit cannot both be cells of a triangulation (since the interiors of these cells intersect). Hence (e j , f r i j ) ∈ τ . Similarly, (e k , f s jk ) ∈ τ and (e i , f t ki ) ∈ τ . Hence, (X rst i jk ) − ⊆ τ . We thus have (X rst i jk ) − ⊆ τ and |X rst i jk ∩ τ | = 3 < |X rst i jk | − 2. By Proposition 2.3, T does not have a flip supported on X rst i jk . Theorem 4.3 follows from Lemma 4.4 if there is some C for which S C is neither empty nor the whole set of triangulations of . We show this in the next section.

Construction of C and Some T ∈ S C
We first consider the regular subdivision S ω where ω : → R is a function such that for all distinct i, j ∈ [4] and −N ≤ r ≤ N . The cells of S ω are described as follows.
The following are true. Proof We first prove Part 1. By Theorem 3.3, a set C ⊆ is a cell of S ω if and only if there is some x ∈ R 4 such that Moreover  Fig. 2.) In this case C(x) is of the form X rst i jk ∪ X suv jkl ∪ D, where D is a simplex affinely independent to X rst i jk ∪ X suv jkl . By Proposition 5.9 (or by an easy check), any triangulations of X rst i jk and X suv jkl can be extended to a triangulation of X rst i jk ∪ X suv jkl , and hence to a triangulation of C(x). Remark 4.8 In terms of mixed subdivisions, the coherent mixed subdivision associated to ω is given by tiling the large permutohedron with smaller unit permutohedra (the cells corresponding to X * ) and pieces of permutohedra (the tiles corresponding to X \ X * ). Each point x ∈ X is the center of the zonotopal tile corresponding to C(x).
We now give a brief overview of the rest of the proof. The goal is to construct a triangulation which refines S ω and which is sufficiently "complicated". The idea is to start with a known triangulation and apply the group action G 3 4 to the triangulations of the cells C(x) in a way that produces another triangulation of . As we will show, by randomly doing this process we can produce triangulations from our original triangulation that cannot be produced through sequences of flips.
For each x ∈ X * , we have an affine isomorphism 3 → (x) given by f (i j) → f be the image of T γ 3 under this isomorphism. We will now choose a random triangulation of every C(x), x ∈ X * , as follows: 1. For each 1 ≤ i < j ≤ 4 and −N ≤ r ≤ N , let g r i j = g −r ji be an independent random element of G 3 4 which is 1 with probability 1/2 and π (i j) with probability 1/2.

Proposition 4.9 For any two x, x ∈ X * , the triangulations T C(x) and T C(x ) agree on the common face of C(x) and C(x ).
Proof The only non-trivial case is when C(x)∩C(x ) contains a circuit X rst i jk . We need to check that T C(x) and T C(x ) agree on this circuit. If X rst i jk ⊆ C(x) ∩ C(x ), then On the other hand, by By Proposition 4.9 and Proposition 4.7 (3), we can thus extend the above triangulations of the C(x) to a full triangulation of . Call this triangulation T .
Let C be the collection of all triangulations T rst i jk ⊆ T with i, j, k ∈ [4] distinct, −N ≤ r, s, t ≤ N , and r + s + t = 0. We prove that C satisfies the hypotheses of Lemma 4.4. We will actually prove the following stronger statement, which we will need in the next section. (For the current proof, we only need the second sentence of (B).) Proof First, note that for any distinct i, j, k ∈ [4] and −N ≤ r, s, t ≤ N with r + s + t = 0, there is some x ∈ X * such that X rst i jk ⊆ (x). Hence C contains a triangulation of X rst i jk . Letting γ = g r i j g s jk g t ki (123), we have By the union bound, the probability that this happens for some distinct i, j, k ∈ [4] and −N ≤ r ≤ N is at most This gives an upper bound on the probability of (A) not happening. We now do the same for (B). Fix T rst i jk ∈ C and γ ∈ 3 4 (i jk). Let H be the set of all x ∈ X * such that X rst i jk ⊆ (x). Note that |H | ≥ N . Let {l} = [4] \ {i, j, k}. We have T rst i jk ∈ C , which happens if and only if g r i j g s jk g t ki (123) ∈ 3 4 (i jk). Suppose we fix g r i j , g s jk , and g t ki such that g r i j g s jk g t ki (123) ∈ 3 4 (i jk). Then for each x ∈ H , it follows from Proposition 4.2 and the definition of γ (x) that γ (x) = γ with probability 1/4. Moreover, since ( The probability that this occurs for some distinct i, j, k ∈ [4], −N ≤ r, s, t ≤ N with r + s + t = 0, and γ ∈ 3 4 (i jk) is thus at most Hence, the probability that either (A) or (B) does not hold is at most which for large enough N (specifically, N ≥ 48) is less than 1.
Thus there exists C which satisfies the hypotheses of Lemma 4.4 and T ∈ S C . There are triangulations of which are not in S C ; for example, a different choice of the g r i j would yield a triangulation which does not contain every element of C . This proves Theorem 4.3.

A Product of Two Simplices with Disconnected Flip Graph
We will use the construction from the previous section to show that the product of two simplices does not in general have connected flip graph. The idea will be to go up one dimension and construct multiple copies of T in different directions in this space.

A Component of the Flip Graph
For each α ∈ 2 5 , construct a finite set α of variables. We will determine the size of this set later. Let 4 := {e 1 , . . . , e 5 } be the standard basis for R 5 , and let n−1 := α∈ 2 5 α be the standard basis for R n , where n = α∈ 2 5 | α |. Let A := 4 × n−1 . We prove the following.

Theorem 5.1 For large enough n, the flip graph of A is not connected.
We first identify a component of the flip graph of A. For distinct i 1 , …, i t ∈ [5] and distinct f 1 , …, f t ∈ n−1 , let and (X , …, f t ∈ (i t i 1 ) , then we call the circuit X f 1 ··· f t i 1 ···i t and the triangulation T f 1 ··· f t i 1 ···i t zonotopal. We now identify a component of the flip graph of A.

Lemma 5.2
For each S ∈ [5] 4 and distinct i, j ∈ S, let C S,i, j be a collection of zonotopal triangulations of the form T Assume that for each S ∈ [5] 4 and distinct i, j ∈ S, and for each T ∈ C S,i, j , the following are true.
such that 2. If i 1 = i, i 2 = j, and i 4 is as above, then there exist In addition, assume the following about the C S,i, j .

For each distinct i, j, l ∈ [5] and f
Let S C be the set of all triangulations T of A satisfying the following.
(i) For each S ∈ [5] 4 , distinct i, j ∈ S, and T (ii) For each S ∈ [5] 4 , distinct i, j ∈ S, and T Then S C is closed under flips.
Proof We need the following two facts about triangulations of A. They are analogous to Propositions 4.5 and 4.6. We defer their proofs to the Appendix.

Proposition 5.3 Let T be a triangulation of A and let T be the result of a flip on
T supported on X = (X + , X − ). Suppose that σ ∈ T and σ / ∈ T , and G(σ ) is connected. Then σ contains a maximal simplex of T + X .

Proposition 5.4 Let T be a triangulation of
Then T does not have a flip supported on X We also need the following two facts about flips on elements of S C . such that T If τ is a maximal simplex in T containing σ , then (X So by Proposition 2.3, T does not have a flip supported on X Proof This is a direct corollary of Property 1 of C S,i, j and Proposition 5.4. We now proceed with the proof. Suppose that T ∈ S C , and let T be the result of a flip on T supported on X = (X + , X − ). We prove that properties (i) and (ii) hold for T . Property (i). Suppose that T T . Without loss of generality, assume that X contains a maximal simplex of T + X . This leaves two cases for X .
Case 1: X has size 4. Thus we can write X = X However, this contradicts Proposition 5.5. So we cannot have |X | = 4.
This contradicts Proposition 5.6. Hence we must have T ∈ T , with variables as defined in (ii). By Proposition 5.3, σ contains a maximal simplex of X . By the same argument as in part (i), we cannot have |X | = 4 or X = X . This leaves as the only possibility.
We show that T cannot have a flip on this circuit. Let l = S \ {i, j, k}. By Property 2 of C S,i, j , there exist By Property (ii), the first of these inclusions implies that and hence This inclusion along with the last two inclusions of (5.1) imply, by Proposition 5.4, that T does not have a flip on X f f 3 f 2 jik , as desired.

Reduction to Zonotopes
Let := be the large 4-permutohedron embedded in A. Suppose we have collections C S,i, j which satisfy the conditions of Lemma 5.2. Let T be a triangulation of . Notice that if Properties (i) and (ii) of Lemma 5.2 hold for T , then they hold for any collection containing T . In particular, if T can be extended to a triangulation T of A, then we will have some T ∈ S C . The next proposition guarantees we can always do this.

Proposition 5.7 If T is a triangulation of , then there is a triangulation T of A with T ⊆ T .
Proof Let S A := S ω A be the regular subdivision of A with ω : A → R defined as follows. For each distinct i, j, k ∈ [5] and f ∈ (i j) , let f,k > 0 be a generic positive real number, and define Then S A contains as a cell, corresponding to x = 0 under the notation of Theorem 3.3. Since the f,k are generic, all other cells of S A can be written as F ∪ D, where F is a face of and D is a simplex affinely independent to F. Hence, any triangulation of the cell can be extended to a refinement of S A which is a triangulation of A.

Construction of a Zonotopal Tiling
We now construct a triangulation of from which we will obtain our collections C S,i, j . For each (i j) ∈ 2 5 , partition (i j) into the following sets: i, j , j,i , and S,i , j ,(i j) for each S ∈ [5] 4 with i, j, ∈ S and distinct i , j ∈ S with {i , j } = {i, j}.
Next, for each distinct i, j ∈ [5], choose an element f * i, j ∈ i, j . For each S ∈ [5] 4 with i, j ∈ S, let S,i, j,(i j) ⊆ i, j be sets such that The sizes of all of these sets will be determined later. Now, we let S :=S ω be the regular subdivision of with ω : → R defined as follows. First, for every distinct i, j ∈ [5], we set Finally, for each S ∈ [5] 4 and distinct i, j, k ∈ S, let 0 < S,i, j,k < 1 be a generic real number. We set We analyze the cells of S . For each S ∈ [5] 4 and distinct i, j ∈ S, let S,i, j :=  [5] 4 are distinct and contain i, j and k ∈ S \ {i, j}, k ∈ S \ {i, j}, and k ∈ S \ {i, j} are distinct.
We will call the cells in (a) and (b) the complex cells of S .
Proof Using the notation of Theorem 3.3, the cell in (a) corresponds to x ∈ R 5 with x i = 0, x j = 1, x l = S,i, j,l for each l ∈ S \ {i, j}, and x k = S ,i, j,k . The cell in (b) corresponds to x ∈ R 5 with x i = 0, x j = 1, x k = S,i, j,k , x k = S ,i, j,k , and x k = S ,i, j,k . Due to the genericness of the S,i, j,k , it can be checked that every cell of S is the union of a face of one of these cells and an affinely independent simplex; we leave the details to the reader.
Thus, in order to give a triangulation of which refines S , it suffices to specify triangulations of the complex cells of S which agree on common faces. To do this, we first specify triangulations of each S,i, j . We will then use a "pseudoproduct" operation to extend these to triangulations of the complex cells.
Fix S ∈ [5] 4 and distinct i, j ∈ S. Let be the large 3-permutohedron defined in equation (4.3). Let ψ : [4] → S be a map such that ψ(1) = i and ψ(2) = j. We now choose the sizes of the S,i, j,(i j ) so that we have an affine isomorphism : → S,i, j given by e k → e ψ(k) for all k ∈ [4] and so that { f r kl } −N ≤r ≤N maps bijectively to S,i, j,(ψ(k)ψ(l)) . We will define : → S,i, j so that f −N 12 maps to f * i, j .
Let T and C be the triangulation of and the collection of triangulations, respectively, constructed in Sect. 4.4 which satisfy Proposition 4.10. Let T S,i, j and C S,i, j be the images of T and C , respectively, under . We thus have a triangulation T S,i, j of each S,i, j .
To extend the T S,i, j to triangulations of the complex cells, we use the following "ordered pseudoproduct" construction. It is proved in the Appendix. Now, fix distinct i, j ∈ [5], and specify an ordering S 1 , S 2 , S 3 of the sets S ∈ [5] 4 which contain i and j. Each complex cell is of the form By Proposition 5.9, we thus have a triangulation T ( We leave it as an exercise to verify that all of these triangulations of the complex cells agree on common faces. 1 Hence, we can extend these triangulations of the complex cells to a triangulation T of . By Proposition 5.9, this triangulation T contains all the triangulations T S,i, j as subcollections.
For each S ∈ [5] 4 and distinct i, j ∈ [n], let D S,i, j be the set of all zonotopal triangulations of the form T We now finally show that the assumptions of Lemma 5.2 hold for C S,i, j and T . Proof We first prove that Properties 1 and 2 hold. Suppose T ∈ C S,i, j , then Properties 1 and 2 follow from Proposition 4.10 (B) by setting the appropriate values of γ . So we may assume T By definition, we thus have T 1 The argument is the same as the proof of Property 1 in the proof Proposition 5.9.
which proves Property 1. The argument for Property 2 is analogous.
We now prove Property 3. Let i, j, l ∈ [5] be distinct and let f 1 ∈ (i j) . Let S = [5] \ {l}. By Proposition 4.10 (A) applied to the triangulation T S,i, j , for any k ∈ S \ {i, j} there exists f 2 ∈ S,i, j,( jk) and f 3 ∈ S,i, j, (ki) such that Thus by definition, T  {i, j,k,l} , let f α ∈ n−1 . Suppose that Then T does not have a flip supported on X Proof Suppose the contrary. In particular, this means (Fig. 3). By assumption, σ 0 ∈ T . By Proposition 6.1, there exists σ 1 ∈ T such that where e, e ∈ {e i , e j , e l }. We prove the following claims.    (Fig. 4). Now, by Proposition 6.1 and the comment afterwards, there exists σ 2 ∈ T such that We prove the following.
Finally, by Proposition 6.1 there exists σ 3 ∈ T such that where e ∈ {e i , e j , e k , e l }.
Case 1: x ∈ σ t \ ρ where t < s. Subcase 1.1: σ t \ {x} is not contained in a facet of t . Then there is some x ∈ t such that σ t := σ t \ {x} ∪ {x } ∈ T t . Then is an element of M containing σ \ {x}, as desired. Subcase 1.2: σ t \ {x} is contained in a facet F of t . Let φ : t → R be a linear functional supporting F on t . Since ρ ⊆ σ t \ {x}, we have ρ ⊆ F, hence φ(ρ) = {b} where b is the minimum of φ on t . We may assume b = 0. 2 Now, we can extend φ to a linear functional φ on 1 ∪ · · · ∪ N by setting φ (v) = φ(v) for all v ∈ t and φ (v) = 0 for all v ∈ r , r = t. 3 Then the face of 1 ∪ · · · ∪ N supported by φ contains σ \ {x}, and this face is proper because it does not contain t . Thus σ \ {x} is contained in a facet of 1 ∪ · · · ∪ N , as desired.
Subcase 2.2: σ ∩ ρ = ρ. This implies s = N and x ∈ σ r for all r . First, suppose there is some t such that σ t \ {x} is not contained in a facet of t , and let t be the largest such number. Then there is some x ∈ t such that σ t := σ t \ {x} ∪ {x } ∈ T t . Thus is an element of M containing σ \ {x}, as desired. 4 Now suppose there is no such t. Then for all r, σ r \ {x} is contained in a facet F r of r . We have x / ∈ F r since σ r is full dimensional in r . For each r , let φ r : r → R be a linear functional supporting F r on r . Let {y} = ρ \ {x}. Then y ∈ F r and x / ∈ F r , so φ r (y) < φ r (x). By appropriately choosing φ r , we may assume φ r (y) = 0 and φ r (x) = 1 for all r . As before, we can then define a linear functional φ on 1 ∪· · ·∪ N such that φ (v) = φ r (v) for all v ∈ r . Then the face of 1 ∪· · ·∪ N supported by φ contains σ \ {x}, and this face is proper. Thus σ \ {x} is contained in a facet of 1 ∪ · · · ∪ N .
is an element of M containing σ \ {x}, as desired. Now suppose that there is some t > s such that σ t ∪ {x } / ∈ T t , and let t be the smallest such number. Let x be a point in T t such that x / ∈ σ t and σ t := σ t ∪ {x } ∈ T t . By definition of t, x = x , and hence x / ∈ ρ. Then is an element of M containing σ \ {x}, as desired. Subcase 3.2: σ s \ {x} is contained in a facet F of s . If ρ ⊆ F, then by the same argument as in Subcase 1.2, σ \ {x} is contained in a facet of 1 ∪ · · · ∪ N . So we may assume ρ F. In particular, this means σ s ∩ ρ = ρ \ {x } for some x ∈ ρ.
First, suppose that there is some t < s such that σ t \ {x } is not contained in a facet of T t , and let t be the largest such number. Then there is some x ∈ T t such that Now suppose that there is no such t. Then for all r , we have that (σ \ {x}) ∩ r is contained in a facet F r of r . We also have F r ∩ ρ = ρ \ {x } for all r : For r < s, this holds because σ r is maximal in r , for r = s, this holds by our original assumption, and for r > s, this holds by definition. Hence, by the argument in the second part of Subcase 2.2, σ \ {x} is contained in a facet of 1 ∪ · · · ∪ N , as desired.
is an element of M containing σ \ {x}, as desired. Suppose σ t / ∈ U . Let F be the facet of t containing σ t . Since ρ ⊆ σ t , but F ∩ ρ = ρ , we must have ρ ⊆ F. By the argument from Subcase 1.2, it follows that σ \ {x} is contained in a facet of 1 ∪ · · · ∪ N , as desired.
This proves that M is the set of maximal simplices of a triangulation T . Suppose that σ s is a maximal simplex of T s . If ρ σ s , then for all r = s we can choose σ r ∈ r such that r <s (σ r \ ρ) ∪ σ s ∪ r >s σ r is in M . If ρ ⊆ σ s , then for all r = s we can choose σ r ∈ r such that ρ ⊆ σ r for all r and