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

Discrete & Computational Geometry, Mar 2018

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 \(\Delta ^4 \times \Delta ^n\) is not connected by flips for large n. Our proof makes use of a non-explicit probabilistic construction.

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

A Zonotope and a Product of Two Simplices with Disconnected Flip Graphs Gaku Liu 0 1 0 Max Planck Institute for Mathematics in the Sciences , 04103 Leipzig , Germany 1 Mathematics Subject Classification 52B11 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 4 × n is not connected by flips for large n. Our proof makes use of a non-explicit probabilistic construction. 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 Zonotope; Zonotopal tiling; Flip graph; Product of two simplices 1 Introduction 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. ff6oo.2rr]aaflloll rnna.,nHaonovdwertevhvieeerwa,u.wtSheaonsrtho[os6w][1pt4hr]oapvtretodhveethdflaittphtahgtertahpflehipfloipgfrgarpa4hp×hofof n 3i2s××notnncoiissnncceoocnntnneeedccttfeeoddr n ≈ 4 · 104. 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, for mulated 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. 2 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. 2.1 Subdivisions and Triangulations Throughout this section, let A ⊂ Rm 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 ∈ (Rm )∗ 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 . Definition 2.1 A polyhedral subdivision, or subdivision, of A is a collection S of cells of A such that 1. If C ∈ S and F is a face of C , then F ∈ S . 2. If C, C ∈ S , then conv(C ) ∩ conv(C ) = conv(F ) where F is a face of C and C . 3. C∈S conv(C ) = conv( A). 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 S [F ] of F by S [F ] := {C ∈ S : C ⊆ F }. 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. 2.2 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 Rm . If X = {x1, . . . , xk } is a circuit, then the elements of X satisfy an affine dependence equation k i=1 λi xi = 0, where λi ∈ R \ {0} for all i , i λi = 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 + = {xi : λi > 0} and X − = {xi : λi < 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: TX+ := {σ ⊆ X : σ X +}, TX− := {σ ⊆ X : σ X −}. Given a subdivision S and a cell C ∈ S , we define the link of C in S as linkS (C ) := {C ∈ S : C ∩ C = ∅, C ∪ C ∈ S }. We can now state the definition of a flip, in the form of a proposition. Proposition 2.2 (Santos [12]) Let T be a triangulation of A. Suppose there is a circuit X = ( X +, X −) contained in A such that 1. TX+ ⊆ T . 2. All maximal simplices of TX+ have the same link L in T . Then the collection T := T \ {ρ ∪ σ : ρ ∈ L , σ ∈ TX+} ∪ {ρ ∪ σ : ρ ∈ L , σ ∈ TX−} 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. 2.3 Regular Subdivisions Let A ⊆ Rm be as before. Let ω : A → R be any function. For a cell C define the lift of C to be the set C ω ⊂ Rm ×R given by ⊆ A, we C ω := {(x , ω (x )) : x ∈ C }. We call a subset F ⊆ Aω a lower face of Aω if either F is empty or there is a linear functional φ ∈ (Rm × R)∗ such that φ (0, 1) > 0 and F is the set of all points which minimize φ on Aω. Then the collection of all C ⊆ A such that C ω is a lower face of Aω is a subdivision of A. We call this the regular subdivision of A with respect to ω, and denote it by SAω. If F is a face of A, then SAω[F ] = SFω|F . Both triangulations of a circuit are regular, as stated below. Proposition 2.4 Suppose ik=1 λi xi = 0 is the affine dependence equation for a circuit X = (X +, X −) with X + = {xi : λi > 0} and X − = {xi : λi < 0}. Let ω : X → R be a function. Then SXω = TX+ if TX− if k i=1 λi ω(xi ) > 0, k i=1 λi ω(xi ) < 0. 2.4 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 := {e1, . . . , em } be the standard basis for Rm and n−1 := { f1, . . . , fn } be the standard basis for Rn. We embed m−1 × n−1 in Rm × Rn by m−1 × n−1 := {(ei , f j ) : i ∈ [m], j ∈ [n]}. Let G := Km,n be the complete bipartite graph with vertex set m−1 ∪ n−1 and edge set {ei f j : i ∈ [m], j ∈ [n]}. We have a bijection (ei , f j ) → ei f j between m−1 ∪ n−1 and the edge set of G. For each cell C ⊆ m−1 × n−1, let G(C ) be the minimal subgraph of G with edge set {ei f j : (ei , f j ) ∈ C }. 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. 3 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]. 3.1 Mixed Subdivisions Let A1, . . . , An be finite subsets of Rm . The Minkowski sum of A1, . . . , An is the set of points Ai = A1 + · · · + An := {x1 + · · · + xn : xi ∈ Ai for all i }. In this paper, we want the phrase “ Ai ” to identify a set of points but also retain the information of what A1, . . . , An are. In other words, Ai will formally mean an ordered tuple ( A1, . . . , An) but by abuse of notation will also refer to the Minkowski sum. A mixed cell of Ai is a set Bi where Bi is a cell of Ai for all i . A mixed cell is fine if all the Bi are simplices and lie in independent affine subspaces. A face of a mixed cell Bi is a mixed cell Fi of Bi such that there exists a linear functional φ ∈ (Rm )∗ such that for all i , either Fi = ∅ or Fi is the set of all points which minimize φ on Bi . Definition 3.1 A mixed subdivision of such that 1. If Bi ∈ S and Fi is a face of 2. If Bi , Bi ∈ S , then conv is a face of Bi and Bi . 3. Bi ∈S conv Bi = conv Ai is a collection S of mixed cells of Bi , then Bi ∩ conv Fi ∈ S . Bi = conv Ai . Fi where Ai Fi 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 Ai whose minimal elements are the fine mixed subdivisions. 3.2 Basic Sums and Zonotopes Let A1, . . . , An be as in the previous section. Let m−1 = {e1, . . . , em } be the standard basis of Rm . We will say that Ai is basic if Ai ⊆ m−1 for all i . Let A1, . . . , An be as in the previous section, and assume additionally that | Ai | = 2 for all i . Then Ai is called a zonotope, and its mixed subdivisions are called zonotopal tilings. A fine zonotopal tiling is called a tight zonotopal tiling. Example 3.2 For all 1 ≤ i < j ≤ m, let Ai j = {ei , e j }. Then m−1 := is a basic zonotope called the (m − 1)-dimensional permutohedron. 1≤i< j≤m Ai j 3.3 Coherent Mixed Subdivisions Let A1, . . . , An be finite subsets of Rm . For each i = 1, . . . , n, let ωi : Ai → R be a function. For a mixed cell Bi of Ai , define the lift Bi ω by Bi ω := Biωi . We call a mixed cell Fi of Ai ω a lower face of Ai ω if there is a linear functional φ ∈ (Rm × R)∗ such that φ (0, 1) > 0 and for each i , either Fi is empty or Fi is the set of points which minimize φ on Aiωi . The collection of all Bi such that Bi ω is a lower face of Ai ω is a mixed subdivision of Ai called the coherent mixed subdivision of Ai with respect to ω. We denote it by S ω . In the case where Ai is basic, we have the following characterization ofASi ω . Ai Theorem 3.3 (Develin and Sturmfels [ 3 ]) Let Ai be basic, and let ωi : Ai → R be functions. For any x = (x1, . . . , xm ) ∈ Rm and Ai , let type(x , ω, Ai ) be the set of all e j ∈ Ai such that x j − ωi (e j ) = max {xk − ωi (ek ) : ek ∈ Ai }. Then Bi is an element of S ω Ai if and only if there is some x ∈ Rm such that for all i , either Bi = ∅ or Bi = type(x , ω, Ai ). 3.4 The Cayley Trick Let A1, . . . , An be as before. Let We define the Cayley embedding of n−1 = { f1, . . . , fn} be the standard basis of Rn. Ai to be the following set in Rm × Rn: n i=1 C Ai := {(x , fi ) : x ∈ Ai }. The Cayley trick says the following. Theorem 3.4 (Sturmfels [16], Huber et al. [ 4 ]) The following are true. 1. C is a bijection between the mixed cells of Ai and the cells of C Ai , and this map preserves facial relations. 2. For any mixed subdivision S of Ai , the collection C(S ) is a subdivision of C Ai . The map S → C(S ) is a poset isomorphism between the mixed subdivisions of Ai and the subdivisions of C Ai . 3. If ωi : Ai → R are functions for i = 1, . . . , n and C(ω) : C Ai → R is defined as C(ω)(x , fi ) = ωi (x ), then C S ω Ai = S CC((ω) Ai ). Example 3.5 If Ai is basic, then C m−1 for all i , then C Ai = m−1 × Ai is a a subset of n−1. Example 3.6 If m−1 is the (m − 1)-dimensional permutohedron (see Example 3.2), then C( m−1) can be written as m−1 × n−1. If Ai = C( m−1) = {(ei , fi j ), (e j , fi j )} ⊂ Rm × R(m2 ). 1≤i< j≤m 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. Remark 3.7 Cayley embeddings of zonotopes are Lawrence polytopes. The connectivity of flip graphs on zonotopal tilings of zonotopes can be equivalently stated as the connectivity of flip graphs on triangulations of Lawrence polytopes. 4 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. 4.1 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 denote the set of all ordered k-tuples S (i1, . . . , ik ) of distinct i1, . . . , ik ∈ S under the equivalence relation (i1, . . . , ik ) ∼ (i2, . . . , ik , i1). We will use (i1 · · · ik ) to denote the equivalence class of (i1, . . . , ik ) in Sk. We write—(i1 · · · ik ) to denote (ik · · · i1). We abbreviate [kn] as nk . 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 {e1, . . . , e4} be the standard basis for R4 and let { fα}α∈ 42 be the standard basis for R6. Then 3 := {(ei , f(i j)), (e j , f(i j))} ⊂ R4 × R6 (i j)∈ 42 (i j k) ∈ 43, we have a circuit X(i jk) = (X(+i jk), X(−i jk)) in relation 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 3 with affine dependence ei , f(i j) − e j , f(i j) + e j , f( jk) − ek , f( jk) + ek , f(ki) − ei , f(ki) and X(+i jk), X(−i jk) defined in terms of this affine relation. Define T(i jk) := TX+(i jk) . Note that X−γ = (Xγ−, Xγ+) and hence T−γ = TX−γ . Remark 4.1 In the zonotope setting, each circuit X(i jk) “corresponds” to a pair of opposite hexagonal facets of the 3-permutohedron, namely the facets generated by {ei − e j , e j − ek , ek − ei }; see Fig. 1. The correspondence is in the following sense: In any tight zonotopal tiling S of the permutohedron, these facets are tiled in the same way. The Cayley embedding of S is a triangulation C(S ) of 3, and the tiling of the aforementioned facets is determined by the triangulation induced by C(S ) on the circuit X(i jk). Fig. 1 Facets (outlined in red) corresponding to the circuit X(i jk) ei − e j ek − ei e j − ek We will now construct eight different tilings of element of 43. Fix some γ = (i j k) ∈ 43. Let ω : 3 3, each indexed by a different → R be the function with ω ei , f(i j) = ω e j , f( jk) = ω ek , f(ki) = 1 and ω (x ) = 0 for all other x ∈ 3. We define T γ3 := S ω3 . It is easy to check (using Theorem 3.3, for example) that T γ3 is a triangulation. Let us take a closer look at T γ3 where γ = (i j k). For each γ ∈ 43, the circuit Xγ is a face of 3, and thus T γ3 contains the triangulation it induces on Xγ . From the definition of ω, we have ω ei , f(i j) − ω e j , f(i j) + ω e j , f( jk) − ω ek , f( jk) + ω ek , f(ki) − ω ei , f(ki) > 0 and if l is the element of [ 4 ] \ {i, j, k}, we have ω ei , f(i j) − ω e j , f(i j) + ω e j , f( jl) − ω el , f( jl) ω e j , f( jk) − ω ek , f( jk) + ω ek , f(kl) − ω el , f(kl) + ω el , f(li) − ω ei , f(li) > 0, ω ek , f(ki) − ω ei , f(ki) + ω ei , f(il) − ω el , f(il) + ω el , f(l j) − ω e j , f(l j) > 0, + ω el , f(lk) − ω ek , f(lk) > 0. Thus, by Proposition 2.4, T γ3 induces the following triangulations on the circuits Xγ : T(i jk), T(i jl), T( jkl), T(kil) ⊆ T γ3 . (4.1) 4.2 A Group Action on The key property of T γ3 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 γ3 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 43. For each γ = (i j k) ∈ 43, we define a function oγ : [43] → 43 by oγ ({i, j, k}) = (i j k), oγ ({i, j, l}) = (i jl), oγ ({ j, k, l}) = ( j kl), oγ ({k, i, l}) = (kil), π(i j)(i j k) = ( j il), π(kl)(i j k) = (i jl). oπ(i j)γ ({i, j, k}) = −oγ ({i, j, k}), oπ(i j)γ ({i, j, l}) = −oγ ({i, j, l}), oπ(i j)γ ({ j, k, l}) = oγ ({ j, k, l}), oπ(i j)γ ({k, i, l}) = oγ ({k, i, l}). where {l} = [ 4 ] \ {i, j, k}. The function oγ is a way to “orient” each element of [ 4 ] with respect to γ . Recalling Eq. (4.1), oγ is defined in such a way that Toγ (S) ⊆ T3γ3 for all S ∈ [43] . It is easy to check that γ is determined by oγ . Now, we will map each α ∈ 42 to a permutation πα : 43 → 43. 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 ∈ 43, we have (4.2) In other words, π(i j) reverses the value of oγ on S ∈ [ 4 ] if {i, j } ⊆ S and leaves it 3 the same otherwise. Let G 3 be the permutation group of 43 generated by all the πα. 4 Proposition 4.2 The following are true. 1. Every element of G 43 is an involution, and G 43 is abelian and transitive on 43. 2. For l ∈ [ 4 ], let Hl be the subgroup of G 3 generated by π(il) for all i ∈ [ 4 ] \ {l}. 4 Let i, j, k ∈ [ 4 ] \ {l} be distinct, and let 43(i j k) be the set of all γ ∈ 43 such that oγ ({i, j, k}) = (i j k). Then 43(i j k) is an orbit of Hl . Proof Since each γ is determined by oγ , we can view G 3 as an action on the set of 4 functions oγ . It is then clear from (4.2) that we can embed G 3 as a subgroup of Z42. 4 This implies that every element of G 3 is an involution and G 3 is abelian. It is also 4 4 easy to check from (4.2) that every element of 43 has orbit of size 8, and hence G 3 4 is transitive. From (4.2), we see that Hl maps 43(i j k) to itself and every element of 43(i j k) has orbit of size 4 under Hl . Since | 43(i j k)| = 4, 43(i j k) is an orbit of Hl . 4.3 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 firj , and we make the identification of variables firj = f j−i r . { firj }1≤i< j≤4,−N ≤r≤N be the standard basis for R6(2N +1). Let We think of the firj as copies of the variable f(i j) defined in Sect. 4.1. Let := 1≤i< j≤4 −N ≤r≤N {(ei , firj ), (e j , firj )} ⊂ R4 × R6(2N +1) (4.3) 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 . For distinct i, j, k ∈ [ 4 ] and for any −N ≤ r, s, t ≤ N , let Xirjskt = ((Xirjskt )+, (Xirjskt )−) be the circuit with affine dependence relation (ei , firj ) − (e j , firj ) + (e j , f jsk ) − (ek , f jsk ) + (ek , fkti ) − (ei , fkti ) and (Xirjskt )+, (Xirjskt )− defined in terms of this relation. Let Tirjkst := TX+irjskt . Lemma 4.4 Let C be a collection of triangulations of the form Tirjkst such that for all Tirjkst ∈ C and {l} = [ 4 ] \ {i, j, k}, there exist 1 ≤ u, v, w ≤ N such that Tirjlv(−u), T jskwl(−v), Tktiul(−w) ∈ C . Let SC be the set of all triangulations of subset. Then SC is closed under flips. The proof of this lemma follows immediately from the following two facts. Proposition 4.5 Let T be a triangulation of such that Tirjkst ⊆ T . Let T be the result of a flip on T which is not supported on Xirjskt . Then Tirjkst ⊆ T . have X − ⊆ Xirjskt . Proof Suppose the flip from T to T is supported on X = (X +, X −). By Proposition 2.2, if σ ∈ T and σ ∈/ T , then σ ⊇ X −. Thus, if Tirjkst T , then we must On the other hand, the only circuits of are of the form Xir sj kt or which contain every element of C as a {(ei , fir j ), (e j , fis j )}, {(e j , fir j ), (ei , fis j )} Xirjskt , and hence Tirjkst ⊆ T . for some i , j , k , r , s , t . Of these circuits, the only circuit whose negative part is contained in Xirjskt is Xirjskt itself. Since X = Xirjskt by assumption, we must therefore have X − Proposition 4.6 Let T be a triangulation of , and suppose that there are distinct i, j, k, l ∈ [ 4 ] and 1 ≤ r, s, t, u, v, w ≤ N such that Tirjlv(−u), T jskwl(−v), Tktiul(−w) ⊆ T . Then T does not have a flip supported on Xirjskt . Proof Consider the cell C : = (ei , fiul ), (el , fiul ), (e j , f jvl ), (el , f jvl ), (ek , fkwl ), (el , fkwl ) . C is a face of , as seen by the linear functional φ such that φ (ea , 0) = p φ (0, fab) = 0 for all a, p −1 if fab = fiul , f jvl , or fkwl , 0 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 {(ei , firj ), (e j , firj )}, {(e j , f jsk ), (ek , f jsk )}, {(ek , fkti ), (ei , fkti )}. Suppose (ei , firj ) ∈ τ . Then since C ⊆ τ , we have (Xirjvl(−u))+ = {(ei , firj ), (e j , f jvl ), (el , fl−iu )} ⊆ τ. On the other hand, since Tirjlv(−u) ⊆ T by assumption, we have (Xirjvl(−u))− ∈ 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 , firj ) ∈ τ . Similarly, (ek , f jsk ) ∈ τ and (ei , fkti ) ∈ τ . Hence, (Xirjskt )− ⊆ τ . We thus have (Xirjskt )− ⊆ τ and |Xirjskt ∩ τ | = 3 < |Xirjskt | − 2. By Proposition 2.3, T does not have a flip supported on Xirjskt . Theorem 4.3 follows from Lemma 4.4 if there is some C for which SC is neither empty nor the whole set of triangulations of . We show this in the next section. 4.4 Construction of C and Some T ∈ SC We first consider the regular subdivision S ω where ω : that → R is a function such ω(ei , firj ) − ω(e j , firj ) = r for all distinct i, j ∈ [ 4 ] and −N ≤ r ≤ N . The cells of S ω are described as follows. Proposition 4.7 Let X be the set of x = (x1, x2, x3, x4) ∈ R4 such that x1 + · · · + x4 = 0 and if i j kl is a permutation of [ 4 ] such that xi ≥ x j ≥ xk ≥ xl , then xi − x j , x j − xk , and xk − xl are integers at most N . Let X ∗ be the set of x ∈ X such that |xi − x j | ≤ N for all i, j ∈ [n]. The following are true. 1. The map C (x ) = {(ei , firj ) : xi − x j ≥ r } is a bijection from X to the maximal cells of S ω. 2. If x ∈ X ∗, then C (x ) is the Cayley embedding of a translated 3-permutohedron. Specifically, C (x ) = (x ) ∪ D, where (x ) := Proof We first prove Part 1. By Theorem 3.3, a set C ⊆ if there is some x ∈ R4 such that is a cell of S ω if and only C = (ei , firj ) : xi − ω(ei , firj ) ≥ x j − ω(e j , firj ) = (ei , firj ) : xi − x j ≥ r = C (x ). Moreover, if C is maximal, then there are pairs (i, j ), (i , j ), (i , j ) such that {ei − e j , ei − e j , ei − e j } is linearly independent and (ei , firj ), (e j , firj ), (ei , fir j ), (e j , fir j ), (ei , fir j ), (e j , fir j ) ⊆ C (4.4) Fig. 2 A portion of the tiling associated to ω for some −N ≤ r, r , r ≤ N . If x ∈ R4 is such that C = C (x ) for this maximal cell C , then (4.4) implies xi − x j = r , xi − x j = r , and xi − x j = r . By the linear independence of {ei − e j , ei − e j , ei − e j }, there is a unique x for which these equalities hold and x1 + · · · + x4 = 0. Moreover, this x is contained in X . Finally, for any x ∈ X , C (x ) is a maximal cell of S ω. This proves Part 1. Part 2 follows immediately. The only nontrivial case of Part 3 is when x satisfies xi ≥ x j ≥ xk ≥ xl for some permutation i j kl of [ 4 ], xi − xk ≤ N , x j − xl ≤ N , and xi − xl > N . (A zonotopal tile corresponding to such a cell is highlighted in yellow in Fig. 2.) In this case C (x ) is of the form Xirjskt ∪ X sjuklv ∪ D, where D is a simplex affinely independent to Xirjskt ∪ X sjuklv. By Proposition 5.9 (or by an easy check), any triangulations of Xirjskt and X sjuklv can be extended to a triangulation of Xirjskt ∪ X sjuklv, 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 to the triangulations 4 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) → fixji −x j . For each γ ∈ 43, let T γ(x) be the image of T γ3 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 girj = g −jir be an independent random element of G 3 which is 1 with probability 1/2 and π(i j) with probability 4 1/2. 2. For each x ∈ X ∗, triangulate (x ) by T γ((xx)), where γ (x ) := gixji −x j . 3. Extend T γ((xx)) uniquely to a triangulation TC(x) of C (x ). Proposition 4.9 For any two x , x ∈ X ∗, the triangulations TC(x) and TC(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 Xirjskt . We need to check that TC(x) and TC(x ) agree on this circuit. If Xirjskt ⊆ C (x ) ∩ C (x ), then xi − x j = xi − x j = r, x j − xk = x j − xk = s, xk − xi = xk = xi = t. On the other hand, by Proposition 4.2 (2), oγ (x)({i, j, k}) depends only on gixji −x j , gxjkj −xk , and gkxik−xi . It follows that oγ (x)({i, j, k}) = oγ (x )({i, j, k}). Thus T γ((xx)) and T γ((xx )) contain the same triangulation of Xirjskt , as desired. By Proposition 4.9 and Proposition 4.7 (3), we can thus extend the above triangu lations of the C (x ) to a full triangulation of . Call this triangulation T . Let C be the collection of all triangulations Tirjkst ⊆ 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).) Proposition 4.10 For large enough N , with probability greater than 0, T and C satisfy the following: (A) For every distinct i, j, k ∈ [ 4 ] and −N ≤ r ≤ N , there exist −N ≤ s, t ≤ N such that Tirjkst ∈ C . (B) For every Tirjkst ∈ C and γ ∈ 43(i j k), there exists x ∈ X ∗ such that Xirjskt ⊆ (x ) and T [ (x )] = T γ(x). In particular, when γ = (i j k), this implies there is some −N ≤ u, v, w ≤ N such that Tirjlv(−u), T jskwl(−v), Tktiul(−w) ∈ C , where {l} = [ 4 ] \ {i, j, k}. triangulation of Xirjskt . Letting γ = girj gsjk gkti (123), we have 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 Xirjskt ⊆ (x ). Hence C contains a Tirjkst ∈ C if and only if oγ ({i, j, k}) = (i j k). (4.5) We first bound the probability that (A) does not hold. Fix distinct i, j, k ∈ [4] and −N ≤ r ≤ N . Let H be the set of all ordered pairs (s, t ) ∈ [−N , N ]2 with r +s+t = 0. Note that |H | ≥ N . For each (s, t ) ∈ H and γ (s, t ) := girj g jk gkti (123), it is easy to s see from the proof of Proposition 4.2 that oγ (s,t)({i, j, k}) = (i j k) with probability 1/2. In fact, this probability does not change if we fix girj , so for all (s, t ) ∈ H these probabilities are mutually independent. Now, from (4.5), there does not exist (s, t ) ∈ H such that Tirjkst ∈ C if and only if oγ (s,t)({i, j, k}) = (i j k) for all (s, t ) ∈ H . The probability this happens is 1 |H| 2 ≤ 2 1 N . 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 Tirjkst ∈ C and γ ∈ 43(i j k). Let H be the set of all x ∈ X ∗ such that Xirjskt ⊆ (x ). Note that |H | ≥ N . Let {l} = [ 4 ] \ {i, j, k}. We have Tirjkst ∈ C , which happens if and only if girj gsjk gkti (123) ∈ 43(i j k). Suppose we fix girj , gsjk , and gkti such that girj gsjk gkti (123) ∈ 43(i j k). 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 (xi − xl , x j − xl , xk − xl ) is different for each x ∈ H , these probabilities are mutually independent for all x ∈ H . Now, for each x ∈ X ∗ we have T [ (x )] = T γ((xx)). Thus the probability that there is no x ∈ H with T [ (x )] = T γ(x) is 3 |H| 4 The probability that this occurs for some distinct i, j, k ∈ [ 4 ], −N ≤ r, s, t ≤ N with r + s + t = 0, and γ ∈ 43(i j k) is thus at most 96(2N + 1)2 3 N . 4 (4.7) 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 ∈ SC . There are triangulations of which are not in SC ; for example, a different choice of the girj would yield a triangulation which does not contain every element of C . This proves Theorem 4.3. 5 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. 5.1 A Component of the Flip Graph A := n−1. We prove the following. For each α ∈ 52, construct a finite set α of variables. We will determine the size of this set later. Let 4 := {e1, . . . , e5} be the standard basis for R5, and let n−1:= α∈ 52 α be the standard basis for Rn, where n = α∈ 52 | α|. Let 4 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 i1, …, it ∈ [5] and distinct f1, …, ft ∈ n−1, let X f1··· ft i1···it = (Xif11······itft )+, (Xif11······itft )− be the circuit in A with affine dependence relation (ei1 , f1) − (ei2 , f1) + (ei2 , f2) − (ei3 , f3) + · · · + (eit , ft ) − (ei1 , ft ) and (Xif11······itft )+, (Xif11······itft )− defined in terms of this relation. Let Ti1f·1···i·t· ft := TX+f1··· ft . i1···it If f1 ∈ (i1i2), f2 ∈ (i2i3), …, ft ∈ f1··· ft zonotopal. We now identify a component of the flip graph of A. triangulation Ti1···it (it i1), then we call the circuit Xif11······itft and the 2. If i1 = i , i2 = j , and i4 is as above, then there exist f1 ∈ and f3 ∈ (i3i4) such that (i1i4), f2 ∈ (i2i4), In addition, assume the following about the CS,i, j . 3. For each distinct i, j, l ∈ [ 5 ] and f1 ∈ ( jk), and f3 ∈ (ki) such that (i j), there exist k ∈ [ 5 ] \ {i, j, l}, f2 ∈ each S ∈ true. Lemma 5.2 For each S ∈ [45] and distinct i , j ∈ S, let CS,i, j be a collection of f1 f2 f3 where i1, i2, i3 ∈ S. Assume that for zonotopal triangulations of the form Ti1i2i3 [45] and distinct i, j ∈ S, and for each Ti1fi12if32 f3 ∈ CS,i, j , the following are Let SC be the set of all triangulations T of A satisfying the following. (i) For each S ∈ [54] , distinct i, j ∈ S, and Ti1fi12if32 f3 ∈ CS,i, j , we have (ii) For each S ∈ [45] , distinct i, j ∈ S, and Ti jfk1 f2 f3 ∈ CS,i, j where k ∈ S \ {i, j }, if {l} = [ 5 ] \ S, then for any f ∈ (il) we have Xifj1kf2 f3 \ {(ei , f1)} ∪ {(ei , f )} ∈ T . Then SC 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 TX+. Proposition 5.4 Let T be a triangulation of A. Let i, j, k, l ∈ [ 5 ] be distinct, and for each α ∈ {2i, j,k,l}, let fα ∈ n−1. Suppose that T j kf(ljk) f(kl) f(l j) , Tkifl(ki) f(il) f(lk) ⊆ T and Then T does not have a flip supported on Xifj(kij) f( jk) f(ki) . We also need the following two facts about flips on elements of SC . Proposition 5.5 Let i, j, l ∈ [ 5 ] be distinct and let f1 ∈ T ∈ SC . Then T does not have a flip supported on Xifj1 f2 . (i j), f2 ∈ (il). Let ∈ C[ 5 ] \ {l},i, j . By property (ii) of SC , it follows that ( jk), and f3 ∈ (ki) σ := Xifj1kf2 f3 \ {(ei , f1)} ∪ {(ei , f2)} ∈ T . supported on Xif11i2fi23f3 . So by Proposition 2.3, T does not have a flip supported on Xifj1 f2 . If τ is a maximal simplex in T containing σ , then (Xifj1 f2 )− ⊆ τ but |Xifj1 f2 ∩ τ | = 2. Proposition 5.6 Let Ti1fi12if32 f3 ∈ CS,i, j . Let T ∈ SC . Then T does not have a flip Proof This is a direct corollary of Property 1 of CS,i, j and Proposition 5.4. We now proceed with the proof. Suppose that T ∈ SC , 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 Ti1fi12if32 f3 ∈ CS,i, j and Ti1fi12if32 f3 T . Without loss of generality, assume that X f1 f2 f3 i1i2i3 \ {(ei1 , f1)} ∈/ T . By Proposition 5.3, Xif11i2fi23f3 \ {(ei1 , f1)} contains a maximal simplex of TX+. This leaves two cases for X . Case 1: X has size 4. Thus we can write X = Xif11i2f2 for some i1, i2 ∈ {i1, i2, i3}, f1 ∈ (i1i2), and f2 ∈ (i1i3), where {i3} = {i1, i2, i3} \ {i1, i2}. However, this contradicts Proposition 5.5. So we cannot have |X | = 4. Case 2: X = X f1 f2 f3 . This contradicts Proposition 5.6. i1i2i3 Hence we must have Ti1fi12if32 f3 ⊆ T , as desired. Property (ii). Suppose that σ := Xifj1kf2 f3 \ {(ei , f1)}∪{(ei , f )} ∈/ 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 = Xifj1kf2 f3 . This leaves X = X jfifk3 f2 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 CS,i, j , there exist f1 ∈ (il ), f2 ∈ ( jl ), f3 ∈ (kl ) such that By Property (ii), the first of these inclusions implies that and hence Xifj1l f2 f1 \ {(ei , f1)} ∪ {(ei , f )} ∈ T X jfifl1 f2 \ {(e j , f )} ∈ T . 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 jfifk3 f2 , as desired. 5.2 Reduction to Zonotopes Let := (i j)∈ 52 f ∈ (i j) be the large 4-permutohedron embedded in A. Suppose we have collections CS,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 ∈ SC . 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 SA := SAω 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 SA 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 SA 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 SA which is a triangulation of A. and and 5.3 Construction of a Zonotopal Tiling We now construct a triangulation of CS,i, j . For each (i j ) ∈ 52, partition from which we will obtain our collections (i j) into the following sets: i, j , j,i , and 5 S,i , j ,(i j) for each S ∈ [ 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 fi∗,j ∈ with i, j ∈ S, let S,i, j,(i j) ⊆ i, j be sets such that i, j . For each S ∈ [45] i, j = S,i, j,(i j) → R defined as Finally, for each S ∈ [45] 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 ∈ [45] and distinct i, j ∈ S, let S,i, j := (i j )∈ 2S f ∈ S,i, j,(i j ) PS,i, j := In addition, for each k ∈ S \ {i, j }, let S,i, j,k := Proposition 5.8 Every cell of S is of the form C ∪ D, where D is a simplex affinely independent to C and C is a face of one of the following. ∈ [54] are distinct and contain i, j 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 ∈ R5 with xi = 0, x j = 1, xl = S,i, j,l for each l ∈ S \ {i, j }, and xk = S ,i, j,k . The cell in (b) corresponds to x ∈ R5 with xi = 0, x j = 1, xk = S,i, j,k , xk = S ,i, j,k , and xk = 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 ∈ [45] 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 ek → eψ(k) for all k ∈ [ 4 ] and so that { fkrl }−N ≤r≤N maps bijectively to S,i, j,(ψ(k)ψ(l)). We will define : → S,i, j so that f1−2N maps to fi∗,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 TS,i, j and CS,i, j be the images of T and C , respectively, under . We thus have a triangulation TS,i, j of each S,i, j . To extend the TS,i, j to triangulations of the complex cells, we use the following “ordered pseudoproduct” construction. It is proved in the Appendix. Proposition 5.9 Let 1, 2, . . . , N , and ρ be cells of such that ρ = {(ei , f ), (e j , f )} for some distinct i, j ∈ [ 5 ] and f ∈ (i j), and for any 1 ≤ r, s ≤ N , we have G( r ) ∩ G( s ) = G(σ ). Let T1, . . . , TN be triangulations of 1, . . . , N , respectively. Let M be the set of all simplices σ ∈ A of the following form: There is an integer 1 ≤ s ≤ N such that σ = where • If r < s, then σr is a maximal simplex of Tr with ρ ⊆ σr . • σs is a maximal simplex of Ts , and if s < N , then ρ := σs ∩ ρ = ρ. • If r > s, then σr is a maximal simplex of Tr [Fr ], where Fr is a facet of r with Fr ∩ ρ = ρ . Then M is the set of maximal simplices of a triangulation T (T1, . . . , TN ) of 1 ∪ · · · ∪ N , and Tr ⊆ T (T1, . . . , TN ) for all 1 ≤ r ≤ N . Now, fix distinct i, j ∈ [ 5 ], and specify an ordering S1, S2, S3 of the sets S ∈ [45] which contain i and j . Each complex cell is of the form F1 ∪ F2 ∪ F3 where for each distinct 1 ≤ r, s ≤ 3, Fr is a face of Sr ,i, j , and G(Fr ) ∩ G(Fs ) = G(ρ) where ρ := {(ei , fi∗,j ), (e j , fi∗,j )}. By Proposition 5.9, we thus have a triangulation T (TS1,i, j [F1], TS2,i, j [F2], TS3,i, j [F3]) of F1∪F2∪F3. 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 TS,i, j as subcollections. For each S ∈ [45] and distinct i, j ∈ [n], let DS,i, j be the set of all zonotopal fi∗, j f2 f3 triangulations of the form Ti jfk1 f2 f3 such that k ∈ S \ {i, j } and Ti jk ∈ CS,i, j . Let CS,i, j := CS,i, j ∪ DS,i, j . We now finally show that the assumptions of Lemma 5.2 hold for CS,i, j and T . Proposition 5.10 The collections CS,i, j satisfy Properties 1–3 of Lemma 5.2. exist f1 ∈ such that Proof We first prove that Properties 1 and 2 hold. Suppose Ti1fi12if32 f3 ∈ CS,i, j . If Ti1fi12if32 f3 ∈ CS,i, j , then Properties 1 and 2 follow from Proposition 4.10 (B) by setting the appropriate values of γ . So we may assume Ti1fi12if32 f3 = Ti jfk1 f2 f3 ∈ DS,i, j . By definition of DS,i, j , we have Ti jfki∗, j f2 f3 ∈ CS,i, j . Hence, by Proposition 4.10 (B), there S,i, j,(il), f2 ∈ S,i, j,( jl), and f3 ∈ S,i, j,(kl), where l = S \ {i, j, k}, By definition, we thus have Ti jfl1 f2 f1 ∈ DS,i, j . Hence Ti jfli∗, j f2 f1 , T jkl f3 f1 f3 f2 f3 f2 , Tkil 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 f1 ∈ (i j). Let S = [5] \ {l}. By Proposition 4.10 (A) applied to the triangulation TS,i, j , for any k ∈ S \ {i, j } there exists f2 ∈ S,i, j,( jk) and f3 ∈ S,i, j,(ki) such that Ti jfki∗, j f2 f3 Thus by definition, Ti jfk1 f2 f3 ∈ DS,i, j ⊆ CS,i, j , which proves Property 3. Proposition 5.11 Properties (i) and (ii) of Lemma 5.2 hold for T with respect to the collections CS,i, j . Proof We first note the following two facts. Proposition 5.12 For any f ∈ (i j), we have {(ei , fi∗,j ), (e j , f )} ∈ T . Proof In our construction of S from ω, we had ω(ei , fi∗,j ) − ω(e j , fi∗,j ) + ω(e j , f ) − ω(ei , f ) ≤ 0 for all f ∈ (i j), with equality if and only if f ∈ i, j . In addition, T is a refinement of a regular subdivision of given by height function ω where ω(e1, f1−2N ) − ω(e2, f1−2N ) + ω(e2, f1r2) − ω(e1, f1r2) < 0 for all r = N . Thus, for all f ∈ (i j), restricting T to the face Xifji∗, j f of yields the triangulation TX−ifji∗, j f , and hence {(ei , fi∗,j ), (e j , f )} ∈ T . fi∗, j f2 f3 Proposition 5.13 Let Ti jk ⊆ T and f1 ∈ (i j). Then Ti jfk1 f2 f3 ⊆ T . Proof Let σ0 := Xifji∗k,j f2 f3 \ {(ei , fi∗,j )} ∈ T . By considering a maximal simplex of T containing σ0, we have σ0 ∪{(e, f1)} ∈ T for some e ∈ {ei , e j }. If e = ei , then {(ei , f1), (e j , fi∗,j )} ⊆ σ0. However, this contradicts Propositions 5.12 and 6.2 in the Appendix. Thus e = e j . Hence, we have σ1 := Xifj1kf2 f3 \ {(ei , f1)} ∈ T . Now, the circuit X := Xifj1kf2 f3 is a face of , so T [X ] is a triangulation of X . Since σ1 ∈ T [X ], this triangulation must be TX+. Thus Ti jfk1 f2 f3 ⊆ T . ∈ CS,i, j by definition. Thus ⊆ Ti jfki∗, j f2 f3 ⊆ T , so by Proposition 5.13, Ti jfk1 f2 f3 ⊆ T . Hence Property (i) holds. Now suppose Ti jfk1 f2 f3 ∈ CS,i, j , and let f ∈ (il) where {l} = [ 5 ] \ S. By Property (i), we have Xifj1kf2 f3 \ {(ei , f1)} ∈ T . By Proposition 6.1 in the Appendix, we have Xifj1kf2 f3 \ {(ei , f1)} ∪ {(ei , f )} ∈ T . Thus Property (ii) holds. Hence, we have collections CS,i, j which satisfy the hypotheses of Lemma 5.2, and by Propositions 5.11 and 5.7, there exists a triangulation T ∈ SC . Clearly SC is not the set of all triangulations of A; for example, there exist triangulations of A which do not contain any triangulations of circuits of size six [ 2 ]. Hence Lemma 5.2 proves Theorem 5.1. 6 Appendix 6.1 Proof of Proposition 5.3 Since σ ∈/ T , by Proposition 2.2 we have σ ⊇ X −. Let τ be a maximal simplex of T containing σ . We must have τ ∈/ T since σ ∈/ T . So by Proposition 2.2, τ must contain a maximal simplex τ of TX+. Now, since σ ∪ τ ⊆ τ , we have that G(σ ∪ τ ) is acyclic. But since σ ⊇ X − and G(σ ) is connected, this can only happen if σ ⊇ τ , as desired. 6.2 Proof of Proposition 5.4 Throughout this section, let A := m−1 × n−1. We need the following two facts. Proposition 6.1 Let T be a triangulation of A. Let σ ∈ T . Suppose that f ∈ and f ∈/ G(σ ). Then there exists e ∈ m−1 ∩ G(σ ) such that σ ∪ {(e, f )} ∈ T . Proof Let σ := m−1 ∩ G(σ ). Let φ : A → R be the linear functional with −1 if e ∈ σ , 0 otherwise. φ (0, f ) = 0 for all f ∈ n−1. Let F be the face of A which minimizes φ. Then σ ⊆ F . Let τ be a maximal simplex of T [F ] containing σ . Then τ ⊇ σ ∪ {(e, f )} for some e ∈ σ . By symmetry of f ∈ n−1 with e ∈ m−1 and m−1. n−1, the above proposition also holds after replacing Proposition 6.2 Let T be a triangulation of A, and suppose X = (X +, X −) is a circuit of A. Then T does not contain both X + and X −. Fig. 3 G(σ0) f(li) e j el Proof The interiors of the convex hulls of opposite parts of a circuit intersect. The next proposition immediately implies Proposition 5.4. Proposition 6.3 Let T be a triangulation of A. Let i, j, k, l ∈ [m] be distinct , and for each α ∈ {2i, j,k,l}, let fα ∈ n−1. Suppose that and T j kf(ljk) f(kl) f(lj) , Tkil f(ki) f(il) f(lk) ⊆ T Then T does not have a flip supported on Xifj(kij) f( jk) f(ki) . Proof Suppose the contrary. In particular, this means Ti jfk(i j) f( jk) f(ki) ⊆ T . Set σ0 = Xifj(li j) f( jl) f(li) \ {(ei , f(i j))} (Fig. 3). By assumption, σ0 ∈ T . By Proposition 6.1, there exists σ1 ∈ T such that σ1 = σ0 ∪ {(e, f( jk))} ∪ {(e , f(kl))}, where e, e ∈ {ei , e j , el }. We prove the following claims. Claim 1: e = e j . Suppose first that e = ei . Then {(ei , f( jk)), (e j , f(i j))} ⊆ σ1. However, we also have {(ei , f(i j)), (e j , f( jk))} ∈ Ti jfk(i j) f( jk) f(ki) ⊆ T by assumption. Thus T contains both parts of the circuit Xifj( jk) f(i j) , which contradicts Proposition 6.2. Now suppose that e = el . Then {(el , f( jk)), (e j , f( jl))} ⊆ σ1. However, we also have {(el , f( jl)), (e j , f( jk))} ∈ T j kf(ljk) f(kl) f(lj) ⊆ T , again a contradiction. So e = e j . Fig. 4 G(σ1) Fig. 5 G(σ2) f(i j ) ei e j el e j el f(li) f(kl) f(li) f(kl) Claim 2: e = el . Suppose first that e = ei . Then {(ei , f(kl)), (el , f(li))} ⊆ σ1. However, we have {(ei , f(li)), (el , f(kl))} ∈ Tkifl(ki) f(il) f(lk) ⊆ T , a contradiction. Now suppose that e = e j . Then {(e j , f(kl)), (el , f( jl))} ⊆ σ1. However, we have {(e j , f( jl)), (el , f(kl))} ∈ T j kf(ljk) f(kl) f(lj) ⊆ T , a contradiction. So e = el (Fig. 4). Now, by Proposition 6.1 and the comment afterwards, there exists σ2 ∈ T such that σ2 = σ1 ∪ {(ek , f )} where f ∈ { f(i j), f( jk), f(il), f( jl), f(kl)}. We prove the following. Thus f = f( jk) (Fig. 5). Claim 3: f = f( jk). The argument goes as follows: If f = f(i j), then {(ek , f(i j)), (e j , f( jk))} ⊆ σ2 but {(ek , f( jk)), (e j , f(i j))} ∈ Ti jfk(i j) f( jk) f(ki) . If f = f(il), then {(ek , f(il)), (el , f(kl))} ⊆ σ2 but {(ek , f(kl)), (el , f(il))} ∈ Tkifl(ki) f(il) f(lk) . If f = f( jl), then {(ek , f( jl)), (e j , f( jk))} ⊆ σ2 but {(ek , f( jk)), (e j , f( jl))} ∈ T j kf(ljk) f(kl) f(lj) . If f = f(kl), then (X jfk(jlk) f(kl) f(lj) )+ ⊆ σ2, but (X jfk(jlk) f(kl) f(lj) )− ∈ T j kf(ljk) f(kl) f(lj) . Fig. 6 G(σ3) e j el f(ki) f(li) f(kl) Finally, by Proposition 6.1 there exists σ3 ∈ T such that σ3 = σ2 ∪ {(e, f(ki))} Claim 4: e = ei . The argument goes as follows. If e = e j , then {(e j , f(ki)), (ek , f( jk))} ⊆ σ3 but {(e j , f( jk)), (ek , f(ki))} ∈ Ti jfk(i j) f( jk) f(ki) . If e = ek , then (Xkfi(lki) f(il) f(lk) )+ ⊆ σ3 but (Xkfi(lki) f(il) f(lk) )− ∈ Tkil f(ki) f(il) f(lk) . If e = el , then {(el , f(ki)), (ei , f(il))} ⊆ Tkifl(ki) f(il) f(lk) . σ3 but {(el , f(il)), (ei , f(ki))} ∈ Thus e = ei (Fig. 6). Now, let τ be a maximal simplex of T containing σ3. Then Xifj(kij) f( jk) f(ki) − ⊆ τ but Xifj(kij) f( jk) f(ki) ∩ τ = 4. This contradicts Proposition 2.3 and our assumption that T has a flip supported on Xifj(kij) f( jk) f(ki) . So T does not have such a flip. 6.3 Proof of Proposition 5.9 Our proof is based on the following. Theorem 6.4 (Rambau [ 8 ]) Let M be a nonempty collection of full-dimensional simplices of a point set A. Then M is the set of maximal simplices of a triangulation of A if and only if 1. There is no circuit X = (X +, X −) of A and simplices τ, τ ∈ M such that X + ⊆ τ , X − ⊆ τ . 2. For any simplex τ ∈ M and facet σ of τ , either σ is contained in a facet of A or there is another τ ∈ M , τ = τ , such that σ ⊆ τ . Let M be as in Proposition 5.9. We prove Properties 1 and 2 of Theorem 6.4 for M . Proof of Property 1 Suppose that X = (X +, X −) is a circuit of we have X + ⊆ σ , X − ⊆ σ for some σ, σ ∈ M . Let N and σ = σ = r<s r<s as in Proposition 5.9, with analogous definitions for σ . Suppose first that X ⊆ r for some 1 ≤ r ≤ N . Then we have X + ⊆ σ ∩ r ⊆ σr and similarly X − ⊆ σr . This contradicts the fact that σr , σr are simplices of a triangulation Tr . So we cannot have this case. It follows that X ⊆ r1 ∪ r2 for some 1 ≤ r1 < r2 ≤ N , and we can uniquely write X = Xr1 ∪ Xr2 \ ρ , where Xr1 = (Xr+1 , Xr−1 ) and Xr2 = (Xr+2 , Xr−2 ) are circuits in r1 and r2 respectively, and Xr+1 = (X + ∩ Xr+2 = (X + ∩ r1 ) ∪ ρ1, r2 ) ∪ ρ2, Xr−1 = (X − ∩ Xr−2 = (X − ∩ r1 ) ∪ ρ2, r2 ) ∪ ρ1, where ρ1, ρ2 are different one-element subsets of ρ. We claim that Xr+1 ⊆ σr1 . First suppose that r1 < s. Then (X + ∩ r1 ) ∪ ρ ⊆ σr1 , so Xr+1 ⊆ σr1 , as desired. So we may assume r1 ≥ s, and hence s < r2 ≤ N . Now, since s < N , we have σ ∩ ρ = ρi for either i = 1 or 2. If i = 1, then Xr+1 ⊆ σr1 and we are done. Suppose i = 2. Then Xr+2 ⊆ σr2 . Since r2 > s, it follows that Xr+2 is contained in a facet F of r2 with F ∩ ρ = ρ2. However, Xr+2 ⊆ F implies Xr2 ⊆ F because F is a face of r2 . This contradicts F ∩ ρ = ρ2. Hence we have Xr+1 ⊆ σr1 . By the same argument, we have Xr−1 ⊆ σr1 . Hence Xr+1 , Xr−1 ∈ Tr1 , a contradiction. This proves Property 1. Proof of Property 2 Let σ = r>s be an element of M as before. Let x ∈ σ , and consider the facet σ \ {x }. We have the following cases. Case 1: x ∈ σt \ ρ where t < s. Subcase 1.1: σt \ {x } is not contained in a facet of t . Then there is some x ∈ such that σt := σt \ {x } ∪ {x } ∈ Tt . Then t r<s r =t (σr \ ρ) ∪ (σt \ ρ) ∪ σs ∪ r>s is an element of M containing σ \ {x }, as desired. Subcase 1.2: σt \ {x } is contained in a facet F of t . Let φ : t → R be a linear functional supporting F on t . Since ρ ⊆ σt \ {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. Case 2: x ∈ ρ. Subcase 2.1: σ ∩ ρ = ρ. Then σ \ {x } is contained in a facet of 1 ∪ · · · ∪ N because f has no neighbors in G(σ \ {x }). Subcase 2.2: σ ∩ ρ = ρ. This implies s = N and x ∈ σr for all r . First, suppose there is some t such that σt \ {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 σt := σt \ {x } ∪ {x } ∈ Tt . Thus r<t r>t is an element of M containing σ \ {x }, as desired.4 Now suppose there is no such t . Then for all r, σr \ {x } is contained in a facet Fr of r . We have x ∈/ Fr since σr is full dimensional in r . For each r , let φr : r → R be a linear functional supporting Fr on r . Let {y} = ρ \ {x }. Then y ∈ Fr and x ∈/ Fr , so φr (y) < φr (x ). By appropriately choosing φr , we may assume φr (y) = 0 and φr (x ) = 1 for all r . As before, we can then define a linear functional φ on 1 ∪ · · · ∪ N such that φ (v) = φr (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 . Case 3: x ∈ σs \ ρ. Subcase 3.1: σs \ {x } is not contained in a facet of s . Then there is some x ∈ such that σs := σs \ {x } ∪ {x } ∈ Ts . First assume that x ∈/ ρ. Then s 2 This is because t is contained in an affine subspace not containing the origin. 3 We can do this because any affine dependence in 1 ∪ · · · ∪ N can be written as a sum of affine dependencies each contained in one of the r . 4 In more detail: by definition of t, if r > t then σr \ {x} is contained in a facet F of r . We must have F ∩ ρ = ρ \ {x} because σr is full-dimensional in r . r<s is an element of M containing σ \ {x }, as desired. Now assume that x ∈ ρ. First, suppose that for all r > s, σr := σr ∪ {x } ∈ Tr . Then (σr \ ρ) ∪ (σs \ ρ) ∪ (σr \ ρ) ∪ σN r<s s<r<N is an element of M containing σ \ {x }, as desired. Now suppose that there is some t > s such that σt ∪ {x } ∈/ Tt , and let t be the smallest such number. Let x be a point in Tt such that x ∈/ σt and σt := σt ∪ {x } ∈ Tt . By definition of t , x = x , and hence x ∈/ ρ. Then (σr \ ρ) ∪ (σs \ ρ) ∪ (σr \ ρ) ∪ σt ∪ r<s s<r<t r>t is an element of M containing σ \ {x }, as desired. Subcase 3.2: σs \ {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 σs ∩ ρ = ρ \ {x } for some x ∈ ρ. First, suppose that there is some t < s such that σt \ {x } is not contained in a facet of Tt , and let t be the largest such number. Then there is some x ∈ Tt such that σt := σt \ {x } ∪ {x } ∈ Tt . Then r<t t<r<s r>s is an element of M containing σ \ {x }, as desired. Now suppose that there is no such t . Then for all r , we have that (σ \ {x }) ∩ r is contained in a facet Fr of r . We also have Fr ∩ ρ = ρ \ {x } for all r : For r < s, this holds because σr 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. Case 4: x ∈ σt \ ρ where t > s. Let ρ = σt ∩ρ. Let U ⊆ Tt be the simplicial complex consisting of all simplices in Tt on the boundary of t . Let U be the subcomplex of U consisting of all simplices in Tt contained in facets F of t with F ∩ ρ = ρ . In particular, we have σt ∈ U . Since U is homeomorphic to a sphere, there is some x ∈ t such that σt := σt \ {x } ∪ {x } ∈ U . If σt ∈ U , then (σr \ ρ) ∪ σs ∪ σt ∪ r<s r>s r =t 1 ∪ · · · ∪ is an element of M containing σ \ {x }, as desired. Suppose σt ∈/ U . Let F be the facet of t containing σt . Since ρ ⊆ σt , 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 N , as desired. σr ∈ ∪ σs ∪ r >s is in M . If ρ ⊆ σs , then for all r = s we can choose σr ∈ r and r such that ρ ⊆ σr for all σ = (σr \ ρ) ∪ (σs \ ρ) ∪ (σr \ ρ) ∪ σN r <s s<r <N is in M . Either way, σs ∈ T . Thus, T1, . . . , TN ⊆ T . Acknowledgements Open access funding provided by Max Planck Society. 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Gaku Liu. A Zonotope and a Product of Two Simplices with Disconnected Flip Graphs, Discrete & Computational Geometry, 2018, 1-33, DOI: 10.1007/s00454-018-9971-6