Some Rigidity Problems in Toric Topology: I

SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I

FEIFEI FAN, JUN MA AND XIANGJUN WANG

Abstract. We study the cohomological rigidity problem of two families of manifolds with torus actions: the so-called moment-angle manifolds, whose study is linked with combinatorial geometry and combinatorial commutative algebra; and topological toric manifolds, which can be seen as topological generalizations of toric varieties. These two families are related by the fact that a topological toric manifold is the quotient of a moment-angle manifold by a subtorus action. In this paper, we prove that when a simplicial sphere satisﬁes some combinatorial condition, the corresponding moment-angle manifold and topological toric manifolds are cohomological rigid, i.e. their homeomorphism classes in their own families are determined by their cohomology rings. Our main strategy is to show that the combinatorial types of these simplicial spheres (or more generally, the Gorenstein∗ complexes in this class) are determined by the Tor-algebras of their face rings. This is a solution to a classical problem (sometimes know as the B-rigidity problem) in combinatorial commutative algebra for a class of Gorenstein∗ complexes in all dimensions > 2.

Contents

1. Introduction2 Acknowledgements4 2. Preliminaries5 2.1. Notations and conventions5 2.2. Face rings and Tor-algebras6 arXiv:2004.03362v7 [math.AT] 17 Nov 2020 2.3. Moment-angle complexes and manifolds8 2.4. Cohomology of moment-angle complexes8 2.5. Quasitoric manifolds and topological toric manifolds 10 2.6. B-rigidity of simplicial complexes 12 2.7. Puzzle-moves and puzzle-rigidity 13

2010 Mathematics Subject Classiﬁcation. Primary 05E40, 13F55, 14M25; Secondary 05E45. The authors are supported by the National Natural Science Foundation of China (Grant Nos. 11801580, 11871284, 11761072). 1 2 F. FAN, J. MA & X. WANG

3. B-rigidity of a class of Gorenstein∗ complexes 14 4. Proof of Theorem 3.14 18 5. Cohomological rigidity of topological toric manifolds 23 6. The case of simplicial 2-spheres 25 6.1. For B-rigidity 25 6.2. For cohomological rigidity of quasitoric 6-manifolds 28 Appendix A. Cohomology of real moment-angle complexes and the second Baskakov-Hochster ring 30 Appendix B. Factor index and its application 32 Appendix C. Proof of Propsition 4.6 34 Appendix D. Proof of Lemma 4.11 and Lemma 4.12 42 Appendix E. Construction of ﬂag spheres satisfying the SCC 45 Appendix F. A generalization of Proposition 3.11 51 References 54

1. Introduction

A central problem in topology is to classify spaces by equivalent relations such as homeomorphism or homotopy equivalence. The cohomological rigidity problem for manifolds asks whether two manifolds are homeomorphic (or even diffeomorphic) or not if their cohomology rings are isomorphic as graded rings. Although answering this question is very difficult even for the simplest case: spheres, and the answer is no in general cases, if we put further conditions on one or both manifolds we can exploit this additional structure in order to show that the desired homeomorphism or diffeomorphism must exist. In this paper, we study the cohomological rigidity problem for two families of manifolds with torus actions: moment-angle manifolds and topological toric manifolds. They play key roles in the emerging field of toric topology, which has many connections with algebraic geometry, commutative algebra and combinatorics, etc. [43] and [21] are good references for related results and problems.

Associated with every ﬁnite simplicial complex K, there is a CW-complex K , called a moment-angle complex, whose topology is determined by the combinatorialZ type of K (see subsection 2.3); if K is a triangulation of a sphere or more generally a generalized homology sphere, is a compact manifold, called a moment-angle ZK SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 3 manifold. In particular, if K is a polytopal sphere, or more generally the underlying sphere of a complete simplicial fan, then admits a smooth structure. ZK Suppose K has m vertices and k[m] = k[x1, . . . , xm] (deg xi = 2) is the polynomial algebra over a commutative ring k. Let k[K] be the face ring of K (see Deﬁnition 2.1). Then the cohomology ring of the moment-angle complex K can be calculated as follows. Z Theorem 1 ([13, Theorem 4.5.4]). The following isomorphism of algebras holds: H∗( ; k) = Tor∗,∗ (k[K], k). ZK ∼ k[m]

The cohomology of K therefore acquires a bigrading, and we may ask the following question aboutZ the cohomological rigidity for moment-angle manifolds. Problem 2. Suppose and are two moment-angle manifolds such that ZK1 ZK2 H∗( ) = H∗( ) ZK1 ∼ ZK2 as graded (or bigraded) rings. Are and homeomorphic? ZK1 ZK1 Although the cohomology ring is known to be a weak invariant even under homotopy equivalence, no example providing the negative answer to this problem, in both graded and bigraded versions, has been found yet. On the contrary, many results support the aﬃrmative answers to the problem (see e.g. [18,6, 33, 16]). Another important class of spaces in toric topology is the so-called topological toric manifolds, which can be seen as the topological generalizations of compact smooth toric varieties. For topological toric manifolds, we can also ask the following question:

Problem 3. Let M1 and M2 be two topological toric manifolds with isomorphic cohomology rings. Are they diﬀeomorphic?

The problem is solved positively for some particular families of topological toric manifolds, such as some special classes of Bott manifolds [20, 19, 17], quasitoric manifolds over a product of two simplices [24], and 6-dimensional quasitoric manifolds over polytopes from the Pogorelov class [11]. The cohomological rigidity problem is also open for general topological toric manifolds. This paper is divided into two parts. In the first part of this paper we fo- cus our attention on moment-angle manifolds corresponding to a special class of Gorenstein∗ complexes and topological toric manifolds related to such sort of simplicial fans. This class consists of Gorenstein∗ complexes which are flag and satisfy the separable circuit condition (see Definition 3.5). The main results of this paper are the following theorems. 4 F. FAN, J. MA & X. WANG

Theorem 4 (Theorem 3.12, Corollary 3.13). Let and 0 be two moment-angle ZK ZK manifolds. Suppose the cohomology of K and K0 are isomorphic as bigraded rings. Then if K is flag and satisfiesZ the separableZ circuit condition, there is a homeomorphism K = K0 , which is induced by a combinatorial equivalence 0 Z ∼ Z K ∼= K . Theorem 5 (Theorem 5.1). Let M and M 0 be two topological toric manifolds. If the underlying sphere of the simplicial fan corresponding to M is flag and satisfies the separable circuit condition, then the following two conditions are equivalent:

∗ ∗ 0 (i) H (M) ∼= H (M ) as graded rings. (ii) M and M 0 are weakly equivariantly homeomorphic.

Theorem4 and Theorem5 are generalizations of cohomological rigity results in [33] and [11], respectively, for simplicial spheres of dimension 2 to all dimensions. By using a result of Masuda [42], we can get an analog of Theorem5 for toric varieties. Theorem 6 (Theorem 5.4). Let X and X0 be two compact smooth toric varieties. Assume the underlying simplicial sphere of the simplcial fan corresponding to X is ﬂag and satisﬁes the separable circuit condition. Then X and X0 are isomorphic as varieties if and only if their cohomology rings are isomorphic.

In §6 we give more discussion for the special case that K is a simplicial 2-sphere. This section can be regarded as an introduction to the second part [32] of this paper.

Acknowledgements. We thank the Institute for Mathematical Science, National University of Singapore, and the organizers of the Program on Combinatorial and Toric Homotopy in summer 2015, where the first and the second authors got financial support, and special thanks to Victor Buchstaber and Nikolay Erokhovets for delightful conversations and insightful comments there. Thanks to Suyoung Choi for enlightening conversations about the cohomological rigidity problem of quasitoric manifolds at Fudan University. Thanks also to the organizers of the Fudan Topology Workshop in January 2016 for giving us an opportunity to discuss. We also thank the Departments of Mathematics, Princeton University and Rider University, and the organizers of the Princeton-Rider Workshop on the Homotopy Theory of Polyhedral Products in May-June 2017, for making the first author present our work and for providing the first and the second authors financial support, and special thanks to FrédéricBosio for helpful comments and for introducing us to the important result of [7] there, which is closely related to the B-rigidity result of our work. SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 5

Many thanks to Nikolay Erokhovets for providing many important references and remarks, as well as proofreading early versions for misprints.

2. Preliminaries

2.1. Notations and conventions. Throughout this paper k is a commutative ring with unit. Let K be an abstract simplcial complex. We denote the geometric realisation of K by K and the vertex set of K by (K); if the cardinality (K) = m, we identify (K| )| with [m] = 1, . . . , m . ByV ∆m−1 we denote the|V simplex| consisting of allV subsets of [m], and{ by ∂∆m−}1 the boundary complex of ∆m−1. A subset I [m] is a missing face of K if I K but J K for all proper subsets J I.⊂ Denote by MF (K) the set of all missing6∈ faces∈ of K. K is called a flag complex⊂ if each of its missing faces consists of two vertices. The link and the star of a face σ K are the subcomplexes ∈ lkσK := τ K : τ σ K, τ σ = ∅ ; { ∈ ∪ ∈ ∩ } st K := τ K : τ σ K . σ { ∈ ∪ ∈ } The join of two simplicial complexes K and K0, where the vertex set (K) is disjoint from (K0), is the simplicial complex V V K K0 := σ σ0 : σ K, σ0 K0 . ∗ { ∪ ∈ ∈ } In particular, if K0 = ∆0 is a point, K K0 is called the cone over K. If K0 = ∂∆1, K K0 is called the suspension of K. ∗ ∗ For a subset I [m], the full subcomplex K K is defined to be ⊂ I ⊂ K := σ K : σ I . I { ∈ ⊂ } Set core (K) := i (K) : stiK = K . The core of K is the subcomplex V { ∈ V 6 } s−1 s−1 core K := Kcore V(K). Then we may write K = core(K) ∆ , where ∆ is the simplex on the set (K) core (K). ∗ V \ V A simplicial complex is said to be pure if all its facets have the same dimension. Let K, K0 be two pure simplicial complexes of the same dimension. Choose two facets σ K, σ0 K0 and fix an identification of σ and σ0 (by identifying their vertices).∈ Then the∈ simplicial complex 0 0 0 K# 0 K := (K K ) σ = σ σ,σ ∪ \{ } is called a connected sum of K and K0. Its combinatorial type depends on the way of choosing the two facets and identifying their vertices. Let (K#K0) denote the set of connected sums of K and K0. If (K#K0) has only oneC combinatorial type we also use the abbreviation K#K0 toC denote a connected sum. A pure simplicial complex is called reducible if it is a connected sum of two simplicial complexes; otherwise called irreducible. 6 F. FAN, J. MA & X. WANG

A simplicial complex K is called a triangulated manifold (or simplicial manifold) if the geometric realization K is a topological manifold. More generally, K is a k- | | homology n-manifold if the link of each i-face of K, 0 6 i 6 n, has the k-homology of an (n i 1)-sphere. − − A simplicial complex K is a generalized k-homology n-sphere if it is a k-homology n-manifold having the k-homology of an n-sphere. (Use “generalized” as the ter- minology “homology sphere” usually means a manifold having the homology of a sphere.) If k = Z it is simply called a generalized homology sphere. A fan is a ﬁnite collection Σ of strongly convex cones in Rn, which means that the origin is the apex, such that every face of a cone in Σ is also a cone in Σ and the intersection of any two cones in Σ is a face of each. A fan Σ is simplicial (resp. nonsingular) if each of its cones is generated by part of a basis of Rn (resp. Zn). A fan Σ is complete if the union of its cones is the whole Rn. A starshaped sphere is a triangulated sphere isomorphic to the underlying complex of a complete simplicial fan. A polytopal sphere is a triangulated sphere isomorphic to the boundary complex of a simplicial polytope. Clearly a polytopal sphere is always a starshaped sphere.

2.2. Face rings and Tor-algebras. Given a simplicial complex K on the set [m].

Deﬁnition 2.1. Let k[x1, . . . xm] be the polynomial ring with m generators. The face ring (also known as the Stanley-Reisner ring) of K is the quotient ring k[K] := k[x , . . . x ]/ , 1 m IK where K is the ideal generated by the monomials xi1 xis for which i1, . . . , is does notI span a simplex of K. ··· { }

Since K is a monomial ideal, the quotient ring k[K] can be graded or multi- I m graded by setting deg xi = 2 or mdeg xi = 2ei, where ei Z is the ith unit vector. ∈ The Koszul complex (or the Koszul algebra) of the face ring k[K] is defined as the m differential Z Z -graded algebra (Λ[y1, . . . , ym] k[K], d), where Λ[y1, . . . , ym] is the exterior algebra⊕ on m generators over k, and⊗ the multigrading and differential is given by mdeg y = ( 1, 2e ), mdeg x = (0, 2e ); i − i i i dyi = xi, dxi = 0. It is known that ∗ H (Λ[y , . . . , y ] k[K], d) = Tork (k[K], k). 1 m ⊗ [x1,...,xm] m Then the Tor-algebra Tork[x1,...,xm](k[K], k) is canonically an Z Z -graded algebra. It can also be seen as a bigraded algebra by setting bideg⊕y = ( 1, 2) and i − SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 7

bideg xi = (0, 2). We refer to Tork[x1,...,xm](k[K], k) as the Tor-algebra of K and simply denote it by Tor(K; k), or Tor(K) if the coefficient ring is clear. Another way to calculate Tor(K) is by using the Taylor resolution for k[K]. This method was introduced by Yuzvinsky [50], and used by Wang and Zheng [49] to calculate the cohomology rings of moment-angle complexes (see subsection §2.3, §2.4). Concretely, let P = MF (K), and let Λ[P] be the exterior algebra over k generated by P. Then the tensor product = Λ[P] k[x1, . . . , xm] can be endowed T ⊗ with a differential as follows. For a generator u f Λ[P] k[x1, . . . , xm] with ⊗ ∈ ⊗ u = σk1 σk2 σki (σkq P), let Su = σk1 σk2 σkq , and define ··· ∈ ∪ ∪ · · · ∪ ∂ u := σ σ σ = σ σ σ σ ; i k1 ···bki ··· kq k1 ··· ki−1 ki+1 ··· kq X i d(u f) := ( 1) ∂iu xS \S f. ⊗ − ⊗ u ∂iu · i Q Here xI k[x1, . . . , xm] denotes the monomial i∈I xi for I [m]. It can be shown that∈ ( , d) is a k[x , . . . , x ]-free resolution of k[K], which⊂ is known as the T 1 m Taylor resolution (cf. [45, §4.3.2]). It follows that H( k k, d) = Tor(K). T ⊗ [x1,...,xm] An easy calculation shows that ( k k, d) is just the cochain complex T ⊗ [x1,...,xm] X i (Λ[P], d), d(u) = ( 1) εi ∂iu, (2.1) − · i where εi = 1 if Su = S∂i(u) and zero otherwise. Define a new product structure ( -product) on Λ[P] by × ( u1 u2 if Su1 Su2 = ∅, u u = · ∩ 1 × 2 0 otherwise, where denote the ordinary product in the exterior algebra Λ[P]. Then (Λ[P], , d) × becomes a differential Z Zm-graded (resp. bigraded) algebra by setting · ⊕ mdeg σ = ( 1, 2σ) (resp. bideg σ = ( 1, 2 σ )) for σ P. − − | | ∈ P m Here we identify σ with the vector ei Z . This makes Tor(K) into a i∈σ ∈ Z Zm-graded (or bigraded) algebra which agrees with the one induced by the Koszul⊕ algebra of the face ring. From (2.1) we can readily get that Lemma 2.2. Let k be a field. Then the number of missing faces of K with j −1, 2j vertices is equal to dimk Tor (K).

Let A = L be a graded connected commutative k-algebra. The socle of A is i>0 the ideal Soc(A) = x A : A+ x = 0 . { ∈ · } 8 F. FAN, J. MA & X. WANG

For a field k, the face ring k[K] is called Gorenstein if dimk Soc(Tor(K)) = 1, or in other words, Tor(K) is a Poincaréduality algebra. We call a simplicial complex K Gorenstein (over a field k) if its face ring k[K] is Gorenstein. Further, K is called Gorenstein∗ if k[K] is Gorenstein and core K = K. There is a combinatorial-topological criterion characterization of Gorenstein∗ complexes as follows. Theorem 2.3 (Stanley [47, Theorem II.5.1]). A simplicial complex K is Goren- stein* (over a field k) if and only if it is a generalized k-homology sphere.

2.3. Moment-angle complexes and manifolds. Moment-angle complexes play a key role in toric topology. They were ﬁrst introduced by Davis and Januszkiewicz [26] and extensively studied in detail and named by Buchstaber and Panov [12]. Further generalizations of moment-angle complexes, now known as polyhedral product, were studied in the work of [2]. Let us recall their construction in the most general way.

m Deﬁnition 2.4. Given a collection of based CW pairs (X,A) = (Xi,Ai) i=1 together with a simplicial complex K on the set [m]. The polyhedral{ product} determined by (X,A) and K is deﬁned to be the CW complex: [ (X,A) = B , ZK σ σ∈K where m ( Y Xi if i σ, B = Y and Y = ∈ σ i i A if i σ. i=1 i 6∈ 2 1 2 1 If (X,A) = (D ,S ), i.e. the pairs (Xi,Ai) are identically (D ,S ), then the 2 1 space K (D ,S ) is referred to as a moment-angle complex and simply denoted Z 1 0 1 0 by K ; in the case (X,A) = (D ,S ), the space R K = K (D ,S ) is referred to as aZreal moment-angle complex. Z Z 2 m Note that K (D ) , the standard coordinatewise action of the m-torus m m mZ ⊂ 2 m m T = R /Z on (D ) induces a canonical T -action on K . Z A result of Cai [15, Corollary 2.10] says that K is a closed orientable topological manifold of dimension m + n if and only if KZis a generalized homology (n 1)- sphere. In this case, we call a moment-angle manifold. In particular, if−K is ZK a starshaped sphere, then K admits a smooth structure (see [13, Chapter 6]). In general, the smoothness ofZ is open. ZK

2.4. Cohomology of moment-angle complexes. The cohomology ring of K was studied rationally by Buchstaber and Panov in [12] and integrally by FranzZ in [36, 37] as well as Baskakov, Buchstaber and Panov in [4]. It turns out that SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 9

∗ H ( K ; k) can be described in terms of the Tor-algebra of the face ring k[K]. The coefficientZ ring k will be fixed throughout the the rest of this subsection, and for simplicity we will omit it from the notation. Theorem 2.5 ([13, Theorem 4.5.4]). The following isomorphism of algebras holds: M H∗( ) = Tor(K),Hp( ) = Tor−i,2J (K), ZK ∼ ZK ∼ −i+2|J|=p m where J = (j1, . . . , jm) Z and J = j1 + + jm. ∈ | | ··· Viewing a subset J [m] as a (0, 1)-vector in Zm whose jth coordinate is 1 if j J and is 0 otherwise,⊂ then there is the following well-known Hochster’s formula. (This∈ formula was obtained by Hochster [38] for field coefficients, and generalized to arbitrary coefficients by Panov [46].) Theorem 2.6 ([13, Theorem 3.2.9]). Tor−i,2J (K) = 0 if J is not a (0, 1)-vector, and for any subset J [m] we have ⊂ −i,2J |J|−i−1 Tor (K) ∼= He (KJ ). −1 Here we assume He (K∅) = k.

L ∗ So Tor(K) is isomorphic to J⊂[m] He (KJ ) as k-modules, and this isomorphism L ∗ endows the direct sum J⊂[m] He (KJ ) with a k-algebra structure. On the other hand, Baskakov [5] directly deﬁned a natural multiplication struc- L ∗ ture on J⊂[m] He (KJ ) by the following rules: p−1 q−1 p+q−1 He (KI ) He (KJ ) He (KI∪J ), for I J = ∅, ⊗ → ∩ which are induced by the simplicial maps KI∪J , KI KJ and isomorphisms of reduced simplicial cochains → ∗ p−1 q−1 p+q−1 Ce (KI ) Ce (KJ ) Ce (KI KJ ); ⊗ → ∗ σ τ σ τ ⊗ 7→ ∪ L ∗ and zero for I J = ∅. We refer to He (KJ ) endowed with this multipli- ∩ 6 J⊂[m] cation structure as the Baskakov-Hochster ring (BHR for abbreviation) of K, and use pJ to denote the projection maps M ∗ ∗ pJ : He (KJ ) He (KJ ). (2.2) → J⊂[m] Theorem 2.7 (Baskakov’s formula, [13, Theorem 4.5.8]). There is a ring isomorphism (up to a sign for products).

∗ M ∗ p M p−|J|−1 H ( K ) = He (KJ ),H ( K ) = He (KJ ), (2.3) Z ∼ Z ∼ J⊂[m] J⊂[m] 10 F. FAN, J. MA & X. WANG where the right side of the isomorphism is the BHR of K. Remark 2.8. The sign for products was omitted in Baskakov’s original paper [5]. This defect was pointed out by Bosio and Meersseman [8, Remark 10.7], and was corrected by them for the case of polytopal spheres. For general simplicial complexes, the sign is given in the proof of [13, Proposition 3.2.10] (see also [34, Theorem 2.15]).

The following lemma is useful for later proofs of our results.

i j ∗ 0 Lemma 2.9. Given α He (KI ), β He (KJ ) with I J = ∅. Suppose α = φ (α ) i ∈ 0∈ ∩ 0 0 for some α He (KI0 ), where I I and φ : KI , KI0 . Let J = (I J) I and 0 ∗ ∈ ⊂ → 0 0 ∪ \ β = ψ (β), where ψ : K 0 , K . Then in the BHR, α β = α β . J → J · · We include the proof of this lemma in AppendixA for the reader’s convenience.

2.5. Quasitoric manifolds and topological toric manifolds. In their pio- neering work [26] Davis and Januszkiewicz suggested a topological generalisation of smooth projective toric varieties, which became known as quasitoric manifolds. A quasitoric manifold is a close 2n-dimensional manifold M with a locally standard action of T n (i.e. it locally looks like the standard coordinatewise action of T n on Cn) such that the quotient M/T n can be identiﬁed with a simple n-polytope P . Let us review their construction.

Let P be a simple n-polytope, = F1,...,Fm the set of facets of P . Given n F { } n a map λ : Z , and write λ(Fi) in the standard basis of Z : F → T n λ(Fi) = λi = (λ1i, . . . , λni) Z , 1 i m. ∈ 6 6 If the matrix λ11 λ1m . ···. . Λ =  . .. .  λ λ n1 ··· nm has the following property:

det(λi1 ,..., λin ) = 1 whenever Fi1 Fin = ∅ in P, (2.4) ± ∩ · · · ∩ 6 then λ is called a characteristic function for P , and Λ is called a characteristic matrix. Let (P, Λ) be a characteristic pair consisting of a simple polytope P and its characteristic matrix Λ. Denote by T := (e2πλ1it, . . . , e2πλnit) T n the circle i { ∈ } subgroup of T n = Rn/Zn. For each point x P , deﬁne a subtorus ∈ Y T (x) := T T n. i ⊂ i: x∈Fi SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 11

Then the quasitoric manifold M(P, Λ) is defined to be M(P, Λ) := P T n/ , (2.5) × ∼ where the equivalence relation is given by (x, g) (x0, g0) if and only if x = x0 and g−1g0 T (x). The action of∼T n on P T n by the∼ right translations descends to a T m-action∈ on M(P, Λ), and the orbit× space of this action is just P . Definition 2.10. Two characteristic pairs (P, Λ) and (P 0, Λ0) are said to be equivalent if P and P 0 are combinatorially equivalent, and Λ = A Λ0 B, where · · A GL(n, Z) and B is a diagonal (m m)-matrix with 1 on the diagonal. Here∈ we identify the set of facets of P and× P 0 so that the characteristic± functions are defined on the same set. We say that two T n-manifolds M and M 0 are weakly equivariantly homeomorphic if there is a homeomorphism f : M M 0 and an automorphism θ : T n T n such that f(t x) = θ(t) f(x) for any x →M and t T n. → · · ∈ ∈ Proposition 2.11 ([26, Proposition 1.8]). There is a one-to-one correspondence between weakly equivariant homeomorphism classes of quasitoric manifolds and equivalence classes of characteristic pairs.

In [26], Davis and Januszkiewicz also generalized the construction of quasitoric manifolds to the cases where the base space is more general than a simple polytope. We introduce such a generalization below. Let K be an n 1 dimensional simplicial complex on [m] and let K0 denote − its barycentric subdivision. For each face σ K (including ∅), let Fσ denote the ∈ geometric realization of the poset K>σ = τ K : τ > σ . Hence, for σ = ∅, 0 { ∈ } 6 Fσ is the subcomplex of K consisting of all simplices of the form σ = σ0 < σ1 < < σ , and F is the cone over K0. The polyhedron P = F together with ··· k ∅ K ∅ its decomposition into “faces” F ∈ will be called a simple polyhedral complex. { σ}σ K In particular, there are m facets Fi,...,Fm of PK , in which Fi is the geometric realization of the star of the ith vertex of K in K0. n Suppose λ : (K) Z , i λi is a characteristic function, i.e. it satisfies the V → 7→ condition that for each (k 1)-face σ = i1, . . . , ik K, the lattice generated − { } ∈ n by λi1 ,..., λik is a k-dimensional unimodular subspace of Z . Let Λ be the { } m characteristic matrix corresponding to λ. Then define M(PK , Λ) := PK T / , where the equivalence relation is defined exactly as in (2.5). × ∼ The case that K is a starshaped (n 1)-sphere in the construction above was − considered in [40]. It turns out that in this case M(PK , Λ) is a compact smooth 2n-dimensional manifold, called a topological toric manifold. This is because it can also be obtained as a quotient of the moment-angle manifold K by a freely and smoothly acting subtorus ([13, Proposition 7.3.12]). Z 12 F. FAN, J. MA & X. WANG

If Λ is deﬁned by the primitive vectors of the rays (one-dimensional cones) of a nonsingular complete fan Σ and K is the underlying sphere of Σ, then M(PK , Λ) is homeomorphic to the compact smooth toric variety corresponding to Σ (cf. [25, §12.2]). So compact smooth toric varieties are contained in the family of topological toric manifolds. The construction of M(PK , Λ) implies that there is also an analogue of Proposition 2.11 for topological toric manifolds. The cohomology ring of a topological toric manifold has a simple expression as follows.

Theorem 2.12 ([40]). Let M = M(PK , Λ) be a topological toric 2n-manifold with Λ = (λij), 1 6 i 6 n, 1 6 j 6 m. Then the cohomology ring of M is given by ∗ H (M) = Z[K]/ , where is the ideal generated by linear forms λi1x1 + + J J ··· λimxm, 1 6 i 6 m. The following result in [22] says that the cohomology of a topological toric manifold M(PK , Λ) determines the Tor-algebra of K. 0 0 Lemma 2.13 ([22, Lemma 3.7]). Let M = M(PK , Λ) and M = M(PK0 , Λ ) be two 2n-dimensional topological toric manifolds. If there is a graded ring isomorphism H∗(M) = H∗(M 0), then (K) = (K0) and there is an isomorphism of bigraded ∼ |V0 | |V | algebras Tor(K) ∼= Tor(K ). 2.6. B-rigidity of simplicial complexes. It is well-known that the combinatorial information of a simplicial complex and the algebraic information of its face ring are determined by each other. Theorem 2.14 (Bruns-Gubeladze [9]). Let k be a ﬁeld, and K and K0 be two simplicial complexes. Suppose k[K] and k[K0] are isomorphic as k-algebras. Then there exists a bijective map (K) (K0) which induces an isomorphism between K and K0. V → V

Note that the Tor-algebra of K is also determined by its face ring. In the converse direction, Buchstaber asked the following question in his lecture note [10]. Problem 2.15. Let K and K0 be simplicial complexes satisfying K = core K and K0 = core K0. Suppose their Tor-algebras are isomorphic as bigraded algebras. 0 When there exists a combinatorial equivalence K ∼= K ? (The reason why assume K = core K and K0 = core K0 is that the Tor-algebras of a simplicial complex and the cone over itself are always the same.) Deﬁnition 2.16. A simplicial complex K (K = core K) is said to be B-rigid if 0 0 0 there is no simplicial complex K ∼= K (K = core K ) such that Tor(K; k) ∼= Tor(K0; k) as bigraded rings for any6 ﬁeld k. Remark 2.17. Although the B-rigidity problem above was originally proposed by Buchstaber in terms of bigraded Tor-algebras, the concept of B-rigidity was SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 13

first defined in [22] using graded isomorphism rather than bigraded isomorphism of cohomology of moment-angle complexes. Here we define the B-rigidity in the bigraded sense, and suggest to call a simplicial complex K (K = core K) strongly B-rigid, if there is no simplicial complex K0 = K (K0 = core K0) such that ∗ ∗ 6∼ H ( K ; k) ∼= H ( K0 ; k) as graded rings for any field k. This make sense since strongZ B-rigidity clearlyZ implies B-rigidity.

In fact, the B-rigid simplicial complexes are rare. For example, letting K1 and K be simplicial complexes and K K be the simplicial complex obtained from 2 1 ∪σ 2 K1 and K2 by gluing along a common simplex σ, the Tor-algebra of K1 σ K2 do not depend on the choice of σ ([34, Proposition 4.5]), but the combinatorial∪ type of K K generally does. 1 ∪σ 2 Intuitively, the more algebraic information Tor(K; k) has, the more likely K is B- rigid, so the most interesting case is the class of Gorenstein∗ complexes since their Tor-algebras have Poincaréduality. However, even in the Gorenstein∗ class the B- ∗ rigid does not hold in most cases. For example, letting K1 and K2 be Gorenstein complexes of the same dimension, the set (K #K ) in general contains more C 1 2 than one combinatorial type, but the Tor-algebras of all elements in (K1#K2) are the same ([34, Theorem 4.2]). An irreducible non-B-rigid exampleC can be found in [31] (see also [7, Proposition 2.4]): there are three different irreducible polytopal 3-spheres with 8 vertices, which have the same Tor-algebras, and their corresponding moment-angle manifolds are all homeomorphic to a connected sum of sphere products [39]. Another interesting example is that there are non-B-rigid polytopal 4-spheres with 8 vertices, whose combinatorial type is determined by the cohomology of quasitoric manifolds over themselves [23]. All non-B-rigid examples above are not flag. Flag non-B-rigid examples can be constructed by puzzle-move operations in the next subsection (see §3 for the reason). On the other hand, the B-rigidity problem is solved positively for some particular classes of Gorenstein∗ complexes, such as joins of boundary complexes of finite simplices and the connected sum of such a complex and the boundary of a simplex [22], as well as some highly symmetric simplicial 2-spheres [18]. Another family of B-rigid 2-spheres was obtained by the authors in [33], and one of the purposes of this paper is to generalize this result to all dimensional Gorenstein∗ complexes (see §3).

Because the topology of a moment-angel complex K is uniquely determined by the combinatorics of K, if K is a B-rigid generalizedZ homology sphere, then the moment-angle manifold K is cohomologically rigid (in the bigraded sense). This is the method we use toZ deal with Problem2.

2.7. Puzzle-moves and puzzle-rigidity. In order to study the B-rigidity problem, Bosio [7] defined two transformations of simple polytopes, which can produce 14 F. FAN, J. MA & X. WANG new polytopes having the same Tor-algebras as the original ones. Let us introduce one of his constructions (in the dual simplicial version and in a more general sense for simplicial spheres) that we will use later. Let K and K0 be two simplicial n-spheres. Suppose L K and L0 K0 0 k ⊂ ⊂ are simplicial (n 1)-spheres such that L ∼= L ∼= ∂∆ Γ, where 1 6 k 6 n, Γ = Sn−k−1. Then− L (resp. L0) separates K (resp. ∗K0) into two simplicial | | ∼ 0 0 0 disks K = K+ L K− (resp. K = K+ L0 K−). We assume there are isomorphisms 0 ∪ 0 ∪ i : K K and i− : K− K . By abuse of notation, we write their restrictions + + → + → − to L also as i+ and i−. −1 k k Consider the automorphism φ = (i−) i+ of L = ∂∆ Γ. If φ fixes ∂∆ , i.e. φ(∂∆k) = ∂∆k, and φ is the identity map◦ when restricted∗ to Γ then we say that we pass from K to K0 by a puzzle-move, denoted (K, L, φ), and K0 is denoted KL,φ. Two simplicial spheres are called puzzle-equivalent if one can be obtained from the other by a sequence of puzzle-moves. Definition 2.18. A simplicial sphere K is called puzzle-rigid if there is no puzzle- equivalent simplicial sphere K0 such that K0 = K. 6∼ Theorem 2.19. [7, Theorem 3.5] Suppose K and K0 are puzzle-equivalent simpli- 0 cial spheres. Then there is a bigraded ring isomorphism Tor(K) ∼= Tor(K ). Remark 2.20. Actually it is easy to generalize the definitions of puzzle-move et al. to Gorenstein∗ complexes, and following Bosio’s proof of Theorem 2.19, we can also get the same result for Gorenstein∗ complexes.

Theorem 2.19 implies that for simplicial spheres (or more generally Gorenstein∗ complexes) B-rigidity implies puzzle-rigidity. Although B-rigidity and puzzle- rigidity are not equivalent in general (see [7, §3.3 ]), Bosio conjectured that they are equivalent for simplicial 2-spheres ([7, Conjecture 1]). See §6 for further discussion about this problem.

3. B-rigidity of a class of Gorenstein∗ complexes

In [33], we proved that the B-rigidity holds for a class of simplicial 2-spheres, whose dual class is also known as the Pogorelov class (see [11]). It is characterized by the flagness condition and the “no -condition” described below. Definition 3.1. An one-dimensional simplicial complex is called a k-circuit if it is a triangulation of S1 and contains k vertices. A simplicial complex K is said to satisfy the no -condition if there is no full subcomplex KI K such that KI is a 4-circuits. ⊂ Theorem 3.2 ([33]). Suppose K is a flag simplicial 2-sphere satisfying the no -condition. Then K is B-rigid. SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 15

A natural question arises of whether these conditions can be applied to higher dimensional simplicial spheres (or even Gorenstein∗ complexes) to get more B-rigid examples. However the following result makes this question meaningless. Theorem 3.3 ([25, Proposition 12.2.2]). Suppose a flag Gorenstein∗ complex (over a field k) of dimension n satisfies the no -condition. Then n 6 3. Remark 3.4. The statement of Theorem 3.3 in [25] is for generalized homology spheres, but the proof extends without difficulty to the Gorenstein∗ case.

Even for the case n = 3, the Gorenstein∗ examples satisfying the conditions in Theorem 3.3 are rather rare. (The only known example is the 600-cell.) So we should change our strategy for higher dimensional Gorenstein∗ complexes. In fact, we notice that only a special combinatorial property, which is essentially weaker than the no -condition for n > 3, is needed in the proof of Theorem 5.1. It is the following condition. Definition 3.5. A flag complex K with m > 3 vertices and K = core K is said to satisfy the separable circuit condition (SCC for short) if for any three different vertices i1, i2, ik with i1, i2 MF (K), there exists a full subcomplex KI with 1 { } ∈ KI = S such that i1, i2 I, ik / I and He0(KJ ) = 0, where J = ik I i1, i2 . | | ∼ ∈ ∈ 6 { }∪ \{ } Example 3.6. The dual polytopal 2-spheres of fullerenes satisfies the SCC. (A fullerene is a simple 3-polytope with only pentagonal and hexagonal facets.) This can be deduced from Proposition 3.11 below and the result of Buchstaber and Erokhovets [14]. Example 3.7. If two flag Gorenstein∗ complexes K and K0 both satisfy the SCC, 0 0 so does K K . Indeed, if i1, i2 MF (K K ), then either i1, i2 MF (K) ∗ 0 { } ∈ ∗ { } ∈ or i1, i2 MF (K ). So we may assume i1, i2 MF (K). The case ik (K) { } ∈ {0 } ∈ ∈ V 0 is trivial, by the SCC on K. For ik (K ), we can choose a vertex j1 (K ) 0 ∈ V 0 ∈ V such that ik, j1 MF (K ). The SCC implies that K is not a suspension (see { } ∈ 0 Proposition 3.8 below), so there exists j2 = ik such that j1, j2 MF (K ). It is easily verified that I = i , i , j , j is the6 desired subset.{ } ∈ { 1 2 1 2} See AppendixE for more examples of flag spheres satisfying the SCC. Proposition 3.8. Let K be a flag Gorenstein∗ complex (over k) of dimension n > 2. Assume K satisfies the SCC. Then for any missing face I MF (K), say n−2 ∈ I = i1, i2 , and any KJ lki1 K lki2 K, we have He (KJ ; k) = 0. Moreover, K is{ not a} suspension. ⊂ ∩

Proof. The coefficient k is omitted throughout the proof. Suppose on the contrary n−2 that He (KJ ) = 0 for some KJ lki1 K lki2 K. Since K is flag, KI∪J = KI KJ , n−1 6 ⊂ ∩ ∗ so that He (KI∪J ) = 0. Set S = (K) I J. Then He0(KS) = 0, by Alexander 6 V \ ∪ 6 16 F. FAN, J. MA & X. WANG duality. (K has Alexander duality because it is a generalized homology k-sphere by Theorem 2.3.) Taking two vertices s1 and s2 to be in different components of KS, the assumption that K satisfies the SCC shows that there exists a subset 1 U (K) with KU ∼= S , such that s1, s2 U, i1 / U and He0(KV ) = 0, where V =⊂ Vi U |s , s| . ∈ ∈ 6 { 1} ∪ \{ 1 2} The subcomplex KU\{s1,s2} clearly has two components. We claim that each component of KU\{s1,s2} must have nonempty intersection with KJ . To see this, we first observe that each component of KU\{s1,s2} has nonempty intersection with

K{i2}∪J , since otherwise there would be a path in KS connecting s1 and s2, contradicting the assumption that s1 and s2 are in diﬀerent components of KS. This fact together with the fact that K lk K implies that i U, since other- J ⊂ i2 2 6∈ wise KU\{s1,s2} would be path-connected. Hence the claim is true. However since

KJ lki1 K, we get He0(KV ) = 0 by the claim, a contradiction. Then we prove the ﬁrst⊂ statement of the proposition.

For the second statement, note that if K were a suspension, i.e. K = KI L ∗ ∗ with I = i1, i2 MF (K), then L would be an (n 1)-dimensional Gorenstein { } ∈ − ∗ complex, and for any j (L), lkjL would be an (n 2)-dimensional Gorenstein n−2 ∈ V − complex, so that He (lkjL) = 0. However the assumption that K is flag implies that lk L ( L lk K lk K6 ) is a full subcomplex of K, contradicting the first j ⊂ ⊂ i1 ∩ i2 statement of the proposition. Remark 3.9. Proposition 3.8 gives a necessary condition for a flag Gorenstein∗ n-complex K to satisfy the SCC: There must be no full subcomplex KI K such n−1 ⊂ that KI is a suspension and He (KI ) = 0. We refer to this condition as the no suspension of codimension one condition6 (NSC for short).

Since the NSC is much easier to be veriﬁed (for computer algorithms) than the SCC, we expect that NSC SCC. So we propose the following conjecture. ⇒ Conjecture 3.10. The SCC and NSC are equivalent for all ﬂag Gorenstein∗ complexes.

It is easy to see that for flag 2-spheres the NSC implies the no -condition. The following result shows that the converse is also true, and therefore Conjecture 3.10 holds for flag 2-spheres. Proposition 3.11 ([33]). Let K be a flag 2-sphere. If K satisfies the no - condition, then K satisfies the SCC.

We will prove a more general version of Proposition 3.11 in AppendixF. As we mentioned in §2.6, there are many non-ﬂag Gorenstein∗ complexes which are not B-rigid. Now let us illustrate why the NSC also plays an essential role for a Gorenstein∗ complex to be B-rigid. SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 17

0 k Suppose that K is a flag n-sphere and K = KL,φ, where L = ∂∆ Γ K is k ∗ ⊂ a simplicial (n 1)-sphere. If k > 1, then KV(L) = ∆ KV(Γ) since K is flag. n−1 − ∗ Thus He (KV(L)) = 0, and so He0( K KV(L) ) = 0 by Alexander duality. It | | − | | follows that in the decomposition K = K+ L K−, one of K+ and K− is just KV(L). However this implies that K0 = K, by the∪ definition of puzzle-moves. 1 1 If k = 1, we only consider the case that ∆ K, for otherwise KV = ∆ 6⊂ (L) ∗ KV(Γ), and we reduce to the previous case. In this case, the flagness of K implies 1 that KV(L) = ∂∆ KV(Γ). So if in addition K satisfies the NSC, then we have n−1 ∗ He (KV(L)) = 0. Hence the previous argument shows that one of K+ and K− is 0 KV(L), and K = K. From this we see that the NSC together with the flagness condition actually guarantees that K is puzzle-rigid, which is a necessary condition for K to be B-rigid. Theorem 3.12. If K is a flag Gorenstein∗ complex (over a field k) satisfying the SCC, then K is B-rigid. As a direct consequence we get the following (bigraded) cohomological rigidity result for moment-angle manifolds.

Corollary 3.13. Let and 0 be two moment-angle manifolds. Suppose the ZK ZK cohomology of and 0 are isomorphic as bigraded rings. Then if K is flag ZK ZK and satisfies the SCC, and 0 are homeomorphic. ZK ZK According to Lemma 2.2 a simplicial complex K is flag if and only if −1, 2j Tor (K) = 0 for j > 3. Thus Theorem 3.12 is an immediate consequence of the following theorem. Theorem 3.14. Let K be a flag Gorenstein∗ complex (over a field k) satisfying the SCC, K0 be a flag complex satisfying K0 = core K0. If there is a graded ring ∼ ∗ = ∗ 0 isomorphism φ : H ( ; k) H ( 0 ; k), then K = K . ZK −→ ZK ∼ Remark 3.15. In the case that dim K = 2, Theorem 3.14 remains true if the flagness condition of K0 is omitted. In other words, a simplicial 2-sphere (in dimension 2, all Gorenstein∗ complexes are simplicial spheres) is strongly B-rigid if it satisfies the condition in Theorem 3.14. To see this, first note that for a ∗ Gorenstein complex of dimension n 1 > 2, there is a well-known formula (the −n+1 rigidity inequality in [41]): f1 > nf0 2 , where f0 and f1 are the numbers of vertices and edges respectively. In particular,− when n = 3, the inequality above turns into identity. Hence we have the formula 2 3 f0 f0 (2n + 1)f0 n + 1 dimk H ( ) = f − + ZK 2 − 1 6 2 2 d2 d = 2n2 (2d 1)n + − , − − 2 18 F. FAN, J. MA & X. WANG where d is the top degree of H∗( ), and the last equality comes from the fact ZK that f0 = d n. This, together with the clear fact that d > 2n + 1 and the − 2 d−1 fact that the function f(n) = 2n (2d 1)n is strictly decreasing for n 6 2 , ∗ ∗ − − 0 implies that if H ( K ) = H ( K0 ) and K is a simplicial 2-sphere (note that K ∗ Z ∼∗ Z 0 is Gorenstein because H ( K0 ) is a poincaréduality algebra), then K is also a simplicial 2-sphere. Finally,Z since whether a simplicial 2-sphere K is flag or not ∗ depends only on some algebraic property of H ( K ) (see [34, §6] or [11, Theorem 4.8]), Theorem 3.14 applies. Z

4. Proof of Theorem 3.14

Throughout this section, the coefficient ring k is a field unless otherwise stated, and the cohomology with coefficients in k will be implicit. First let us recall some notions in commutative algebra. Definition 4.1. Let A = L Ai be a nonnegatively graded commutative con- i>0 nected k-algebra. The nilpotence length of A, denoted by nil(A), is the greatest integer n (or ) such that ∞ (A+)∗n := A+ A+ (n factors) = 0. ··· 6 Definition 4.2. Let R be a commutative ring. For an element r R, the annihilator of r is defined to be ∈ ann(r) := a R a r = 0 . { ∈ | · } L i In particular, if R = i R is a graded commutative ring, then the annihilator of degree k of r is defined to be ann (r) := a Rk a r = 0 . k { ∈ | · } Notational Convention. In what follows we will often not distinguish the co- ∗ L ∗ homology ring H ( K ) of K and the BHR He (KJ ) of K because of the Z Z J⊂[m] k isomorphism (2.3), so that an element ξJ He (KJ ) can also be seen as an el- |J|+k+1 ∈ ement in H ( K ). In particular, if K is flag, then for any ω MF (K), 0 3 Z 3 ∈ k = He (Kω) H ( K ), denote ω a generator of this subgroup of H ( K ). ∼ ⊂ Z e Z ∗ Lemma 4.3. Let K be a simplicial complex with m vertices, R = H ( K ). Given + ∗k Z a homogenous cohomology class [c] R. Suppose [c] (R ) and 0 = pJ ([c]) i ∈ ∈ 6 ∈ He (KJ ) for some J [m]; here pJ is the projection map in (2.2). Then i k 1. ⊂ > − + ∗k + ∗k Proof. It is easy to see that [c] (R ) implies pJ ([c]) (R ) . Thus the statement of Lemma 4.3 is an immediate∈ consequence of the∈ following multiplication rule of BHR: p q p+q+1 He (KI ) He (KI ) He (KI ∪I ). 1 ⊗ 2 → 1 2 SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 19

Lemma 4.4. Let K be an n-dimensional Gorenstein* complex with m vertices, R = H∗( ). Then nil(R) n + 1. If K is flag in addition, then nil(R) = n + 1, ZK 6 and (R+)∗(n+1) = Rm+n+1 = He n(K). Proof. The inequality is a direct consequence of Lemma 4.3. To prove the equality for the flag case, we use an induction on the dimension n. The case n = 0 is trivial. Suppose inductively that this is true for dimK < n. Take L to be the link of a vertex of K. Then L is a (n 1)-dimensional Gorenstein* complex, and since K is − flag, L is a full subcomplex of K, i.e., L = KI for some I [m]. So by induction, ∗ + ∗n n−1 ⊂ nil(H ( L)) = n and (H ( L)) = He (KI ) ∼= k. Let J = [m] (L). Then by AlexanderZ duality, Z \V ∗ 0 n−1 He (KJ ) = He (KJ ) = He (KI ) = k. Since R is a poincaréduality algebra, by seeing R as the BHR of K we have n−1 0 n m+n+1 He (KI ) He (KJ ) = He (K) = R . · >n The last equality follows from the fact that He (KI ) = 0 for any proper subset I ( [m] since K is Gorenstein*. Then from the definition of nilpotence length, we immediately get nil(R) > n + 1. So nil(R) = n + 1 since we already have the inequality nil(R) 6 n + 1. It remains to prove the last statement. First, the same inductive argument as before shows that Rm+n+1 (R+)∗(n+1). On the other hand, if ξ R and ξ / m+n+1 ⊂ ∈ ∗ ∈ R , then there exists a proper subset I ( [m] such that 0 = pI (ξ) He (KI ). Therefore ξ / (R+)∗(n+1) by Lemma 4.3. This gives the inverse6 inclusion∈ relation (R+)∗(n+1) ∈Rm+n+1. ⊂ Corollary 4.5. If the hypotheses in Theorem 3.14 is satisfied, then K0 is also a Gorenstein* complex of the same dimension and with the same number of vertices as K.

Proof. Since K is Gorenstein*, we have ∗ dimk Soc(Tor(K)) = dimk Soc(H ( )) = 1, ZK 0 where the left equality comes from Theorem 2.5. Hence dimk Soc(Tor(K )) = 1, and then K0 is a Gorenstein* complex too. The fact that dim K0 = dim K and (K) = (K0) is an immediate consequence of Lemma 4.4. |V | |V | Proposition 4.6. Let K and K0 be two simplicial complexes, and suppose K is ∼ ∗ = ﬂag and satisﬁes the SCC. If there is a graded ring isomorphism φ : H ( K ) ∗ Z −→ H ( 0 ), then: ZK 0 (a) For any subset I (K) such that He (KI ) = 0 and I 6 3, there exists 0 ⊂ V 0 0 0 6 | | J (K ) such that φ(He (KI )) He (K ). The converse is also true, i.e. for ⊂ V ⊂ J 20 F. FAN, J. MA & X. WANG

0 0 0 −1 0 0 0 any J (K ) such that He (KJ ) = 0 and J 6 3, φ (He (KJ )) He (KI ) for some⊂ VI (K). 6 | | ⊂ ∈ V 0 0 0 (b) In (a), if I = 3 and φ(He (KI )) He (KJ ), then for any ω MF (KI ), 0 | | 0 0 ⊂ ∈ φ(ωe) = ωe for some ω MF (KJ ). ∈ 0 0 (c) If KI K is an n-circuit, then there exists an n-circuit KJ K such that 1 ⊂ 1 0 ⊂ φ(He (KI )) = He (KJ ). Since the proof of Proposition 4.6 involves many complicated algebraic arguments, we include it in AppendixC. Notation. In view of Proposition 4.6, the isomorphism φ induce a bijection be- 0 0 tween MF (K) and MF (K ). We denote this map by φM : MF (K) MF (K ). → 4 Remark 4.7. It is easy to see that for any simplicial complex K, H ( K ) is L 0 Z isomorphic to the subgroup |I|=3 He (KI ) (under the isomorphism in Theorem 0 2.5). If He (KI ) = 0 with I = 3 then KI has the form of (one isolated vertex 6 | | and one edge) or (three isolated vertices). By comparing the Betti numbers of these two cases and applying Proposition 4.6(a), we immediately get the following

0 Corollary 4.8. Under the hypotheses of Proposition 4.6, suppose He (KI ) = 0 for 0 0 0 6 0 some I (K), I = 3, and suppose φ(He (KI )) He (KJ ) for some J (K ). Then K⊂ V K0 . | | ⊂ ⊂ V I ∼= J 0 Lemma 4.9. Under the hypotheses of Theorem 3.14, suppose KI (resp. KJ ) is p 0 a Gorenstein* complex of dimension p, and 0 = ξI He (KI ) (resp. 0 = ξJ p 0 q 0 −1 0 6 ∈q 6 ∈0 He (KJ )). Then φ(ξI ) He (KJ ) (resp. φ (ξJ ) He (KI )) for some J (K ) (resp. I (K)) and q∈ p. Especially, if p dim∈ K 1 then q = p. ⊂ V ⊂ V > > − Proof. For simplicity we only prove the lemma in one direction since the converse ∗ 0 ∗ can be proved in the same way. Let R = H ( K ) and R = H ( K0 ). By lemma 4.4, ξ (R+)∗(p+1), so φ(ξ ) (R0+)∗(p+1). HenceZ it follows fromZ Lemma 4.3 that I ∈ I ∈ M >p 0 φ(ξI ) He (K ). ∈ J J⊂V(K0) Now suppose MF (K ) = ω , . . . , ω and suppose φ(ξ ) = Pk ξ0 , where 0 = I { 1 t} I i=1 Ji 6 ξ0 Hqi (K0 ), q p, J = J for i = j. Let ω0 := φ (ω ) for 1 s t. Since Ji e Ji i > i j s M s 6 6 ∈ 6 6 0 KI is Gorenstein*, ωes is a factor of ξI for 1 6 s 6 t. It follows that ωes is a factor of ξ0 for each J , and so ω0 MF (K0 ) for all 1 s t, 1 i k. Ji i s ∈ Ji 6 6 6 6 −1 0 Pl pj Suppose φ (ξ ) = ξI , where 0 = ξI He (KI ), Ii = Ij for i = j. By J1 j=1 j 6 j ∈ j 6 6 using Lemma 4.3 again we have p p for 1 j l since ξ0 (R0+)∗(p+1). The j > 6 6 J1 same reasoning shows that ω is a factor of ξ for all 1 s t,∈1 j l. Hence es Ij 6 6 6 6 MF (K ) MF (K ). This implies that I I as K = core K . Since ξ (resp. I ⊂ Ij ⊂ j I I I SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 21

∗ ξIj ) has cohomological degree p + I + 1 (resp. pj + Ij + 1) in H ( K ), it follows −1| |0 p | | Z that pj = p and I = Ij . So φ (ξ ) He (KI ). Applying this argument for all | | | | J1 ∈ ξ0 we get φ−1(ξ0 ) Hp(K ) for 1 i k. Note that Hp(K ) = k. Hence we Ji Ji e I 6 6 e I ∼ ∈ q 0 0 must have k = 1 in the formula for φ(ξI ), i.e., φ(ξI ) He (K ) for some J (K ) ∈ J ⊂ V and q > p. The second statement is a consequence of Corollary 4.5 for dimension reasons. Remark 4.10. From the proof of Lemma 4.9 we can see that for the Gorenstein∗ subcomplex KI in Lemma 4.9, the restriction of the map φM to MF (KI ) MF (K) is ⊂ 0 φM : MF (K ) MF (K ). |MF (KI ) I → J Next we state the following two lemmas whose proofs we defer to AppendixD, again because of their complexity.

Lemma 4.11. If p > n 2 (n = dim K) in Lemma 4.9, then the map φM MF (KI ) : MF (K ) MF (K0 ) is− a bijection. | I → J Lemma 4.12. Under the hypotheses of Theorem 3.14, let L be the link of a vertex ∗ ∗ 0 of K. Then for a nonzero element [L] He (L) ∼= k, we have φ([L]) He (L ), where L0 is the link of some vertex of K0∈. ∈

Corollary 4.13. In the notation of Lemma 4.12, suppose KI L is an (n 2)- ⊂n−2 − dimensional Gorenstein* subcomplex. Then for 0 = ξI He (KI ), we have n−2 0 0 6 ∈ 0 0 φ(ξI ) He (KJ ) for some J (L ). The converse is also true: If KJ L is ∈ ⊂ V 0 n−2 ⊂0 an (n 2)-dimensional Gorenstein* subcomplex, then for 0 = ξJ He (KJ ), we −−1 0 n−2 6 ∈ have φ (ξ ) He (KI ) for some I (L). J ∈ ⊂ V

p 0 Proof. By Lemma 4.9 φ(ξI ) He (KJ ) for some J [m] and p > n 2. Note that ∈ 0 ⊂ 0 − 0 ξI is a nontrivial factor of [L], so ξJ is a nontrivial factor of [L ], thus J ( (L ). 0 ∗ V It follows that p 6 n 2 since L is a (n 1)-dimensional Gorenstein complex, and then p = n 2. The− converse can be proved− in the same way. − Lemma 4.14. Let K be a ﬂag simplicial complex on [m]. Then K is a suspension if and only if there is a missing face ω MF (K) (ω H3( ; k)) such that ∈ e ∈ ZK dimk(ann3(ωe)) = 1.

Proof. ( ) If K is a suspension, then K = Kω K[m]\ω for some ω MF (K). It is easily⇒ veriﬁed that ann(ω) = k ω. ∗ ∈ e · e 22 F. FAN, J. MA & X. WANG

( ) Suppose ω = i1, i2 is a missing face such that dimk(ann3(ωe)) = 1. As in the⇐ proof of Lemma C.2{ , the} subspace M V := H0(K ) ωe e µ µ∈MF (K), µe6∈ann3(ωe) satisfies ann (ω) V = H3( ). Hence from the hypothesis that dim (ann (ω)) = 3 e ωe K k 3 e ⊕ 0 Z 1 and the obvious fact He (Kω) ann3(ω) we immediately get that if µ = ω, then ⊂ e 6 µe ωe = 0. This implies that µ ω = ∅ for any µ = ω, µ MF (K). In other words,· 6 if i [m] ω, then i, i ∩, i, i K. This is6 equivalent∈ to saying that K ∈ \ { 1} { 2} ∈ is a suspension, since K is flag. Proof of Theorem 3.14. Since K satisfies the SCC, K is not a suspension by Propo- ∗ sition 3.8. From Lemma 4.14, Proposition 4.6(a) and the fact that H ( K ) = ∗ 0 Z ∼ H ( K0 ), it follows that K is not a suspension either. Thus a vertex of K (resp. K0)Z is uniquely determined by its link in K (resp. K0). Then by using Corollary 4.5 and Lemma 4.12, we can construct a bijection between the vertex sets of K and K0, ψ : (K) (K0), ψ(i) = i0, which is induced by the isomorphism φ. V → V Next we will show that ψ is a simplicial map. Since K and K0 are both flag, it suffices to verify that ψ( i , i ) = i0 , i0 K0 whenever i , i K. First note { 1 2} { 1 2} ∈ { 1 2} ∈ that if σ = i1, i2 K, then lkσK is a (n 2)-dimensional Gorenstein* complex. Let I = (lk{ K).} Since ∈ K is flag, − V σ lk K = K = lk K lk K. σ I i1 ∩ i2 n−2 0 Let 0 = ξI He (KI ). Then from Corollary 4.13 it follows that φ(ξI ) = ξJ n−2 6 0 ∈ ∈ He (KJ ) for some 0 0 0 K lki0 K lki0 K . J ⊂ 1 ∩ 2 0 0 0 0 0 0 0 −1 0 0 0 0 So if ω = i1, i2 K , then Kω0∪J = Kω0 KJ . Let ω = φM (ω ), ξ = ωe ξJ , −1 0 { } 6∈ 0 ∗ n−1 · ξ = φ (ξ ). It is clear that ξ = 0, so 0 = ξ = ωe ξI He (Kω∪I ). The fact that µ0 ω0 = 0 for all µ0 MF (K0 6) together6 with the· fact∈ that the map e · e 6 ∈ J 0 φM : MF (K ) MF (K ) |MF (KI ) I → J is a bijection (Lemma 4.11) implies that µe ωe = 0 for all µ MF (KI ). However this · 6 ∈ 0 0 0 forces K ∪ to be K K , contradicting Proposition 3.8. So we get i , i K . ω I ω ∗ I { 1 2} ∈ Thus ψ induces a simplicial injection ψ¯ : K K0. Now let us finish the proof by showing that ψ¯ is also a surjection, i.e. ψ−1(→σ0) K for any σ0 K0. It suffices 0 ∈ 0∈ 0 0 0 to prove this for 1-faces since K and K are both flag. Suppose σ = i1, i2 K . 0 { } ∈ Then by the same reasoning as before lkσ0 K is a (n 2)-dimensional Gorenstein* 0 0 0 − 0 complex and lk 0 K = lk 0 K lk 0 K is a full subcomplex. Let J = (lk 0 K ) and σ i1 i2 σ 0 n−2 0 ∩ −V1 0 let 0 = ξJ He (KJ ). Applying Corollary 4.13 again we have φ (ξJ ) = ξI n−2 6 ∈ ∈ He (KI ) for some KI lki K lki K. It follows that KI∪{i ,i } = KI K{i ,i }. ⊂ 1 ∩ 2 1 2 ∗ 1 2 SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 23

So we must have i , i K, otherwise it would contradict Proposition 3.8. The { 1 2} ∈ theorem has been proved.

5. Cohomological rigidity of topological toric manifolds

The following theorem is the main result of this section. 0 0 Theorem 5.1. Let M = M(PK , Λ) and M = M(PK0 , Λ ) be two topological toric 2n-manifolds. Assume K is ﬂag and satisﬁes the SCC. Then the following two conditions are equivalent:

=∼ (i) there is a cohomology ring isomorphism h : H∗(M; Z) H∗(M 0; Z). (ii) M and M 0 are weakly equivariantly homeomorphic. −→ Remark 5.2. For the case n = 3, the Theorem is also obtained by Buchstaber et al. [11]. Their proof was carried out on the Koszul algebras of H∗(M) = Z[K]/ ∗ 0 0 0 J and H (M ) = Z[K ]/ . Here we give another proof by means of the Taylor J 0 0 resolution of Z[K]/ and Z[K ]/ . J J Lemma 5.3. In the notation of Theorem 2.12, if K is flag and there is a linear form aixi + ajxj , i = j, ai, aj Z, then K is a suspension. ∈ J 6 ∈ Proof. Since the characteristic matrix Λ satisfies (2.5), the assumption in this lemma implies that any facet of K contains i or j. We claim that lkiK = lkjK, and then the lemma follows since K is flag and K = Sn−1. To see this, let σ be | | ∼ 0 any facet of lkiK. Then σ is an (n 2)-face, and so lkσK = S . This, together with the fact that any facet of K contains− i or j implies that σ lk K. Hence the ∈ j claim is true. h Proof of Theorem 5.1: If Z[K]/ = H∗(M) H∗(M 0) = Z[K0]/ 0 is an isomorphism, then by Lemma 2.13, J(K) = (K−→0) and there is anJ isomorphism of ∼ |V | |V | Tor-algebras: φ : Tor∗,∗(K) = Tor∗,∗(K0). Thus K = K0 by Theorem 3.12. −→ ∼ Set m = (K) , and use the abbreviated notation Z[m] for Z[x1, . . . , xm]. It is known that|V φ is| induced by a chain map (up to chain homotopic) between the Z[m]/ -projective resolution of Z[K]/ and the Z[m]/ 0-projective resolution of 0 J 0 J J Z[K ]/ , as shown in the following diagram. (Here we use the Taylor resolution, J i.e. / Z[K]/ , where is the chain complex defined in §2.2.) T J → J T f f f 2 / L 1 / 0 / P1 = (Z[m]/ )ω P0 = Z[m]/ Z[K]/ ··· ω∈MF (K) J J J

φ1 φ0 h

f 0 f 0 f 0 2 / 0 L 0 1 / 0 0 0 / 0 0 P1 = (Z[m]/ )ω0 P0 = Z[m]/ Z[K ]/ ··· ω0∈MF (K0) J J J 24 F. FAN, J. MA & X. WANG

0 0 in which P1 (resp. P1) is a free Z[m]/ - module (resp. Z[m]/ - module) with J0 0 J basis eω : ω MF (K) (resp. eω0 : ω MF (K ) ). Then using formula (2.1) and applying{ ∈ Proposition} 4.6(a){ to all ﬁelds∈ of ﬁnite} characteristics, we see that 0 0 φ1(eω) = eω0 for some ω MF (K ). ± ∈ Q It is clear that f1(eω) = xω := i∈ω x¯i, wherex ¯i is the coset of xi in Z[m]/ . Since the diagram is commutative, we have J

Y 0 Y φ (¯x ) = φ (x ) = φ f (e ) = f φ (e ) = x 0 := x¯ . 0 i 0 ω 0 1 ω 1 1 ω ± ω j i∈ω j∈ω0 Thus from the fact that Z[m]/ = Z[m]/ 0 = Z[m n] is a UFD, we can deduce J ∼ J ∼ − 0 0 that if i ω MF (K), then φ0(¯xi) = x¯j for some j ω MF (K ). Further- more, since∈ K∈ = core K, any vertex of±K belongs to a∈ missing∈ face. Combining these facts together, we see that for each i [m], h(¯x ) = x¯ for some j [m]. ∈ i ± j ∈ Next we will show that h induces a bijection between the vertex set (K) and (K0), and this map is actually a simplicial isomorphism. That is, defineVh¯ :[m] V ¯ → [m] to be such that h(¯xi) = x¯h¯(i). We first verify that h is well defined, i.e. if i = j ± 0 0 6 0 thenx ¯i = x¯j in both Z[K]/ and Z[K ]/ . Indeed, otherwise xi xj or . However6 this± implies that KJor K0 is a suspensionJ by Lemma 5.3±, contradicting∈ J J Proposition 3.8. So h¯ is well defined and it is clearly a bijection. To see that h¯ Q is a simplicial map, notice that if σ K, then 0 = xσ = i∈σ x¯i k[m]/ . So ∈ ¯ 0 6 ¯ ∈ J 0 = h(xσ) = xh¯(σ), which means that h(σ) K . Thus h is a simplicial injection, 6 ± 0 ∈ and then it is a bijection since K ∼= K . Reordering (K0) if necessary we may assume h¯ :[m] [m] is the identity V 0 → map. Now let N and N Zm be the sublattices generated by the row vectors of ⊂ 0 m the characteristic matrices Λ and Λ resp. Then we have (Z[K]/ )2 = Z /N and ∼ 0 0 m 0 Jm = m 0 (Z[K ]/ )2 = Z /N . So h induces a lattice isomorphism h : Z /N Z /N . This, togetherJ with the fact that h(¯x ) = x¯ implies that there is−→ a matrix i ± i A GL(n, Z) and a diagonal (m m)-matrix B with 1 on the diagonal such that∈ Λ = A Λ0 B. Hence the theorem× follows by Proposition± 2.11. · · We end this section by a special version of Theorem 5.1 for toric varieties. Theorem 5.4. Let X and X0 be two compact smooth toric varieties. Assume the underlying simplicial sphere of the simplcial fan corresponding to X is flag and satisfies the SCC. Then X and X0 are isomorphic as varieties if and only if their cohomology rings are isomorphic.

Proof. The ‘if’ part is straightforward. For the ‘only if’ part, we use the result of Masuda [42, Theorem 1.1 and Corollary 1.2]. That is, two compact smooth toric varieties X and X0 are isomorphic as varieties if and only if their equivariant cohomology algebras are weakly isomorphic, i.e., there is a ring isomorphism Φ : SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 25

∼ ∗ = ∗ 0 ∗ n HT (X) HT (X ) together with an automorphism γ of T = (C ) such that −→∗ ∗ ∗ ∗ Φ(uα) = γ (u)Φ(α) for any u H (BT ) and α HT (X), where γ denotes the automorphism of H∗(BT ) induced∈ by γ. ∈ ∗ ∗ 0 0 If H (X; Z) ∼= H (X ; Z), then X and X are equivariantly diﬀeomorphic by Theorem 5.1. Since an equivariantly diﬀeomorphism obviously induces a weakly isomorphism of cohomology algebras, the result follows.

6. The case of simplicial 2-spheres

In this section we give further discussion of the B-rigidity problem of simplicial 2-spheres, and the cohomological rigidity problem of topological toric manifolds of dimension 6. Note that in dimension 2 any simplicial sphere is polytopal. So any topological toric 6-manifolds is also a quasitoric 6-manifold. Unless otherwise stated, K denotes a simplicial 2-sphere and P denotes a simple 3-polytope in this section.

6.1. For B-rigidity. First let us consider the case that K is flag. In this case suppose L = KI K is a 4-circuit full subcomplex, and suppose MF (L) = ω , ω . Then there⊂ are two puzzle-moves (K, L, φ ) and (K, L, φ ) such that φ { 1 2} 1 2 i (i = 1, 2) exchanges the two vertices of ωi and fixes the vertices of the other missing face. From the discussion of puzzle-moves preceding Theorem 3.12, we know that if K is flag, then any nontrivial puzzle-move is of this form. However if L is the link of a vertex in K, then these two puzzle-moves are both trivial. In this case, we refer to L as a simple 4-circuit (or simple ). Let be the class consisting of simplicial 2-spheres which are flag and satisfy the conditionQ that any 4-circuit full subcomplex is simple, or dually, simple 3-polytopes which are flag and satisfy the condition that any 4-belt bounds a 4-gonal facet. (For k 4, a k-belt in a simple 3-polytope is a cyclic sequence F ,...,F of > { i1 ik } facets in which only two consecutive facets (including Fik , Fi1 ) have nonempty intersection. For a 3-belt we assume additionally that Fi1 Fi2 Fi3 = ∅.) ∩ ∩ is known as the class of simple 3-polytopes with strongly cyclically 4-connected graphsQ in [3], and it is also called the almost Pogorelov class in [28]. (Remark: as shown in [28], there is an one-to-one correspondence between almost Pogorelov polytopes and right-angled polyhedra of finite volume in hyperbolic space H3, which were studied by Andreev [1].) In the recent paper [30], Erokhovets proved that whether a simplicial 2-sphere K belongs to is determined by the cohomology ring H∗( ). Q ZK We have seen that if K , there are no nontrivial puzzle-moves on K. So in the spirit of Bosio’s conjecture∈ Q [7, Conjecture 1] that B-rigidity coincides with puzzle-rigidity for simplicial 2-spheres, we may ask the following question. 26 F. FAN, J. MA & X. WANG

Problem 6.1. If K , is K (strongly) B-rigid? ∈ Q We will show in [32] that the answer to this question is yes. For the special case that K and (K) 6 11, this was also obtained by Erokhovets [30]. Now let us look at∈ Q some examplesV of . Q Example 6.2. Let K be an arbitrary simplicial 2-sphere, and K0 be the barycentric subdivision of K. Then K0 . The ﬂagness of K0 is obvious. Set ∈ Q 0 (K ) = vσ : σ K, σ = ∅ . V { ∈ 6 } 0 0 Note that any vσ, vτ with σ = τ is a missing face of K . Let KI be a 4-circuit, { } 4 | | | |0 and suppose I = vσi i=1. Since KI has exactly two missing faces, there are no three vertices in I{ such} that their corresponding simplices in K have the same dimension. So all the possible dimensions of σi, 1 6 i 6 4, writing in increasing order, are (0, 0, 1, 1), (0, 0, 1, 2), (0, 0, 2, 2), (0, 1, 1, 2), (0, 1, 2, 2), (1, 1, 2, 2). It is straightforward, though tedious, to show that (0, 0, 2, 2) is the only possible case. Assume σ and σ (resp. σ and σ ) are 0-faces (resp. 2-faces), then σ , σ 1 2 3 4 1 2 ⊂ σ3, σ4, which means that τ = σ1 σ2 is the common edge of σ3 and σ4. Hence we 0 0 ∪ have KI = lkvτ K .

Recall some definitions in [18]: Let Γ be the boundary sphere of a 3-polytope which is not necessarily simplicial or simple. Suppose F1,...,Fm is the set of facets of Γ, and suppose e , . . . , e is the set of edges{ of Γ. Define} ξ (Γ) to be { 1 n} 1 the simplicial 2-sphere obtained from Γ by replacing every Fi by the cone over its boundary. Obviously, ei is also an edge of ξ1(Γ) for 1 6 i 6 n. Define ξ2(Γ) to be the simplicial 2-sphere obtained from ξ1(Γ) by doing stellar subdivision operations at every ei (see Figures1 and2).

Example 6.3. i) Let P be a flag simple 3-polytope, Γ = ∂P . Then ξ1(Γ) . Indeed, if there is I MF (ξ (Γ)) with I = 3, then it is easy to see that I ∈(Γ). Q ∈ 1 | | ⊂ V However since P is simple, ξ1(Γ)I has to be the boundary of a triangular face of Γ, but this contradicts the assumption that P is flag. So ξ1(Γ) is flag. On the other hand, suppose ξ (Γ) is a 4-circuit. Similarly we have I (Γ) and ξ (Γ) is the 1 I ∈ V 1 I boundary of a quadrangular face of Γ, which means that ξ1(Γ)I is the link of the center of this quadrangular face.

ii) ξ2(∂P ) for any 3-polytope P . This can be veriﬁed in the same way as in Example 6.2.∈ The Q dual simple polytopes of this type are exactly the ideal almost Pogorelov polytopes deﬁned in [30], where it is shown that whether K is of this ∗ type is determined by the cohomology ring H ( K ). In [29], Erokhovets further proved that all ideal almost Pogorelov polytopesZ are strongly B-rigid. SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 27

Now we move on to the case of non-ﬂag simplicial 2-spheres. Note that a simplicial 2-sphere K is non-ﬂag if and only if K is reducible or K = ∂∆3. Ignoring the trivial case K = ∂∆3, the following result provides a necessary condition for a reducible simplcial sphere to be B-rigid.

∗ Proposition 6.4 ([34, Theorem 4.2]). Let K1 and K2 be two Gorenstein complexes of the same dimension, and suppose K (K1#K2). If K is B-rigid, then K is the only element in (K #K ). ∈ C C 1 2 This proposition can also be deduced from Bosio’s result [7, Theorem 3.5] (The- orem 2.19), since all elements in (K1#K2) are actually puzzle-equivalent. Com- bining this proposition with the conclusionC of [18, Lemma 2.3] and [35, Theorem 3], we immediately get the following theorem, which is an analog of [18, Theorem 1.2]. 4 4 4 S.CHOIS.CHOI AND J.S.KIMS.CHOI AND J.S.KIM AND J.S.KIM

COMBINATORIALFigure RIGIDITY 1. C8, ξ1(C8) and ξ2(C8). 5 FigureFigure 1. CFigureCOMBINATORIAL8COMBINATORIAL, 1.COMBINATORIALξ1C(C88, 1.)ξ1 andC( RIGIDITYC RIGIDITY8 RIGIDITY8, )ξξ21 and((CC88).)ξ2 and(C8).ξ2(C8). 555

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Let LetPP1P1(resp.1(resp.(resp.PP2P2)2) be) be be a a apolytope polytope polytope in in in(P(P(#P#Q#Q)Q) obtained) obtained obtained by by by a polytope12a polytope obtaineda polytope obtained by obtained attaching by attaching by a attachingface a of faceQi a of1 facecontainingQi of1 containingQi C1CC containinga,b anda,b aand facea,b a ofand facePi. a of facePi. of Pi. identifyingidentifyingidentifyingFF1F1(resp.1(resp.(resp.FF2F2)2) with) with withFF.F. Then. Then ThenPP−1P1and1andandPP2P2−can2cancan not not not be− be be the the{ the same, same, same,} which{ which which is} is is a{ a a } ThenLemmaQThen 2.3.= QQThenIfℓ satisfies=(PQQ#ℓcontradictionQcontradictionsatisfiescontradiction=) theQ=ℓ 1 satisfiesfor condition to tothe toirreducible(P(P( condition#P#Q# theQ)Q))= (== polytopes1. condition). 1. 1. Thus Thus Thus ( PP).PisPisis face-transitive face-transitiveand face-transitive ( ).Q, then and and and one so so so isof is isQQPQbyby,by the the the same same same |C | |C|C|C | | |∗ ∗ ∗ Q Itis regular is sufficientIt and is sufficient theIt otheris toargument.argument. sufficientargument. construct is face-transitive. to construct toQ construct′ Q′(P1#Q′(P1##P(ℓP)1# which#Pℓ)# which doesPℓ) which not does satisfy not does satisfy notthe satisfy the the LetLetLet type( type( type(PP)P) =) = = (x,y,z (x,y,z (x,y,z)∈) and) and and C type( type( type(∈QQ)Q) C =) = = ( a,b,c ( a,b,c∈(a,b,c). C).). We We We can can can identify identify identify a a aface face faceFF′F′of′ofofPPP conditionProof. conditionIf P (iscondition). not face-transitive, ( ).withwithwith ( a a facea). face faceFF then′′F′′of′′ofofQQPQinininhas the the the followingtwo following following faces two two twoF ways:1 ways: ways:, F2 identify identifywith identify different the the the vertices vertices vertices types. of of of degree degree degreex,y,zx,y,zx,y,z ∗ ∗ inininFFFwith∗withwith (1) (1) (1) the the the vertices vertices vertices of of of degree degree degreea,b,ca,b,ca,b,cinininFFFandandand (2) (2) (2) the the the vertices vertices vertices of of of degree degree degreec,b,ac,b,ac,b,a LetAssumeF beAssume anyℓ face=Assume of 3.ℓQ.= Since Let′ ′ 3.′ℓP1= SinceP(resp. 3.= P SincePT2)4# be=T a4PT polytope#4#T=4T,4T# we4 in#T′′4T can′′,′′(4P# we#TQ assume4) can, obtained we assume can that by assumeP that1 = P thatT14.= LetPT14.= LetT4. Let inininFF′′F′′respectively.′′respectively.respectively. Then Then Then the the the resulting resulting resulting polytope polytope polytopeC has has has a a uniquea unique unique 3-belt 3-belt 3-belt with with with ve ve verticesrticesrtices of of of identifying F1 (resp. F2) with F . Then P1 and P2 can not be the same, which is a R R(P1#PR(2P)1# andP(2Pdegreedegree)1degreea,b,c# andPxx+2x+)+aaa,b,ca andbe22,y2,y,y++ the+bba,b,cb be22,z unique2,z,z++ the+ccc be22 uniquein2 in in 3-belt the the the the first first first unique case 3-belt case of case andR and and.x 3-beltx of+ Notex++ccRc 22.,y2,y,y thatof+ Note++bbRb 22.,zR2,z,z that+ Note++hasaaa 22R2 at thathas leastR athas 7 least at 7 least 7 contradiction to (P #Q) = 1. Thus−−− P −is−− face-transitive−−− and so is Q by−− the− same−−− −−− ∈C ∈C |C∈C inin{in the the| the second second second{} case. case. case. Since{ Since} Since two two two polytopes} polytopes polytopes are are are the the the same, same, same, we we we have have havexx+x++aaa 2=2=2=xx+x++ccc 222 faces.argument.faces. Sincefaces. thereSince are thereSinceorororxx only+x++a area therea 2= 62=2= only faceszz+ arez++aaa 6 only in2. faces2.2. Thus ThusR Thus 6containinga ina faces=a==cRcorcororcontainingxx inx===zRz.z. Sincean. Sincecontaining Since edgeaaa anbbb of edgeccandc theandand an−x−x− ofx 3-belt edgeyy they zz,−z of, we3-belt−, we− wea,b,c the 3-belta,b,c, we a,b,c, we , we −−− −−− ≥≥≥ ≥≥≥ ≥≥≥ ≥≥≥ { {} {} } canLet findcan type( a find facePcan) = aF (x,y,z find faceofhavehaveRhave) aF and faceawhichaof=a==b type(bR=b=F=ccwhichor doescofQororx)xx=R ===ynoty (which=ya,b,c does==zz,z, which contain, which). which not does We implies implies implies contain can notany that thatidentify that contain eithersuch either either any aPP edge.P faceor suchororQ anyQQisFisis regular.′ edge.Let regular.of regular.suchPQ edge.′LetbeQ a′ LetpolytopebeQ a′ polytopebe a polytope inwith(R ain face#P(3FR)′′in obtained#ofPQ(3Rin) obtained# thePAA3 byfollowingAprism)prismprism obtained attachingisis byis the the thetwo product attaching product product ways: byF of ofattachingand of identify an an annn-gonFn-gon a-gonand face and the and andF an vertices ana ofan interval.andface interval. interval.P3. aof of Then A faceA degree AbipyramidPbipyramidbipyramid3. of theThenx,y,zPis3is verticesis the. the the theThen dual dual dual vertices of of of the of F verticesform of F form of F form inCF ′ withC (1) theC verticesaa prism.a prism. prism.of degree Let Let LetBBnBa,b,cnndenotedenotedenotein the theF the′′ bipyramid bipyramid bipyramidand (2) with thewith withn verticesnnvertices.vertices.vertices. of See See degreeSee Figure Figure Figurec,b,a 3. 3. 3. a 3-belta 3-belt of Qa′. 3-belt of ClearlyQ′. of ClearlyQ′′.does ClearlyQ′ notdoesQ satisfy′ notdoes satisfy the not condition satisfy the condition the ( condition). ( ). ( ). in F ′′ respectively. Then the333 resulting polytope has a unique 3-belt with vertices of TheTheThe usual usual usual definition definition definition is is isthat that thatPPPisisis face-transitive face-transitive face-transitive if if forif for for any any any two two two faces faces faces∗FF1F1and1andandF∗F2F2of2ofofPPPtheretherethere∗ is is is degreeAssumex +Assumea ℓ 2,yAssume4.+ bℓ Letanan2an,z automorphism4. automorphismR automorphism+ℓ Letc 24.(R inP on on Let1 on theP#PPsendingsendingsending first(RP1F#F case#FtoPtoto(FℓFP andF.. It1.# It#) It is isxP notisbe not+ℓ not difficultc difficult a difficult1#) polytope2P be,y toℓ to to see+ see a1 seeb that) thatpolytope that be2 our our,z our satisfying adefinition+ definition definition polytopea 2 satisfying is is equivalentis equivalent equivalent the satisfying condition the condition the condition ≥ ≥ ∈C≥ ∈C ∈C111 2−2 2 − − (in). the Since( second).R Since− case.(has). SinceRℓ−hastototo this.2 two this. this.R 3-belts,ℓ polytopeshas−2 3-belts,ℓ there are2 3-belts, the isthere same, a unique isthere we a have unique edge− isx a+ uniquea edgea,b−2= xcontained edge+a,bc − 2containeda,b incontained all 3-beltsin all 3-beltsin all 3-belts ∗or x + a∗ 2= ∗z + a −2. Thus−a = c −or x = z. Since a b c and{−x }y{ z,−} we{ } of R.of Let−RF.of LetbeR aF.− face Letbe aF of facebeR which a of faceR which does of R notwhich does contain≥ not does≥ contain nota,b contain≥.a,b Let≥ Q.a,b′ Let Q.(R′ Let#PQ(ℓR)be′ #P a(ℓR)be#P aℓ)be a have a = b = c or x = y = z, which implies that either P or Q is{ regular.} { } { ∈C} ∈C ∈C polytopepolytope obtainedpolytope obtained by obtained attaching by attaching byF attachingandF aand faceF a ofand facePℓ. a of Then facePℓ. of ThenQP′ ℓdoes. ThenQ′ notdoesQ satisfy′ notdoes satisfy the not satisfy the the conditionA prismcondition (iscondition). the product ( ). (of). an n-gon and an interval. A bipyramid is the dual of a prism. Let∗Bn denote∗ the∗ bipyramid with n vertices. See Figure 3.

Let3 PLetbeP aLet polytopebeP a polytopebe which a polytope which is not which is necessarily not is necessarily not simplicial. necessarily simplicial. The simplicial.first-subdivision The first-subdivision The first-subdivision, , , The usual definition is that P is face-transitive if for any two faces F1 and F2 of P there is denotedan automorphismdenotedξ1(Pdenoted on),ξP1 ofsending(PP),ξ1is ofF(1Pto theP),F2is. of simplicial Itthe isP notis difficultsimplicial the polytope to simplicial see that polytope our obtained definition polytope obtained is equivalentfrom obtainedP frombyP fromaddingbyP adding oneby adding one one vertexto this.vertex at thevertex at center the at center of the each center of face each of and face each connecting and face connecting and it connecting to all it tovertices all it toverticesof all the verticesof face. theof The face. the The face. The second-subdivisionsecond-subdivisionsecond-subdivision(or barycentric(or barycentric(or subdivisionbarycentric subdivision) subdivision denoted) denotedξ2)(P denoted),ξ of2(PP),isξ of2( thePP),is simplicial of theP is simplicial the simplicial polytopepolytope obtainedpolytope obtained from obtainedP frombyP fromaddingbyP adding oneby adding vertex one vertex at one the vertex at center the at center of the each center of face each of and face each and face and connectingconnecting itconnecting to all it tovertices all it tovertices of all the vertices of face the and of face the all and face mid-points all and mid-points all of mid-points the of edg thees of edgof the thees edgof face. thees of face. the face. See FiguresSee FiguresSee 1 and Figures 1 2. and 1 2. and 2. 28 F. FAN, J. MA & X. WANG

Theorem 6.5. Let K be a reducible simplicial 2-sphere. If K is B-rigid, then K = T4#T4#T4 or K = K1#K2, where K T ,O ,I , 1 ∈ { 4 6 12} K T ,O ,I , ξ (C ), ξ (C ), ξ (D ), ξ (D ) B : n 7 . 2 ∈ { 4 6 12 1 8 2 8 1 20 2 20 } ∪ { n ≥ }

The notations in the theorem are adopted from [18], in which T4, C8, O6, D20 and I12 are the boundary spheres of the five Platonic solids: the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron respectively, subscripts indicating vertex number; Bn is the suspension of the boundary of an (n 2)-gon, called a bipyramid (see Figure3). − Remark 6.6. We will shown in [32] that the converse of Theorem 6.5 is also true, i.e. all the reducible simplicial spheres K appearing in Theorem 6.5 are actually B-rigid, which classify all reducible B-rigid simplicial 2-spheres. 6.2. For cohomological rigidity of quasitoric 6-manifolds. Let be the Pogorelov class, i.e. the class of simple flag 3-polytopes without 4-belts,P or dually, simplicial 2-spheres which are flag and satisfy the no -condition. By the Four Color Theorem we know that there is at least one quasitoric manifold over any simple 3-polytope. So Theorem 5.1 applies efficiently to 6-dimensional quasitoric manifolds over polytopes from . However, the applicability of Theorem 5.4 for (complex) 3-dimensional toric varietiesP is more limited since a simplicial 2-sphere from usually can not be obtained from a nonsingular fan. (A few examples of simplicialP spheres from , which are obtained from nonsingular fans, were produced P by Suyama [48].) Especially, any P has no Delzant realizations in R3 (full dimensional lattice simple polytopes whose∈ P normal fans are nonsingular are called Delzant), so that there are no smooth projective toric varieties over any P . This is because a Delzant 3-polytope must have at least one triangular or∈ quadrangularP face by the result of Delaunay [27]. One might expect that Theorem 5.1 is also true for quasitoric 6-manifolds over simple polytopes from , but this is not the case in general. A simplest counterex- Q ample over the simple polytope dual to B6 or B7 can be constructed as follows. (Note that B and B are the only two bipyramids that belong to .) 6 7 Q Example 6.7. Consider the family of quasitoric manifolds over a cube (dual to B6) with characteristic matrices 0 0 0 F1 F2 F3 F1 F2 F3  1 0 0 1 0 0   0 1 0 0 1 k  0 0 1 0− 0 1 − 0 where Fi,Fi is a pair of opposite faces for 1 6 i 6 3. For each k Z, the corresponding{ } quasitoric manifold is S2 H , where H is the Hirzebruch∈ surface × k k SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 29

CP ( (k) C), i.e. the associated projective bundle of the sum of a trivial line O ⊕ ⊗ bundle C and the kth power (k) = γ k of the canonical line bundle γ over CP 1. The homeomorphism classesO of these manifolds only depend on the parity of k: ( S2 S2 if k is even, CP ( (k) C) = × O ⊕ CP 2#CP 2 otherwise, where # denotes the connected sum, and CP 2 is CP 2 with the reverse orientation, but they are equivariantly distinct for diﬀerent k. The case for quasitoric manifolds over the product of a 5-gon and an interval, which is dual to B7, is similar, by the classiﬁcation result [?, p. 553] for quasitoric manifolds over polygons.

Even if we exclude the case of K = B6,B7, Theorem 5.1 may still not hold for K , which can be seen as follows. Let Λ be the characteristic matrix of a quasitoric∈ Q manifold M over a simple polytope P with facet set = F ,...,F . ∈ Q F { 1 m} Suppose F1 is a quadrangular face, and suppose the column vectors of Λ for F1 and the four surrounding faces are as in Figure4. Let M 0 be another quasitoric 0 0 0 manifold over the same P with characteristic matrix Λ such that Λ (F1) = (0, p , 1) 0 0 0 and Λ (Fi) = Λ(Fi) for i = 1. We claim that M and M are homeomorphic if p p mod 2. 6 ≡ First let us recall the conception of adjacent connected sum of manifolds (cf. [44]). Let Y and Z be closed manifolds of dimension m, X be a closed manifold of dimension n < m and suppose that X Dm−n is embedded in both Y and Z, i.e. X is a submanifold with trivial tubular× neighborhood. Remove from Y and Z the interiors of the embedded copies of X Dm−n, obtaining Y¯ and Z¯, each with boundary homeomorphic to X Sm−n. Fix× a homeomorphism f : X Sm−n X Sm−n. The adjacent connected× sum of Y and Z along X is deﬁned× to be → × Y # Z := Y¯ Z.¯ X,f ∪f In particular, if f is the identity map, we simply denote it by Y #X Z. In our case, we consider a hyperplane H R3, whose associated half-space will ⊂ be denoted H+ and H−, intersecting P but not at any vertex of P , such that (P H+) (P ) = (F1). It is easy to see that P+ := P H+ is a cube and V ∩ ∩ V V + − ∩ P− := P H− = P . Denote F and F the new facets of P and P−, respectively, ∩ ∼ 0 0 + i.e. the facets corresponding to P H. Deﬁne two characteristic matrices Λ+ and 0 0 ∩ Λ+ on P+ (resp. Λ− and Λ− on P−) by Λ (F +) = Λ0 (F +) = (0, 1, 0), and + 0 + 0 − Λ (F H ) = Λ0 (F H ) = Λ(F ) for F ; + i ∩ + + i ∩ + i i ∈ F − 0 − resp. Λ−(F0 ) = Λ−(F0 ) = (0, 1, 0), and 0 − Λ−(F H ) = Λ (F H−) = Λ(F ) for F . i ∩ + − i ∩ i i ∈ F 30 F. FAN, J. MA & X. WANG

F2 (1, 0, 0)

F F3 F1 5 (0, p, 1) (0, 1, 0) (0, 1, 0) −

F4 ( 1, 0, 0) −

Figure 4. Local coloring of Λ

Let Y,Z be the quasitoric manifolds deﬁned by the characteristic pair (P+, Λ+) and (P−, Λ−), respectively, and π+ : Y P+, π− : Z P− be the projections. It is → −1 →+ −1 − 2 2 easy to see that the characteristic submanifolds π+ (F0 ) = π− (F0 ) = S S , and their tubular neighborhoods in Y and Z are both S2 S2 D2. Then by× formula 0 0× × (2.5) we have M = Y #S2×S2 Z. Similarly let Y ,Z be the quasitoric manifolds 0 0 0 0 0 0 deﬁned by (P+, Λ+) and (P−, Λ−), then M = Y #S2×S2 Z . Since Λ− = Λ−, we have Z = Z0. If p p0 mod 2, then Example 6.7 says that Y = Y 0, and by the ≡ ∼ 0 result of Orlik-Raymond [?, §5] there is a homeomorphism ϕ : Y Y such that the following diagram is commutative. →

i S2 S2 D2 0 / Y × × id ϕ i0 S2 S2 D2 0 / Y 0 × × 0 −1 + where i0 and i0 are the embeddings of the tubular neighborhoods of π+ (F0 ) and π0−1(F +), respectively. It follows that M M 0 if p p0 mod 2. However, M is + 0 ∼= equivariantly homeomorphic to M 0 if and only if p = ≡p0 by Proposition 2.11. Remark 6.8. Although Theorem 5.1 dose not hold for all quasitoric 6-manifolds over polytopes from , it can be true if the characteristic matrix Λ satisﬁes some mild assumptions (seeQ [32] for details). On the other hand, we may still expect the cohomological rigidity holds for all quasitoric 6-manifolds over polytopes from , and indeed this is the case, as we will show in [32]. Q

Appendix A. Cohomology of real moment-angle complexes and the second Baskakov-Hochster ring

In this appendix, ﬁrst we brieﬂy review a work of Cai [15] describing the cohomology ring of a real moment-angle complex R K . Let K be a simplicial complex Z SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 31 with m vertices. As we know, the cohomology group of R K is isomorphic to the BHR as k-module, in the following way (see [2]): Z

p M p−1 H (R K ; k) = He (KJ ; k). Z ∼ J⊂[m]

∗ To describe the ring structure of H (R K ), let T (u1, . . . , um; t1, . . . , tm) be the diﬀerential graded tensor k-algebra with 2Zm generators such that

deg ui = 1, deg ti = 0; d(ti) = ui, d(ui) = 0.

Let R be the quotient of T (u1, . . . , um; t1, . . . , tm) under the following relations:

uiti = ui, tiui = 0, uitj = tjui, titi = ti, u u = 0, u u = u u , t t = t t , i i i j − j i i j j i for i, j = 1, . . . , m and i = j. Denote by uI (resp. tI ) the monomial ui1 uik (resp. t t ), where I =6 i , . . . , i , i < < i , is a subset of [m]. Let··· be i1 ··· ik { 1 k} 1 ··· k I the ideal of R generated by all square-free monomials uσ such that σ is not a face of K. Clearly, as k-module, R/ is freely spanned by the square-free monomials I u t \ , where σ I [m] and σ K. σ I σ ⊂ ⊂ ∈ Theorem A.1 ([15]). There is a graded ring isomorphism ∗ H(R/ , d) = H (R K ). I ∼ Z In view of Theorem A.1, a natural question is that how to describe the multi- ∗ plication rules of H (R K ) in terms of the full subcomplex of K? We answer this question by slightly modifyingZ the multiplication rules of the BHR. 0 I0 p−1 For subsets I I [m], let ϕI : KI0 , KI . Given two elements α He (KI ) q−1 ⊂ ⊂ → ∈ and β He (KJ ) in the BHR of K, deﬁne the ?-product of α and β to be ∈ J\(I∩J) ∗ p+q−1 α ? β := α (ϕ ) (β) He (KI∪J ), · J ∈ where ‘ ’ means the product in the BHR of K. · L ∗ We refer the k-module I⊆[m] He (KI ; k) with the ?-product to the second Baskakov-Hochster ring (BHRII for abbreviation) of K. Proposition A.2. Let K be a simplicial complex. Then the BHRII of K is iso- ∗ morphic to H (R K ). Z p−1 q−1 Proof. First we prove the special case that α He (KI ) and β He (KJ ) are ∈ ∈ represented by monic monomials in R/ , i.e., α = uσ1 tI\σ1 , β = uσ2 tJ\σ2 . Thus p−1 qI−1 α = [σ1] He (KI ) and β = [σ2] He (KJ ). The ring structure of R/ shows that ∈ ∈ I ( 0 if σ2 I = ∅, uσ1 tI\σ1 uσ2 tJ\σ2 = ∩ 6 · uσ1∪σ2 t(I∪J)\(σ1∪σ2) otherwise. 32 F. FAN, J. MA & X. WANG

p+q−1 An easy calculation shows that this is just α ? β He (KI∪J ). ∈ For the general case that α and β are polynomials, we can get the desired formula by applying the argument above to each monomial term. Proof of Lemma 2.9. According to the deﬁnition of ?-product, α0 ? β = α0 β0. On II ∗ · the other hand, since BHR is isomorphic to the cohomology ring H (R K ), which is a skew commutative ring, it follows that α0 ? β = ( 1)pqβ ? α0 = ( Z1)pqβ α = α β. − − · · Appendix B. Factor index and its application

Definition B.1. Let A be an algebra over a field k. Given a nonzero element α A, if a k-subspace V A satisfies the condition that for any nonzero element v ∈ V , v is a factor of α⊂(i.e., there exists u A such that vu = α), then V is ∈ ∈ called a factor space of α in A. Denote by α the set of all factor spaces of α, and define the factor index of α in A to be F

ind(α) = max dimk(V ) V . { | ∈ Fα} L i k Let A = i A is a graded k-algebra, 0 = α A. If V A is a factor space 6 ∈ ⊂ k of α in A, then V is called a k-factor space of α in A. Denote by α the set of all k-factor spaces of α. The k-factor index of α in A is deﬁned to beF k ind (α) = max dimk(V ) V k { | ∈ Fα} Example B.2. (i) If A is the polynomial algebra k[x] with k = C and deg x = 1, then for any 0 = f A with deg f = n, ind(f) = ind (f) = 1 for k n and 6 ∈ k 6 indk(f) = 0 for k > n. This follows from the fundamental theorem of algebra. Ld i d (ii) If A = i=1 A is a Poincar´eduality algebra, then for 0 = α A and k 6 d, k 6 ∈ indk(α) = dimk A .

In the following, we always assume k to be a ﬁeld, and the cohomology with coeﬃcients in k will be implicit.

Lemma B.3. Let K be a simplicial complex with m (m > 3) vertices. Then for any cohomology class ξ Hp( ), p 4, we have ∈ ZK > m (i) ind3(ξ) 6 2 m. − m (ii) For m > 4, ind3(ξ) = 2 m if and only if K is the boundary of an m-gon and ξ Hm+2( ). − ∈ ZK Proof. We prove (i) by induction on m, starting with the case m = 3 which is easily veriﬁed. For the induction step m > 3, let e be the number of edges of K, and let e be the number of edges containing i (K) as a vertex for 1 i m. i ∈ V 6 6 SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 33

m p Suppose on the contrary that ind3(ξ) > 2 m for some ξ H ( K ) with p > 4. 3 − ∈ Z Since dimk H ( K ) equals the number of two-element missing faces of K, we have the following inequalityZ m 3 m m < ind (ξ) dimk H ( ) = e. 2 − 3 6 ZK 2 −

Pm It follows that e < m. Combining this with the relation 2e = i=1 em we see that ei < 2 for some 1 6 i 6 m.

Without loss of generality, assume em < 2, and let I = (K) vm . Then by [34, Proposition 4.3 and Proposition 4.4] we have a ring isomorphismV \{ }

∗ ∗ He ( K ) = (He ( K ) Λ[y]) B, deg y = 1, (B.1) Z ∼ Z I ⊗ ⊕ where B is a k-algebra with trivial product structure. Moreover, let φ : KI K ∗ ∗ ∗ Z → Z be the natural inclusion map. Then φ : He ( K ) He ( KI ) is the projection to ∗ 3 Z → Z 3 (He ( KI ) 1, 0). If V ξ , then by (B.1) an element α V H ( K ) has the Z ⊗ ∈ F 3 3 ∈ 2⊂ Z form α = α1 1 + α2, where α1 H ( KI ), α2 B , since H ( K ) = 0 holds for ⊗ ∈ Z ∈ ∗ Z any simplicial complex K. We claim that dimk φ (V ) = dimkV . This is because ∗ ∗ φ (α) = α1 and the summand B has trivial product structure. If φ (ξ) = 0, then from the functorial property of cohomology ring we know that φ∗(V ) is6 a factor ∗ p ∗ space of φ (ξ) H ( KI ). If φ (ξ) = 0, then we can write ξ = ξ0 y, since the assumption∈ impliesZ that ξ can be written as the product of two elements⊗ of nonzero degree but any element of B can not. It is easily veriﬁed that φ∗(V ) is p−1 a factor space of ξ0 H ( KI ). So in either case, there is a nonzero element 0 ∗ ∈ Z 0 m m−1 ξ H ( KI ) such that ind3(ξ ) > ind3(ξ) > 2 m > 2 (m 1). By induction,∈ Z this is a contradiction. − − − ∗ The ‘if’ part of (ii) is an immediate consequence of the fact that H ( K ) is a Poincar´eduality algebra under the assumption and the equality Z 3 m dimk H ( ) = MF (K) = m. ZK | | 2 −

3 m For the ‘only if’ part of (ii), since H ( K ; k) > ind3(ξ) = 2 m there are at m Z − least 2 m two-element missing faces of K, hence K has at most m edges, − Pm i.e. m > e. The formula 2m > 2e = i=1 em implies that ei = 2 for each 1 6 i 6 m, otherwise ei < 2 for some i, and by the argument in the proof of (i), m−1 m we would have ind3(ξ) 6 2 (m 1) < 2 m, a contradiction. Hence as an easy combinatorial consequence,− −K is the boundary− of a m-gon, and so ξ Hm+2( ). ∈ ZK 34 F. FAN, J. MA & X. WANG

i Corollary B.4. Let K be a simplicial complex. Given an element 0 = ξ H ( K ), P i−|Jk|−1 6 ∈ Z and write it as ξ = ξk, 0 = ξk He (KJ ) under the isomorphism 6 ∈ k i M i−|J|−1 H ( K ) = He (KJ ). Z ∼ J⊂V(K) i−2 Suppose i > 6 and ind3(ξ) = 2 (i 2). Then KJk is the boundary of an (i 2)-gon for each J above. − − − k ∗ ∗ ∗ Proof. Consider the ring homomorphisms φ : H ( K ) H ( K ), which are k Z → Z Jk induced by the inclusions φk : K K . It is easily veriﬁed that if α is a factor Z Jk → Z of ξ in H∗( ), then φ∗(α) is a factor of ξ in H∗( ). It follows that if V 3, K k k KJk ξ ∗ Z 3 Z ∈ F then φ (V ) and dimk φ (V ) = dimkV . Thus k ∈ Fξk k i 2 ind (ξ ) − (i 2). (B.2) 3 k > 2 − − i−2 On the other hand, if i > 6, then 2 (i 2) > 0. So ξk can be written as the product of two elements of nonzero degree,− − and then the multiplication rules of BHR imply that i J 1 1, i.e. J i 2. According to Lemma B.3 (i), − | k| − > | k| 6 − J ind (ξ ) | k| J i−2 (i 2). (B.3) 3 k 6 2 − | k| 6 2 − −

Combining (B.3) with the inequality (B.2) we have Jk = i 2 and ind3(ξk) = i−2 (i 2). Lemma B.3 (ii) then shows that K is the| boundary| − of an (i 2)-gon 2 − − Jk − for each k.

Appendix C. Proof of Propsition 4.6

Lemma C.1. Let K be a simplicial complex. Assume KI is a full subcomplex 0 0 satisﬁes He (KI ) = 0. Then for any vertex i I and 0 = ξI He (KI ), there is a 6 ∈ 6 ∈ ∗ two-element missing face ω MF (KI ) containing i such that φ (ξI ) = 0, where φ : K , K is the inclusion.∈ 6 ω → I

Proof. Suppose KI1 ,...,KIk are the path-components of KI . Then we can write Pk P ξI = aj l , aj k. The assumption that ξI = 0 implies that the j=1 · l∈Ij { } ∈ 6 coefficients aj’s are not all equal. Thus if we assume i Is, then we can take a component K with a = a . Pick a vertex j I , then∈ω = i, j MF (K ) is It s 6 t ∈ t { } ∈ I the desired missing face. Lemma C.2. Let K be a flag complex which satisfies the SCC. Suppose we are given a sequence of different subsets I1,I2,...,Ik (K) (k > 2) satisfying I1 = 0 ⊂ V | | = Ik 3 and He (KI ) = 0 for 1 j k. Take a nonzero element ξj ··· | | 6 j 6 6 6 ∈ SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 35

0 ∗ ∗ He (KIj ) H ( K ) for each 1 6 j 6 k. Then for x = ξ1 + ξ2 + + ξk H ( K ), we have ⊂ Z ··· ∈ Z

dimk(ann(x)) < min dimk(ann(ξ )) 1 j k . { j | 6 6 }

Proof. We need to show that dimk(ann(x)) < dimk(ann(ξj)) for all j, and without loss of generality we only prove the inequality for ξ1. ∗ First we deﬁne a subspace Vξ1 H ( K ) as follows. For each subset J [m], ∗ ⊂ Z ⊂ let AJ = ann(ξ1) He (KJ ), and take a complementary subspace Vξ1,J of AJ in ∗ ∩ L He (KJ ). Then let Vξ = Vξ ,J . We claim that Vξ ann(ξ1) = 0 and 1 J⊂[m] 1 1 ∩ V ann(ξ ) = H∗( ). To see this, for any 0 = η V , write η = P η , ξ1 ⊕ 1 ZK 6 ∈ ξ1 i 0 = ηi Vξ1,Ji , Jp = Jq for p = q. By the construction of Vξ1,Ji , we have 0 = 6 ∈ ∗ 6 6 6 ξ1 ηi He (KI1∪Ji ). The multiplication rules of the BHR show that I1 Ji = ∅ · ∈ ∩ for all such Ji, so I1 Jp = I1 Jq for p = q, and this implies that ξ1 η = 0. So V ann(ξ ) = 0. The∪ rest6 of the∪ claim can6 be seen from the fact that· 6 ξ1 ∩ 1 ∗ M M He ( K ) = (Vξ ,J AJ ) and AJ ann(ξ1). Z 1 ⊕ ⊂ J⊂[m] J⊂[m]

Next we show that for any 0 = η V , x η = 0. As before write η = P η , 6 ∈ ξ1 · 6 i 0 = ηi Vξ1,Ji . We will prove that pI1∪J1 (x η) = 0, where pI1∪J1 is the projection 6 ∈ ∗ · 6 map in (2.2). Since 0 = ξ1 η1 He (KI ∪J ) and I1 J1 = I1 Jq for q = 1, if 6 · ∈ 1 1 ∪ 6 ∪ 6 p ∪ (x η) = 0 there would be p, q = 1 such that I J = I J and ξ η = 0. I1 J1 · 6 p ∪ q 1 ∪ 1 p · q 6 Combining this with the fact that I1 = Ip, we get I1 Jp = ∅. However, this 6 ∩ 6 implies that ξ1 ηp = 0 by the multiplication rules of the BHR, contradicting the fact that η V· . p ∈ ξ1,Jp Remark C.3. Actually the arguments above shows a more general fact. That is, ∗ for any simplicial complex K, an element x = ξ1 + ξ2 + + ξk H ( K ) with ∗ ··· ∈ Z 0 = ξj He (KI ), Ii = Ij if and only if i = j, has the property that 6 ∈ j dimk(ann(x)) min dimk(ann(ξ )) 1 i k . 6 { i | 6 6 } Now suppose the hypotheses of the lemma is satisﬁed. If we can ﬁnd an element

λ ann(ξ1), such that for any nonzero element η Vξ1 k λ, x η = 0, then we can∈ get the desired inequality ∈ ⊕ · · 6

dimk(ann(x)) < dimk(ann(ξ1)). The rest of the proof will be to ﬁnd such an element. To do this, we start with establishing a few preliminary facts. Choose a vertex i2 I2, i2 I1. By Lemma C.1 there is a missing face ω MF (KI2 ) containing ∈ 6∈ ∗ 0 ∗ ∈ i2 such that 0 = f (ξ2) He (Kω), where f is the homomorphism induced by the inclusion map 6 f : K , ∈K . Choosing a vertex i I I , the assumption of K ω → I2 1 ∈ 1 \ 2 36 F. FAN, J. MA & X. WANG satisfying the SCC shows that there is an subset I [m] such that ω I, i1 / I, 1 0 ⊂ ⊂ ∈ KI ∼= S and He (KJ ) = 0, where J = i1 I ω. It is easy to see that there |are| simplicial inclusions:6 { } ∪ \

φ1 φ2 φ3 K , K \ , K ,K ∩ , K ; ω → I2 J → I2 I2 I → I2 ψ1 ψ2 ψ3 K \ , K \ , K ,K \ , K . I I2 → J I2 → J I ω → J 0 ∗ Let ξJ He (KJ ) be an element such that ψ (ξJ ) = 0. We claim that ∈ 3 6 ξ ψ∗(ξ ), ξ ψ∗ ψ∗(ξ ) = 0. (C.1) 2 · 2 J 2 · 1 ◦ 2 J 6 To deduce (C.1), let h : K , K ∪ . Then we have I → J I2 h∗(φ∗(ξ ) ξ ) = φ∗ φ∗(ξ ) ψ∗(ξ ). (C.2) 2 2 · J 1 ◦ 2 2 · 3 J ∗ ∗ ∗ 0 ∗ 0 Since 0 = f (ξ2) = φ1 φ2(ξ2) He (Kω) = k, 0 = ψ3(ξJ ) He (KI\ω) = k ∗ 6 ◦ ∈ ∼ 6 ∗ ∗ ∈ ∗ ∼ and H ( KI ) is a Poincar´eduality algebra, we have φ1 φ2(ξ2) ψ3(ξJ ) = 0. It Z ∗ ∗ ◦ ∗ · ∗ 6 follows that φ2(ξ2) ηJ = 0, and so ξ2 ψ2(ξJ ) = 0 since ξ2 ψ2(ξJ ) = φ2(ξ2) ξJ by · 6 · 6 · h0 g · Lemma 2.9. Furthermore, note that h is the composition KI , KI2∪I , KI2∪J , ∗ ∗ ∗ ∗ ∗ ∗ → → so h (ξ2 ψ2(ξJ )) = 0 implies g (ξ2 ψ2(ξJ )) = ξ2 ψ1 ψ2(ξJ ) = 0. Thus (C.1) holds. · 6 · · ◦ 6 Now we can choose the desired element λ ann(ξ ) in diﬀerent cases. ∈ 1 0 (i) I1 = 2. In this case, I2 = ω. Let λ = ξJ He (KJ ). Since I1 J = ∅, | | 1 ∈ ∩ 6 λ ann(ξ1). (C.1) says that 0 = ξ2 λ He (KI ∪J ). Then from the fact that ∈ 6 · ∈ 2 I = I for p = q we have p ∪ (x λ) = 0, and so x λ = 0. p 6 q 6 I2 J · 6 · 6 It remains to show that x (η + λ) = 0 for any 0 = η V . To see this, write · 6 6 ∈ ξ1 l X η = η , 0 = η V . (C.3) i 6 i ∈ ξ1,Ji i=1

It suﬃces to show that p ∪ (ξ η ) = 0 for any 1 j k, 1 i l. Since I2 J j · i 6 6 6 6 ξ1 ηi = 0 we have I1 Ji = ∅, and so i1 Ji. This, together with the fact · 6 ∩ 6∈ that I2 J = i1 I and Ij = 2, implies that if pI2∪J (ξj ηi) = 0 then Ji I and I ∪J ={ 1.} But ∪ this would| | give K = D1 since K · = S6 1, contradicting⊂ i Ji ∼ I ∼ | \ |∗ | | | | 0 = ηi He (KJ ). Hence λ is the desired element. 6 ∈ i (ii) I = 3 and I I = 2, say I = i , i , i , I = i , i , i , and we may | 1| | 1 ∩ 2| 1 { 1 3 4} 2 { 2 3 4} assume ω = i2, i3 . First we consider the case i4 I, i.e. I2 J = ∅. In this { } 6∈ ∩ case, take λ = ξJ , then we can prove that λ is the desired element by the same reasoning as in case (i), noting that I2 J = I1 I, and I1 Ji = ∅ for each Ji in (C.3). ∪ ∪ ∩

∗ ∗ 0 For the case i4 I, i.e. I2 I, let λ = ψ ψ (ξJ ) He (KI\I ). First let ∈ ⊂ 1 ◦ 2 ∈ 2 us show that λ ann(ξ ). Note that K \ has two components, say K and ∈ 1 I ω 1 SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 37

∗ 0 K2. Since 0 = ψ3(ξJ ) He (KI\ω), we may assume i1, v MF (K) for all 6 P ∈ { } ∈ 0 v (K1), and v represents the cohomology class ξJ He (KJ ), so ∈ V v∈V(K1){ } ∈ that λ = P v . From the multiplication rules of the BHR it follows v∈V(K1)\{i4}{ } 1 that ξ1 λ He (KJ∪{i }) can be represented by · ∈ 3 X [α] = a σ , where a k,U = (K ) i . i · i i ∈ V 1 ∪ { 3} σi∈KU ,|σi|=2

However it is easy to see that KJ∪{i3} deformation retract to KJ\V(K1). This implies 1 ∗ that ξ1 λ is a zero class in He (KJ∪{i3}) since i ([α]) = 0 for i : KJ\V(K1) KJ∪{i3}. So λ ·ann(ξ ). → ∈ 1 1 On the other hand, since 0 = ξ2 λ He (KI ) by (C.1), it follows that pI (x λ) = 0 (I = I for p = q), and so 6x λ· =∈ 0. It remains to show that x (η + ·λ) =6 0 p 6 q 6 · 6 · 6 for any 0 = η Vξ1 . Suppose on the contrary that x (η + λ) = 0 for some η = P η , 06 = η∈ V . From the discussion in the third· paragraph of the proof i 6 i ∈ ξ1,Ji of this lemma we know that p ∪ (x η) = 0, so the fact that x (η + λ) = 0 I1 J1 · 6 · and λ ann(ξ1) implies that there exists ξp (p = 1) such that pI1∪J1 (ξp λ) = 0, and therefore∈ I J = I (I I ). However6 this together with the fact· 6 that 1 ∪ 1 p ∪ \ 2 I1 (I I2) = ∅ shows that I1 Ip, and so I1 = Ip since Ip = I1 , contradicting the∩ assumption\ of the lemma. So⊂ λ is the desired element.| | | | (iii) I = 3 and I I 1. Let | 1| | 1 ∩ 2| 6 ( ∗ 0 ψ (ξJ ) He (KJ\I ) if I1 (I I2) = ; λ = 2 2 ∅ (C.4) ∗ ∗∈ 0 ∩ \ ψ ψ (ξJ ) He (KI\I ) otherwise. 1 ◦ 2 ∈ 2 Then in both cases λ ann(ξ ) by the multiplication rules of BHR. If I (I ∈ 1 1 ∩ \ I2) = ∅ (resp. I1 (I I2) = ∅), we have pI2∪J (x λ) = ξ2 λ = 0 (resp. ∩ \ 6 · · 6 p ∪ (x λ) = ξ λ = 0) by (C.1) and the fact that I = I for p = q. So if I2 I · 2 · 6 p 6 q 6 x (η + λ) = 0 for any 0 = η Vξ1 , then λ is the desired element. We will show this· by reducing6 to case (ii).6 The∈ argument will be carried out for both cases of (C.4) simultaneously.

Since p ∪ (x λ) = 0 (resp. p ∪ (x λ) = 0), if I2 J · 6 I2 I · 6 l X x (η + λ) = 0 for some η = η , 0 = η V , · i 6 i ∈ ξ1,Ji i=1 there would be ξ and η such that p ∪ (ξ η ) = 0 (resp. p ∪ (ξ η ) = 0). p q I2 J p · q 6 I2 I p · q 6 If I1 (I I2) = ∅, then I1 I2 J because of the assumption I1 I2 6 1 (resp.∩ I \ I I), so p = 1.6⊂ Moreover,∪ since i I J but i | I ∩,J |(resp. 1 6⊂ 2 ∪ 6 1 ∈ 2 ∪ 1 6∈ 2 q I1 (I I2) = ∅ but I1 Jq = ∅), p = 2. From the fact that I2 J = Ip Jq (resp. I ∩ I =\ I 6 J ) we see∩ that 6 ∪ ∪ 2 ∪ p ∪ q I I J I (resp. I I I I ). (C.5) p \ 2 ⊂ \ 2 p \ 2 ⊂ \ 2 38 F. FAN, J. MA & X. WANG

As before we know that p ∪ (x η) = 0, so there would be ξ (r = 1, 2) such I1 Jq · 6 r 6 that pI1∪Jq (ξr λ) = 0 (r = 2 is because I1 I2 J when I1 (I I2) = ∅, and I I I), and· therefore6 6 I (J I ) = I6⊂ J∪ (resp. I ∩(I \I ) = I J ). 1 6⊂ 2 ∪ r ∪ \ 2 1 ∪ q r ∪ \ 2 1 ∪ q Combining this with (C.5) and the fact that Ip Jq = ∅ we have Ip I2 I1. So ∩ \ ⊂ 1 6 Ip I2 6 2 since p = 1, 2. If Ip I2 = 2, then Ip I1 = 2, since Ip I2 I1 and |p =\ 1.| So we can reduce6 to case| \ (ii)| by substituting| ∩ I |for I . \ ⊂ 6 p 2 On the other hand, the fact that I J = I J (resp. I I = I J ) and 2 ∪ p ∪ q 2 ∪ p ∪ q Ip Jq = ∅ implies that ∩ (J I ) J = I I (resp. (I I ) J = I I ). (C.6) \ 2 \ q p \ 2 \ 2 \ q p \ 2 Furthermore, since Ir (J I2) = ∅ and Ir (J I2) = I1 Jq (resp. Ir (I I2) = ∅ and I (I I ) = I ∩ J \), we have J I∪= (\I I ) ∪(J I ), and∩ so \ r ∪ \ 2 1 ∪ q \ 2 1 \ r ∪ q \ r I I = (J I ) J (resp. I I = (I I ) J ) (C.7) 1 \ r \ 2 \ q 1 \ r \ 2 \ q because of I1 Jq = ∅. Hence by (C.6) and (C.7), if Ip I2 = 1, we can reduce ∩ | \ | to case (ii) again by replacing I2 by Ir. The cases above exhaust all possibilities. Thus we complete the proof of Lemma C.2. Now we start the proof of Propsition 4.6.

Proof of (a). We separate the proof of (a) into two cases. (a)-(i) I = 2. In this case, we only prove the ﬁrst half statement of (a) since | | L 0 3 the converse is an immediate consequence. Note that He (KI ) = H ( K ), |I|=2 Z 0 so since φ is a graded isomorphism, for a nonzero element ξI He (KI ) we have ∈ X 0 0 0 0 φ(ξI ) = ξ , ξ He (K ). J J ∈ J |J|=2 The assertion says that there is exactly one term ξ0 = 0 in the formula above. J 6 Suppose on the contrary that φ(ξ ) = Pk ξ0 with k > 1 and ξ0 = 0 for 1 I i=1 Ji Ji 6 0 0 ∗ 6 i k. Since H (K ) = k and ξ ’s are linearly independent in H ( 0 ), one of 6 e I ∼ Ji K ξ0 , say ξ0 , satisﬁes φ−1(ξ0 ) = ξ + Ps ξ (s 1) up to multiplicationZ by a Ji J1 J1 I j=1 Ij > nonzero element of k, in which I = Ij MF (K) and ξIj = 0 for 1 6 j 6 s. Since ∗ ∗ 6 ∈ 6 H ( ) = H ( 0 ), according to Lemma C.2 and Remark C.3 we have ZK ∼ ZK k X dim (ann(ξ )) = dim (ann( ξ0 )) dim (ann(ξ0 )) k I k Ji 6 k J1 i=1 s (C.8) X = dimk(ann(ξI + ξIj )) < dimk(ann(ξI )), j=1 This is a contradiction, so the conclusion holds for I = 2. | | SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 39

Now we can use the conclusion of (a)-(i) to prove (c), which is a intermediate step to deal with the case I = 3 of (a). | |

Proof of (c). Suppose KI is an n-circuit and MF (KI ) = ω1, . . . , ωt (clearly n 1 { } t = 2 n). Let cI be a generator of He (KI ) = k. According to Corollary B.4, −P 0 0 0 0 1 0 φ(cI ) = c , where K is an n-circuit of K and 0 = c He (K ). It is easy Ji Ji 6 Ji ∈ Ji to see that ωej (1 6 j 6 t) is a factor of cI . The conclusion of (a) shows that φ(ω ) = ω0 for some ω0 MF (K0). It follows that ω0 is a factor of c0 for each c0 , ej ej j ∈ ej Ji Ji and therefore ω0 MF (K0 ). Since K0 is a n-circuit, MF (K0 ) = n n = t, j Ji Ji Ji 2 thus we have MF∈(K0 ) = ω0 . . . , ω0 for each K0 . The| fact that| a simplicial− Ji 1 t Ji complex is uniquely determined{ by its} missing face set shows that there is exactly one c0 in the formula for φ(c ), which is just the statement of (c). Ji I Now we go back to the proof of (a) for another case. 0 Pk 0 (a)-(ii) I = 3. For any 0 = ξI He (KI ), we can write φ(ξI ) = ξ , in | | 6 ∈ i=1 Ji which 0 = ξ0 H0(K0 ), J = J for i = j, J = 3 for 1 i k, by Remark Ji e Ji i j i 6 6 4.7. The6 first half∈ of (a) is equivalent6 to6 saying| that| k 1 in the formula above. ≡ 0 Suppose on the contrary that k > 2 for some ξI = 0. Then If He (KI ) = k we can get a contradiction in the same way as in the proof6 of (i). So we only consider the 0 case He (KI ) = k k. In this case, first let us prove the following statement: ⊕ −1 0 0 φ (ξ ) He (KI ) for all 1 i k. (C.9) Ji ∈ 6 6 0 It suffices to show that there exists a basis ξ1, ξ2 of He (KI ) such that φ(ξi) 0 0 0 { } ∈ He (K ) for some Ji (K ), i = 1, 2. Here we allow J1 = J2. Ji ⊂ V 0 To see this, take θ1 He (KI ) to be a nonzero element such that ∈ 0 dimk(ann(θ1)) = m1 := max dimk(ann(θ)) 0 = θ He (KI ) , { | 6 ∈ } 0 and take θ2 He (KI ) to be a nonzero element such that ∈ 0 dimk(ann(θ2)) = m2 := max dimk(ann(θ)) θ He (KI ), θ k θ1 . { | ∈ 6∈ · } 0 0 0 If there exists Ji (K ) such that φ(θi) He (K ), i = 1, 2, then θ1, θ2 is ⊂ V ∈ Ji { } the desired basis. If φ(θ ) = Pp θ0 (p 2), where 0 = θ0 H0(K0 ), J = J 1 i=1 i > i e Ji p q for p = q, then either there exists a θ0 such that φ−1(θ0)6 = θ∈+ Pq ξ (q 6 1, 6 i i 1 j=1 Ij > Ij = I) up to multiplication by a nonzero element of k, which is impossible because of the6 inequality (C.8); or there are two θ0 and θ0 such that ξ = p φ−1(θ0 ) and i1 i2 1 I ◦ i1 ξ = p φ−1(θ0 ) span H0(K ), and ξ , ξ k θ . Actually it can be shown that 2 I i2 e I 1 2 1 p φ−1◦(θ0 ) = 0 (j = 1, 2) for any J = I, since6∈ · otherwise by applying (C.8) again J ij we◦ would have 6

dimk(ann(θ1)) < dimk(ann(ξ1)), dimk(ann(ξ2)), 40 F. FAN, J. MA & X. WANG contradicting the maximality of m . Thus ξ , ξ is the desired basis. 1 { 1 2} 0 0 0 The remaining case is that φ(θ1) He (K ) for some J (K ), and φ(θ2) = ∈ J ⊂ V Pp θ0 (p 2) with 0 = θ0 H0(K0 ), J = J for i = j. By the same reasoning i=1 i > i e Ji i j there are two θ0 and θ06 such∈ that ξ = p 6 φ−1(θ0 )6 and ξ = p φ−1(θ0 ) span i1 i2 1 I i1 2 I i2 0 ◦ ◦ He (KI ). Since ξ1 and ξ2 are linearly independent, one of them is not in k θ1, say ξ k θ . Then p φ−1(θ0 ) = 0 for any J = I since otherwise by (C.8) we· would 1 6∈ · 1 J ◦ i1 6 have dimk(ann(θ2)) < dimk(ann(ξ1)), contradicting the maximality of m2. Hence we can choose θ , ξ as the desired basis. So statement (C.9) is true. { 1 1} Returning now to the proof of (a)-(ii), consider the case that KI is three isolated points. (C.9) implies that k 2 in the expression φ(ξ ) = Pk ξ0 . So we only 6 I i=1 Ji need to exclude the possibility of k = 2. Suppose I = i1, i2, i3 and k = 2 is the case in the formula above. Then without loss of generality{ we} may assume the cohomology classes φ−1(ξ0 ) have the following form: Ji

−1 0 0 φ (ξ ) = ai,1 i1 + ai,2 i2 He (KI ), ai,1 = ai,2 k, i = 1, 2. Ji { } { } ∈ 6 ∈ This assumption is reasonable because if φ−1(ξ0 ) (or φ−1(ξ0 )) = a( i + i ) we J1 J2 { 1} { 2} can use i1, i3 instead of i1, i2 , and then the fact that a i3 is chain homotopic { } 0 { } −1 0 − { } to a( i1 + i2 ) in He (KI ) shows that φ (ξ ) satisﬁes the requirement for i = 1, 2. { } { } Ji Let ω = i1, i2 . Because K satisﬁes the SCC, there is a subset U (K) such { } 1 0 ⊂ V that ω U, i3 / U, KU = S and He (KW ) = 0, where W = i3 U ω. Let cU ⊂ ∈ | | ∼ 6 { } ∪ \ be a generator of H1(K ) k. We have shown in (c) that φ(c ) = c0 H1(K0 ) e U ∼= U V e V 0 0 0 ∈ for some circuit KV K . Notice that cU = ωe ξU\ω for some ξU\ω He (KU\ω) = k, 0 ⊂ · ∈ 0 ∼0 so cV = φ(ωe) φ(ξU\ω). According to the conclusion of (i), φ(ωe) = ωe for some ω 0 · 0 0 0 ∈ MF (KV ), therefore 0 = pV \ω0 φ(ξU\ω) He (KV \ω0 ). Let ξV \ω0 = pV \ω0 φ(ξU\ω). 6 ◦ −∈1 0 ◦ The same reasoning shows that p \ φ (ξ 0 ) = ξ \ . U ω ◦ V \ω U ω The fact that a = a implies that f ∗(φ−1(ξ0 ) ξ ) = 0, where f : K i,1 i,2 Ji U\ω U −1 0 6 · −6 1 0 0 → K ∪ , then φ (ξ ) ξ \ = 0 for i = 1, 2. It follows that φ (ξ ξ 0 ) = 0 since I U Ji · U ω 6 Ji · V \ω 6

−1 0 0 −1 0 −1 0 −1 0 p ∪ φ (ξ ξ 0 ) = p ∪ (φ (ξ ) φ (ξ 0 )) = φ (ξ ) ξ \ . I U ◦ Ji · V \ω I U Ji · V \ω Ji · U ω The second equality comes from the fact that φ−1(ξ0 ) H0(K ). Hence we have Ji e I 0 0 0 ∈ ξJ ξ 0 = 0, Ji (V ω ) = ∅ for i = 1, 2. i · V \ω 6 ∩ \ On the other hand, let

θ = (a a )φ−1(ξ0 ) (a a )φ−1(ξ0 ) I 2,1 − 2,2 J1 − 1,1 − 1,2 J2 0 = (a1,2a2,1 a1,1a2,2)( i1 + i2 ) (a1,1a2,2 a1,2a2,1) i3 He (KI ). − { } { } ∼ − { } ∈ SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 41

0 It is easy to see from He (KW ) = 0 that 0 = θI ξU\ω. Combining this with the 6 0 · fact that J1 = J2 and Ji (V ω ) = ∅ for i = 1, 2, we obtain 6 ∩ \ 0 0 p ∪ \ 0 φ(θ ξ \ ) = (a a ) ξ ξ 0 = 0. J1 (V ω ) ◦ I · U ω 2,1 − 2,2 · J1 · V \ω 0 0 This is a contradiction since a a = 0 by assumption and ξ ξ 0 = 0. Thus 2,1 − 2,2 6 J1 · V \ω 6 k must equal to one and we have proved the ﬁrst half of (a)-(ii). As a preliminary step to showing the second half of (a)-(ii), we need to prove statement (b).

0 ∗ Proof of (b). Let ξI He (KI ) be an element such that f (ξI ) = ωe, where f : ∈ 0 0 0 Kω KI is the inclusion, and set ξJ = φ(ξI ) He (KJ ). Suppose I ω = ik . Then→ by the assumption that K satisﬁes the SCC∈ there is a subset \U {(K}) 1 0 ⊂ V such that ω U, ik / U, KU = S and He (KW ) = 0, where W = ik U ω. ⊂ ∈ | | ∼ 6 { } ∪ \ Let c be a generator of H1(K ) k, and suppose φ(c ) = c0 H1(K0 ) for U e U ∼= U V e V 0 0 0 ∈ 0 some circuit KV K ; ξU\ω be a generator of He (KU\ω) = k; ξW He (KW ) be ⊂ ∗ ∼ ∈ an element such that g (ξ ) = ξ \ , where g : K \ K is the inclusion. Then W U ω U ω → W ξ ξ \ = ω ξ by Lemma 2.9. I · U ω e · W 0 We have seen in the proof of the ﬁrst half statement of (a)-(ii) that ξV \ω0 := 0 0 p \ 0 φ(ξ \ ) = 0 and ξ ξ 0 = 0. Thus we have V ω ◦ U ω 6 J · V \ω 6 0 0 p ∪ \ 0 φ(ω ξ ) = p ∪ \ 0 φ(ξ ξ \ ) = ξ ξ 0 = 0. J (V ω ) ◦ e · W J (V ω ) ◦ I · U ω J · V \ω 6 This implies that ω0 J. Note that this argument is valid for all ω MF (K ). ⊂ ∈ I Then we get the desired result.

−1 0 Pk Returning now to the second half of (a)-(ii). Suppose φ (ξJ ) = j=1 ξIj , where 0 0 = ξIj He (KIj ), Ii = Ij for i = j. Then the ﬁrst half of (a)-(ii) implies that 6 ∈ 0 0 6 6 0 0 φ(ξIj ) He (KJ ) for all 1 6 j 6 k. It follows that k 6 2 since dimk He (KJ ) 6 2. If 0 0∈ 0 H (KJ ) = k the statement obviously holds. So we only consider the case that KJ −1 0 is three isolated points. Suppose on the contrary that φ (ξJ ) = ξI1 + ξI2 for some 0 0 0 0 nonzero element ξ He (K ). (b) shows that φM(ω) MF (K ) for any ω J ∈ J ∈ J ∈ MF (KI1 ) MF (KI2 ). On the other hand, since I1 = I2, MF (KI1 ) MF (KI2 ) > ∪ 0 6 | ∪ | 3 = MF (K ) . It follows that φM(MF (K ) MF (K )) = MF (K ). | J | I1 ∪ I2 J Suppose J = j , j , j . Then without loss of generality we may assume the { 1 2 3} cohomology classes φ(ξIi ) have the form: 0 0 φ(ξI ) = ai,1 j1 + ai,2 j2 He (K ), ai,1 = ai,2 k, i = 1, 2. i { } { } ∈ J 6 ∈ Let ω0 = j , j and ω = φ−1(ω0). We may suppose ω MF (K ). Applying the { 1 2} M ∈ I1 SCC on K to ω and I1, and using the same notations as in the proof of the ﬁrst half of (a)-(ii), i.e. ω U (K), I U = ω, W = I U ω, K = S1, ⊂ ⊂ V 1 ∩ 1 ∪ \ | U | ∼ 42 F. FAN, J. MA & X. WANG

0 1 0 0 φ(cU ) = cV He (KV ), ξV \ω0 := pV \ω0 φ(ξU\ω) etc., and using a similar argument we can deduce∈ that ◦

0 p ∪ φ(ξ ξ \ ) = φ(ξ ) ξ 0 = 0 for i = 1, 2. J V ◦ Ii · U ω Ii · V \ω 6 0 Thus J (V ω ) = ∅ and ξIi ξU\ω = 0, Ii (U ω) = ∅ for i = 1, 2. ∩ \ · 6 ∩ \ 0 ∗ Take ξW He (KW ) to be an element such that g (ξW ) = ξU\ω, where g : K K ∈is the inclusion. Then we have ξ ξ = ω ξ by Lemma 2.9 and U\ω W I1 U\ω e W the fact→ that f ∗(ξ ) = ω, where f : K K is· the inclusion.· So I1 e ω → I1 0 0 = p ∪ φ(ξ ξ \ ) = p ∪ φ(ω ξ ) = ω p ∪ \ 0 φ(ξ ). 6 J V ◦ I1 · U ω J V ◦ e · W e · (J V ) ω ◦ W 0 0 It follows that He (K(J∪V )\ω0 ) = 0. This means that j3, v MF (K) for any 6 0 { } ∈ vertex v of one of the two components of KV \ω0 . Let θ0 = (a a )φ(ξ ) (a a )φ(ξ ) J 2,1 − 2,2 I1 − 1,1 − 1,2 I2 0 0 = (a1,2a2,1 a1,1a2,2)( j1 + j2 ) (a1,1a2,2 a1,2a2,1) j3 He (K ). − { } { } ∼ − { } ∈ J 0 0 It follows by an easy calculation that θ ξ 0 = 0. Since I = I and I (U ω) = J · V \ω 1 6 2 2 ∩ \ ∅, we have I2 I1 U, and then 6⊂ ∪ −1 0 0 p ∪ φ (θ ξ 0 ) = (a a )ξ ξ \ = 0. I1 U ◦ J · V \ω 2,1 − 2,2 I1 · U ω But a a and ξ ξ \ are both nonzero, and we have a contradiction. The 2,1 − 2,2 I1 · U ω proof is ﬁnished.

Appendix D. Proof of Lemma 4.11 and Lemma 4.12

Proof of Lemma 4.11. It is clear that φM MF (KI ) is an injection, so it remains to | 0 prove that it is surjective. Assuming MF (K) = ω1, . . . , ωk and φM(ωi) = ωi 0 { } for 1 6 i 6 k, ﬁrst we claim that for any ωi φM(MF (KI )), if there is an 0 0 0 0 ∈ 0 ωj MF (KJ )(ωj = ωi) such that ωi ωj = ∅, then ωj φM(MF (KI )). ∈ 6 ∩ 6 ∈ p 0 Under this assumption, since ωei is a factor of ξI He (KI ) = k, ωei is a factor 0 0 0 0 0 ∈∗ 0 0 0 of ξJ = φ(ξI ), so ξJ = ωi ξ \ 0 for some ξ \ 0 He (K \ 0 ). Set W = ωi ωj. e · J ωi J ωi ∈ J ωi ∪ 0 0 0 ∗ 0 0 It is easy to see that there is an element ξW He (KW ) such that f (ξW ) = ωei, 0 0 ∈ 0 ∗ 0 where f : K 0 K is the inclusion. Let U = J W and let ξ = g (ξ 0 ), ωi W U J\ωi 0 → 0 0 0 \ 0 0 0 where g : K K 0 . By Lemma 2.9, ξ ξ = ω ξ 0 = ξ . Thus ξI = U J\ωi W U ei J\ωi J −1 0 −1 →0 0 · 0 ·0 0 φ (ξW ) φ (ξU ). The assumption that ωi = ωj and ωi ωj = ∅ implies that · 6 −1 0 0∩ 6 W = 3, so according to Proposition 4.6(a), φ (ξW ) He (KI0 ) for some I0 I, |I |= 3. Then by Proposition 4.6(b) and Corollary 4.8∈ we have ⊂ | 0| 0 ω φM(MF (K )) φM(MF (K )). j ∈ I0 ⊂ I SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 43

Hence the claim is true. As a consequence, if we deﬁne two subsets of (K0) as V [ 0 [ 0 V1 := ωi,V2 := ωi, 0 0 0 ωi∈φM(MF (KI )) ωi∈MF (KJ )\φM(MF (KI ))

0 0 0 0 0 0 then V1 V2 = ∅, and KJ = KV KV since K is flag and KJ = core KJ . ∩ 1 ∗ 2 Next we show that for any ω MF (K) such that ω0 MF (K0 ), we have i ∈ i ∈ V2 ωi I = ∅. Otherwise ωi I = 1 since ωi MF (KI ), and we may assume ∩ | ∩ | 6∈ ωi I = i . Let ωj MF (KI ) be such that i ωj (such ωj always exists since ∩ { } ∈ 0 ∈ KI = core KI ). Then ωi ωj = 3 and He (Kωi∪ωj ) = 0. So by Proposition 4.6(b) 0 0 | ∪ | 0 0 6 0 0 we have ωi ωj = 3. Hence ωi ωj = ∅. However the fact that ωj V1, ωi V2 | ∪ | ∩ 6 ⊂ ⊂ and V1 V2 = ∅ gives a contradiction. ∩ Finally, we prove the lemma by showing that V2 = ∅. Suppose on the contrary that there is an ω0 MF (K0 ). Then the missing face ω MF (K) satisfies i ∈ V2 i ∈ ωi I = ∅ as we showed in the previous paragraph. Suppose ωi = i1, i2 , then ∩ p{+1 } KI lki K lki K, since otherwise KI∪ω = KI Kω and therefore He (KI∪ω ) = 6⊂ 1 ∩ 2 i ∗ i i 6 0, contradicting the SCC by Remark 3.9 (note that p + 1 > n 1). Thus without loss of generality we may assume that K lk K. This implies− that there is I 6⊂ i1 a vertex i3 I such that ωk := i1, i3 MF (K). Clealy ωi ωk = 3 and 0 ∈ { } ∈ | 0 ∪ 0 | He (Kωi∪ωk ) = 0. Hence it follows by Proposition 4.6(b) that ωi ωk = 3. This 6 0 0 0 | ∪0 | 0 0 implies that ∅ = ω ωi V2, and then ω V1 = ∅ since KJ = KV KV . 6 k ∩ ⊂ k ∩ 1 ∗ 2 Letting ωl MF (KI ) be such that i3 ωl and applying Proposition 4.6(b) again ∈ 0 0 ∈ 0 0 for Kωk∪ωl we have ωk ωl = 3, i.e. ωk ωl = ∅. However this contradicts the 0 | ∪0 | ∩ 6 fact that ω V1 and ω V1 = ∅. Thus the proof is finished. l ⊂ k ∩

Proof of Lemma 4.12. Suppose (K) = [m], dim K = n, and suppose L = lkmK, (L) = [l]. L is clearly a GorensteinV ∗ complex of dimension n 1. So from V l+n ∗ − Theorem 2.7 we have [L] H ( K ). Let R = H ( K ). For the ﬁrst step, we will show that if there is a∈k-vectorZ space Rl+n+1Z (R+)∗n such that for any F ⊂ ∩ 0 = ξ and every ω MF (L), ωe is a factor of ξ, then dimk 6 m l 1. Moreover,6 ∈ F the upper bound∈ m l 1 can be reached. F − − − − Suppose 0 = ξ Rl+n+1 (R+)∗n. If there is a subset I [m] such that 6p ∈ ∩ n ⊂ 0 = pI (ξ) He (KI ), then p > n 1 by Lemma 4.3, but He (KI ) = 0 if and only if 6 ∈ n m+n+1 − 6 I = [m] and He (K) R , so p = n 1 and then I = l + 1; if in addition ωe is a factor of ξ for every⊂ ω MF (L), then− (L) I.| Actually| there are exactly m l 1 subsets satisfying∈ these conditions:V ⊂ − − I = (L) j , l + 1 j m 1. j V ∪ { } 6 6 −

The conditions Ij = l + 1 and (L) Ij are straightforward. Furthermore, it is easily veriﬁed that| |L lk K = lkVK and⊂ j, m MF (K) for l + 1 j m 1. ∩ j j Ij { } ∈ 6 6 − 44 F. FAN, J. MA & X. WANG

Thus by Proposition 3.8

n−2 n−1 He (lkjKI ) = He (lkjKI ) = 0 for l + 1 j m 1. j j 6 6 − From the Mayer-Vietoris sequence

n−2 n−1 n−1 n−1 He (lkjKI ) He (KI ) He (stjKI ) He (L) · · · → j → j → j ⊕ n−1 He (lkjKI ) → j → · · · n−1 n−1 we immediately obtain that He (KI ) = He (L) = k. It follows that if j ∼ F satisfies the property we mentioned, then dimk m l 1. F 6 − − Now let M n−1 = He (KI ). F j l+16j6m−1 Then dimk = m l 1. We will show that satisfies the condition in the first paragraph ofF the proof.− − For any ω MF (L), letF U = [l] ω and let W = U j ∈ \ j ∪ { } for l + 1 6 j 6 m 1. Since lkjKWj lkjK L, then by applying Proposition 3.8 n−−2 ⊂ ∩ again, we have He (lkjKWj ) = 0. Now from the Mayer-Vietoris sequence

n−2 n−2 n−2 n−2 He (KW ) He (stjKW ) He (KU ) He (lkjKW ) · · · → j → j ⊕ → j ∗ n−2 h n−2 it follows that He (KWj ) He (KU ) is an epimorphism, where h : KU , KWj . −→ n−2 → Since ωe is a factor of [L], there is an element ξU He (KU ) such that ωe ξU = [L]. n−2 ∗ ∈ ∗ · Letting ξW He (KW ) be such that h (ξW ) = ξU , we have g (ω ξW ) = ω ξU = j ∈ j j e· j e· [L], where g : L, K . So 0 = ω ξ Hn−1(K ) = k, and then ω is a factor of Ij e Wj e Ij ∼ e → n−1 6 · ∈ any nonzero element of He (KIj ). As an easy consequence, satisfies the desired condition. F We already see from Corollary 4.5 that (K0) = (K) = m, and from Lemma n−1 0 |V 0 | |V | 4.9 that φ([L]) He (KJ ) for some J (K ), J = l. Suppose on the contrary 0 ∈ ⊂ V0 | | that KJ is not the link of a vertex of K . We will show that for every vertex i (K0) J, there exists a nonzero element ξ0 Hn−1(K0 ), where J = i J, Ji e Ji i such∈ V that \ω0 is a factor of ξ0 for every ω0 MF∈(K0 ) = φ (MF (L)) (by{ Lemma} ∪ e Ji J M 0 ∈ 0 4.11). Hence if we define to be the vector space spanned by ξJi : i (K ) J , 0 F { 0 ∈0 V \ } then dimk = m l, and it is easy to verify that for any 0 = ξ and every 0 F − 0 0 6 ∈ F ω φM(MF (L)), ωe is a factor of ξ . But this will give a contradiction because any∈ isomorphism preserves algebraic properties. Now let us find such an nonzero element ξ0 for all i (K0) J. Note that Ji 0 0 ∈ V \ the assumption that KJ = lkiK implies that there is a vertex, say j J, such 0 6 0 ∈ 0 that i, j MF (K ). Since KJ = core KJ , we can choose an ω MF (KJ ) such { }0 ∈ 0 0 0 0 ∈ ∗ 0 0 that j ω . Let W = i ω and let ξW He (KW ) be such that q (ξW ) = ωe , ∈ 0 0 { } ∪ ∈ −1 0 0 where q : K 0 , K . According to Proposition 4.6(a), φ (ξ ) = ξU He (KU ) ω → W W ∈ SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 45 for some U [m], U = 3. Moreover, from Proposition 4.6(b) and Lemma 4.11 ⊂ | | −1 0 we can deduce that U [l] = φM (ω ) =: ω. Since ωe is a factor of [L], there is an n−2 ∩ element ξV He (KV ), where V = [l] ω, such that ωe ξV = [L]. It is easy to see that ∈ \ · p φ(ξ ξ ) = p (ξ0 φ(ξ )) = ξ0 p φ(ξ ), Ji ◦ U · V Ji W · V W · X ◦ V in which X = J W = J ω0. Let f : K0 , K0 . We also have i \ \ J → Ji f ∗(ξ0 p φ(ξ )) = ω0 p φ(ξ ) = p φ(ω ξ ) = ξ0 = 0. W · X ◦ V e · X ◦ V J ◦ e · V J 6 These formulae imply that p φ(ξ ξ ) = 0. Ji ◦ U · V 6 n−1 Suppose U ω = k (clearly l < k < m). Since ξU ξV He (KI ) \ { } · ∈ k ⊂ F (remember Ik = k [l]), ωe is a factor of ξU ξV for every ω MF (L) as we { } ∪ 0 · 0 ∈ 0 have seen. It follows that for every ω φM(MF (L)) = MF (K ), ω is a factor of ∈ J e φ(ξ ξ ), and so it is also a factor of p φ(ξ ξ ) Hn−1(K0 ). Thus we get the U V Ji U V e Ji desired· element ξ0 := p φ(ξ ξ ). The◦ proof· of∈ Lemma 4.12 is finished. Ji Ji ◦ U · V

Appendix E. Construction of flag spheres satisfying the SCC

The aim of this appendix is to construct higher dimensional simplicial spheres, which are ﬂag and satisfy the SCC, from lower dimensional simplicial spheres.

Construction E.1. Given an n > 3 dimensional simplicial polytope P with m ∗ vertices, let P be the dual simple polytope, and for each q-face σ of P , let Fσ be the dual n q 1 face of P ∗. Suppose there is a triangulation T of the (n 2)- skeleton of −P ∗ such− that the vertex sets (T ) = (P ∗). For each face σ of P define− V V a subcomplex Lσ of T to be ( τ T : T F if dim σ > 0, L := { ∈ | τ | ⊂ σ} σ τ T : T ∂F if dim σ = 0. { ∈ | τ | ⊂ σ} In other words, Lσ T is the triangulation of Fσ for dim σ > 0 and Li is the ⊂ 0 triangulation of ∂Fi for each vertex i [m]. Let K,K be two copies of ∂P , and m 0 m ∈ 0 let vi i=1 and vi i=1 be their vertex sets respectively (vi and vi correspond to the{ same} vertex{i }[m]). Define a simplicial complex ∈ m [ 1 [ [ 0 L := ( ∆ L ) ( σ L ) ( σ L 0 ), i ∗ i ∪ ∗ σ ∪ ∗ σ i=1 σ∈K σ0∈K0 1 0 where ∆i is the simplex on vi, vi . It is easily verified that L is a triangulation of Sn−1 D1, and the two components{ } of the boundary ∂L are K and K0. Now adding two× more vertices u and u0, and define a triangulation of Sn by := ( u K) L ( u0 K0). EP { } ∗ ∪ ∪ { } ∗ 46 F. FAN, J. MA & X. WANG

Proposition E.2. In the notation of Construction E.1, suppose the simplicial (n 1)-sphere ∂P is flag and not a suspension, and suppose there is a flag triangulation− ∗ ∗ T of the (n 2)-skeleton of P such that (T ) = (P ). Then P is flag and satisfies the SCC.− V V E Remark E.3. (a) Note that if n = 3 in the proposition above, then the assumption that T is flag is automatically satisfied. Since in this case the 1-skeleton of (P ∗) is clearly a simplicial complex, and from the assumption that ∂P is flag, we can readily deduce that there are no 3-circuits in T . (b) For n = 4, there are also many examples satisfying the hypothesis of Propo- sition E.2. For example, let P1, P2 be two polygons with n1 and n2 (n1, n2 > 4) vertices, respectively, and let P be the dual of P P . Then ∂P = ∂P ∂P is 1 × 2 1 ∗ 2 obviously flag and not a suspension. There is an unique triangulation K1 of P1 (and similarly K2 of P2) determined by the following rules: (K1) = (P1) = [n1]; i, j ∂K if and only if j i 1 mod n ; 1, i, i + 1 :V 2 i nV 1 is the { } ∈ 1 | − | ≡ 1 {{ } 6 6 1 − } set of 2-faces of K1. Let K be the canonical triangulation of K K , i.e. 1 × 2 K := σ σ σ : σ K , σ K , { ⊂ 1 × 2 1 ∈ 1 2 ∈ 2 and i i0 implies j j0 for any two pairs (i, j), (i0, j0) σ . 6 6 ∈ } Let T be the restriction of K to the 2-skeleton of P1 P2. It is not hard to check × 4 that T is flag. So by Proposition E.2, P , as a triangulation of S , is flag and satisfies the SCC. E For a general flag simple n-polytope P (n > 3), we do not know whether there is a flag triangulation T of the (n 2)-skeleton of P such that (T ) = (P ). Here we propose the following more general− question: V V

Question. If P is a simple flag n-polytope, is there always a flag triangulation Tk of the k-skeleton of P such that (T ) = (P ) for all n > 3 and 2 k n 1? V k V 6 6 − (c) From Proposition 3.11 and (a), (b) above, we see that for n = 2, 3, 4, there are infinitely many simplicial n-spheres which are flag and satisfies the SCC. Hence by Example 3.7, we can actually construct infinitely many simplicial n-spheres of this type for all n > 2. Proof of Proposition E.2. First we check that is flag. Suppose on the contrary EP that σ is a missing face of P with σ > 3. It follows from the construction of that u, u0 σ. The assumptionE | that| T is flag implies that σ (T ). So EP 6∈ 6⊂ V we may assume σ (K) = ∅, by the symmetry of P . If σ (K) = 1, say ∩ V 6 E | ∩0 V | σ (K) = vi , let τ = σ vi . It is easy to see that τ vi (Li). Since ∩ V { } \{0 } ⊂0 { } ∪ V τ and ( ) 0 = v L , it follows that τ v L . But this gives P P {vi}∪V(Li) i i i i that∈ Eσ , aE contradiction. On∗ the other hand, if σ ∈ (K∗) 2, the fact that ∈ EP | ∩ V | > SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 47

0 0 vi, vj P if and only if i = j implies that σ (K ) = ∅. Setting τ = σ (T ) { } ∈ E ∩ V ∩ V and ω = σ (K) (clearly ∅ = τ, ω P ), we have τ Li for any vi ω. Thus τ L , and∩ then V σ = τ ω 6 , a contradiction∈ E again.∈ ∈ ∈ ω ∪ ∈ EP Next we verify that satisfies the SCC case by case. Suppose i , i EP { 1 2} ∈ MF ( P ) and ik ( P ) i1, i2 . We need to find I ( P ) such that ( P )I ∼= 1 E ∈ V E \{ } ⊂ V E | E | S , i1, i2 I, ik / I and He0(( P )J ) = 0, where J = ik I i1, i2 . ∈ ∈ E 6 { } ∪ \{ } 0 (i)( i1, i2) = (u, u ). (ia) i (K) (or (K0)). Since K is not a suspension, there exists a missing k ∈ V V face vi, vj MF (K) such that ik = vi, vj and vi, ik MF (K). It is easy{ to verify} ∈ that I = u, v , v , v0, v6 0 , u0 satisfies{ the} above ∈ condition. { i j i j } (ib) ik (T ). Choose a missing face vi, vj MF (K). Note that Li Lj = ∈ V 0 { } ∈ 0 ∩ ∅, so vi, ik , vi, ik MF ( P ) or vj, ik , vj, ik MF ( P ). Thus we can{ take }I {= u,} v , ∈ v , v0, vE0 , u0 . { } { } ∈ E { i j i j } (ii) i = u, i = v0 (K0). 1 2 i ∈ V 0 0 0 0 0 (iia) ik = u2. Choose a vertex vj such that vi, vj MF (K ). Suppose KU 0 0 0 { }0 ∈ 1 is a path connecting vi and vj in K , i.e., KU = D with end points 0 0 | | ∼ vi, vj. Note that such a full subcomplex always exists, since if we assume every edge of K0 has length 1, then the shortest path, called a geodesic, 0 0 0 connecting vi and vj in K is a full subcomplex. Let I = u1, vi, vj U. It is easy to see that I is the desired subset. { } ∪ 0 0 0 (iib) ik (K ). Since K is not a suspension, there exists a vertex vj = ik, such∈ V that v0, v0 MF (K0). Then let I = u, v , v , v0, v0 , u0 . 6 { i j} ∈ { i j i j } (iic) ik (T ). As in (ib) we may choose a missing face vi, vj MF (K) ∈ V 0 0 0 { } ∈ and take I = u, vi, vj, vi, vj, u . (iid) i (K). First{ we consider the} case i = v . Since K is not a suspen- k ∈ V k 6 i sion, we can take a vertex vj = vi such that vj, ik K. If vi, vj K, then I = u, v , v , v0, v0 , u0 6is the desired subset.{ } 6∈Otherwise,{ take} 6∈σ to { i j i j } be a facet of K containing vi, vj , and take a vertex vs = ik such that v , v K (such a v exists{ because} K is not a suspension).6 Then let { j s} 6∈ s ( u, v , v ,F , v0, v0 if v0, v0 K0, I = { j s σ i s} { i s} ∈ u, v , v ,F , v0, v0 , u0 if v0, v0 K0. { j s σ i s } { i s} 6∈

Recall that Fσ (T ) is the vertex dual to σ, and note that ik,Fσ since v , i ∈ VK and then i σ. { } 6∈ EP { j k} 6∈ k 6∈ On the other hand, if ik = vi, take σ K to be a facet containing vi, and let τ = σ v . Since K is ﬂag and not∈ a suspension, there exists a vertex \{ i} vj (K) (lkτ K) such that vi, vj MF (K). Since vj lkτ K, there∈ V must\V be a vertex v τ such{ that} ∈v , v MF (K){ as}K 6∈is ﬂag. s ∈ { j s} ∈ 48 F. FAN, J. MA & X. WANG

0 It is easy to see from vi, vs σ and vj / σ that Fσ, vi , Fσ, vs 0 ∈ ∈ { }0 {0 0 } ∈ P and Fσ, vj , Fσ, vj / P . Then I = u, vj, vs,Fσ, vi, vj, u is a Esatisfactory{ subset.} { } ∈ E { } (iii) i = u, i (T ). Let σ K and σ0 K0 be the facets dual to i . 1 2 ∈ V ∈ ∈ 2 0 (iiia) ik = u . Choose two vertices vi, vj σ. Take vs to be a vertex such that 0 ∈ 0 0 vi, vs MF (K), and take KU to be a geodesic connecting vj and vs { } ∈ 0 0 0 with U σ = vj . Such U exists since KV is path-connected, where ∩ 0 {0 } 0 V = (K ) σ vj . It is easy to verify that I = u, vi, vs, i2 U has theV desired\{ property.\{ }} { } ∪ (iiib) i (K0). Assume i = v0 . Choose two vertices v , v σ such that k ∈ V k k i j ∈ vi, vj = vk. Since K is not a suspension, there exists a vertex vs such that 6 v , v MF (K) and v = v . Let { i s} ∈ s 6 k ( u, v , v , i , v0 , v0 if v0 , v0 K0, I = { i s 2 j s} { j s} ∈ u, v , v , i , v0 , v0 , u0 if v0 , v0 K0. { i s 2 j s } { j s} 6∈ (iiic) i (T ). Let τ K be the facet dual to i . Choose two vertices k ∈ V ∈ k vi, vj σ with vi τ, and take vs to be a vertex such that vi, vs MF (K∈). Then we6∈ can take I as follows: { } ∈ ( u, v , v , i , v0 , v0 if v0 , v0 K0, I = { i s 2 j s} { j s} ∈ u, v , v , i , v0 , v0 , u0 if v0 , v0 K0. { i s 2 j s } { j s} 6∈ (iiid) ik (K). If ik σ, we can take a vertex vs σ such that vs, ik MF∈( VK), and take6∈ a vertex v = i such that ∈v , v MF ({K), since} ∈ j 6 k { j s} ∈ K is ﬂag and not a suspension; or if ik σ, then by the same reasoning as in the second paragraph of (iid) there∈ exist v σ and v σ i j 6∈ s ∈ \{ k} such that ik, vj , vj, vs MF (K). In either case, choose a vertex v σ i{, v (dim} { σ }2), ∈ and let i ∈ \{ k s} > ( u, v , v , i , v0, v0 if v0, v0 K0, I = { j s 2 i j} { i j} ∈ u, v , v , i , v0, v0 , u0 if v0, v0 K0. { j s 2 i j } { i j} 6∈

(iv) i (K), i (K0). Assume i = v , i = v0 (clearly i = j). 1 ∈ V 2 ∈ V 1 i 2 j 6 0 (iva) ik = u (or u ). If vi, vj MF (K), take KU to be a geodesic con- { } ∈ 0 0 0 necting vi and vj, and let I = u , vi, vj U. On the other hand, if 0 {0 } ∪ vi, vj K, let I = vi, vj, vi, vj . (ivb) {i }(K ∈0) (or (K)).{ If i = v0,} then let k ∈ V V k 6 i ( v , v , v0, v0 if v , v K, I = { i j i j} { i j} ∈ u, v , v , v0, v0 , u0 if v , v K. { i j i j } { i j} 6∈ SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 49

0 For the case ik = vi, take σ K to be a facet such that vi σ, vj / σ. Choose a vertex v σ v ∈, and let ∈ ∈ s ∈ \{ i}  v , v ,F , v0 , v0 if v , v K and v0 , v0 K0, { i j σ j s} { i j} ∈ { j s} ∈  u, v , v ,F , v0 , v0 if v , v K and v0 , v0 K0, I = { i j σ j s} { i j} 6∈ { j s} ∈ v , v ,F , v0 , v0 , u0 if v , v K and v0 , v0 K0, { i j σ j s } { i j} ∈ { j s} 6∈  u, v , v ,F , v0 , v0 , u0 if v , v K and v0 , v0 K0. { i j σ j s } { i j} 6∈ { j s} 6∈ 0 (ivc) ik (T ). If vi, vj K, then either vi, ik , vi, ik MF ( P ) or v ∈, i V , v0 , i { MF}( 6∈ ). Thus we can{ let } { } ∈ E { j k} { j k} ∈ EP I = u, u0, v , v , v0, v0 . { i j i j} On the other hand, if vi, vj K, let τ K be the facet dual to ik. Then if v τ or v {τ, we} can ∈ just let I∈= v , v , v0, v . Otherwise, i 6∈ j 6∈ { i j i j} take vs to be a vertex such that vi, vs MF (K) and take σ K to be a facet such that v , v σ and{ F =}i ∈. Then we can let ∈ i j ∈ σ 6 k ( u, v , v ,F , v0 , v0 if v0 , v0 K0, I = { i s σ j s} { j s} ∈ u, v , v ,F , v0 , v0 , u0 if v0 , v0 K0. { i s σ j s } { j s} 6∈

0 0 0 (v) i1, i2 (T ). Let σ1, σ2 K and σ1, σ2 K be the facets dual to i1 and i2, respectively.∈ V Assuming every∈ edge of a∈ connected simplicial complex K has length 1, define the distance d (v , v ) of two different vertices v , v (K) K i j i j ∈ V to be the length of a geodesic connecting vi and vj in K, and dK (vi, vi) = 0 for all v (K). Furthermore, define the distance of two subcomplexes i ∈ V ∅ = L1,L2 K to be 6 ⊂ d (L ,L ) := min d (v, u) v L , u L . K 1 2 { K | { } ∈ 1 { } ∈ 2}

0 0 0 0 0 0 (va) i = u (or u). If d 0 (σ , σ ) = 0, take a vertex v σ σ . Since k K 1 2 s ∈ 1 ∩ 2 σ1 = σ2, we can choose two vertices vi σ1, vj σ2 such that vi σ2, v 6 σ . Hence i , v , i , v MF ( ∈), and therefore∈ we can let6∈ j 6∈ 1 { 1 j} { 2 i} ∈ EP ( v , v , i , i , v0 if v , v K, I = { i j 1 2 s} { i j} ∈ u, v , v , i , i , v0 if v , v K. { i j 1 2 s} { i j} 6∈ 0 0 0 On the other hand, if dK0 (σ1, σ2) = l > 0, suppose KU is a path of length 0 0 0 0 l with U σ1 = vs,U σ2 = vt. Choose vertices vi σ1, vj σ2 such that v ∩σ , v σ . Then∩ we can let ∈ ∈ i 6∈ 2 j 6∈ 1 ( v , v , i , i U if v , v K, I = { i j 1 2} ∪ { i j} ∈ u, v , v , i , i U if v , v K. { i j 1 2} ∪ { i j} 6∈ 50 F. FAN, J. MA & X. WANG

0 0 (vb) ik (K ) (or K). Assume ik = vk. This situation is similar to (va). ∈ V 0 0 0 0 0 0 n−1 Set V = (K ) ik , τ1 = σ1 V , τ2 = σ2 V . Then KV = D . 0 V 0 \{ } ∩ 0 0 0 ∩ | | ∼ If d 0 (τ , τ ) = 0, take a vertex v τ τ . Since i , i MF (T ), KV 1 2 s 1 2 1 2 this means σ σ σ 2. Hence∈ there∩ exist vertices{ v} ∈σ v , | 1 ∩ 2| 6 | 1| − i ∈ 1 \{ k} vj σ2 vk such that vi σ2, vj σ1, and as in the previous case we can∈ let \{ } 6∈ 6∈ ( v , v , i , i , v0 if v , v K, I = { i j 1 2 s} { i j} ∈ u, v , v , i , i , v0 if v , v K. { i j 1 2 s} { i j} 6∈ 0 0 On the other hand, if d 0 (τ , τ ) = l > 0, we have σ σ 1. Suppose KV 1 2 1 2 6 K0 K0 is a path of length l with U τ 0 = v0 and| ∩U | τ 0 = v0. Since U ⊂ V ∩ 1 s ∩ 2 t σ1 = σ2 > 3 and σ1 σ2 6 1, we can take a vertex vi σ1 vk, vs and| | a vertex| | v σ| ∩v , v| such that v σ , v σ .∈ Then\{ let } j ∈ 2 \{ k t} i 6∈ 2 j 6∈ 1 ( v , v , i , i U if v , v K, I = { i j 1 2} ∪ { i j} ∈ u, v , v , i , i U if v , v K. { i j 1 2} ∪ { i j} 6∈ 0 0 0 0 (vc) ik (T ). Let σ K be the facet dual to ik, and let V = (K ) σ , 0 ∈ V 0 0 ∈0 0 V \ τ1 = σ1 V , τ2 = σ2 V . Clearly KV is path-connected. ∩0 0 ∩ 0 0 0 If d 0 (τ , τ ) = 0, take a vertex v τ τ , and vertices v σ , v σ KV 1 2 s 1 2 i 1 j 2 ∈ ∩ 0 ∈ ∈ such that vi σ2, vj σ1. It is obvious that ik, vs MF ( P ), so we can let 6∈ 6∈ { } ∈ E ( v , v , i , i , v0 if v , v K, I = { i j 1 2 s} { i j} ∈ u, v , v , i , i , v0 if v , v K. { i j 1 2 s} { i j} 6∈ 0 0 0 0 On the other hand, if d 0 (τ , τ ) = l > 0, suppose K K is a path KV 1 2 U V 0 0 0 0 ⊂ 0 0 of length l and U τ1 = vs and U τ2 = vt. Since l > 0, τ1 τ2 = ∅, and then v0 / σ0 , ∩v0 σ0 . Choose∩ a vertex v σ v and∩ a vertex s ∈ 2 t 6∈ 1 i ∈ 1 \{ s} vj σ2 vt (requiring vi = vj σ1 σ2 whenever σ1 σ2 = ∅), and let∈ \{ } ∈ ∩ ∩ 6  v , i , i U if v = v , { i 1 2} ∪ i j I = v , v , i , i U if v , v K, { i j 1 2} ∪ { i j} ∈  u, v , v , i , i U if v , v K. { i j 1 2} ∪ { i j} 6∈ 0 0 Since one of the two components of ( P )I\{i1,i2} is KU and ik, vp MF ( ) for any v0 U, I is the desiredE subset. { } ∈ EP p ∈

(vi) i1 (K), i2 (T ). Assume i1 = vi, i2 = Fσ = Fσ0 for some facet σ K (σ0∈ VK0). Clearly,∈ V v σ since v ,F . ∈ ∈ i 6∈ { i σ} 6∈ EP (via) ik = u. Let l = dK (vi, σ), and suppose KU is a path of length l connecting v and σ, v = U σ. Choose a vertex v0 σ0 v0 , then i j ∩ s ∈ \{ j} SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 51

let ( U, F , v0, v0 if v0, v0 K0, I = { σ i s} { i s} ∈ U, F , v0, v0 , u0 if v0, v0 K0. { σ i s } { i s} 6∈ 0 0 0 0 (vib) ik = u . Choose a vertex vj σ, and a vertex vs σ vj . Let 0 0 0 ∈ 0 ∈ \{ } V = ( (K ) σ ) vs . It is clear that KV is path-connected. Take 0 V \ ∪ { } 0 0 0 KW to be a geodesic in KV connecting vi and vs. Then let ( v , v ,F ,W if v , v K, I = { i j σ } { i j} ∈ u, v , v ,F ,W if v , v K. { i j σ } { i j} 6∈ 0 0 0 0 0 0 (vic) ik (K ). Choose a vertex vj σ ik , and a vertex vs σ vj, ik (dim∈ VK 2). If i = v0, then let∈ \{ } ∈ \{ } > k 6 i  v , v ,F , v0, v0 if v , v K and v0, v0 K0, { i j σ i s} { i j} ∈ { i s} ∈  u, v , v ,F , v0, v0 if v , v K and v0, v0 K0, I = { i j σ i s} { i j} 6∈ { i s} ∈ (E.1) v , v ,F , v0, v0 , u0 if v , v K and v0, v0 K0, { i j σ i s } { i j} ∈ { i s} 6∈  u, v , v ,F , v0, v0 , u0 if v , v K and v0, v0 K0. { i j σ i s } { i j} 6∈ { i s} 6∈ For the case i = v0, since K = ∂∆n, there is a facet τ K satisfying k i 6 ∈ vi τ and τ σ > 2, so that Fτ ,Fσ MF ( P ). Choose a vertex 0 ∈ 0 0 | \ 0 | 0 {0 } ∈ E vt τ (σ vi ), where τ K is the facet corresponding to τ, then let∈ \ ∪ { } ∈  v , v ,F ,F , v0, v0 if v , v K and v0, v0 K0, { i j τ σ t s} { i j} ∈ { t s} ∈  u, v , v ,F ,F , v0, v0 if v , v K and v0, v0 K0, I = { i j τ σ t s} { i j} 6∈ { t s} ∈ v , v ,F ,F , v0, v0 , u0 if v , v K and v0, v0 K0, { i j τ σ t s } { i j} ∈ { t s} 6∈  u, v , v ,F ,F , v0, v0 , u0 if v , v K and v0, v0 K0. { i j τ σ t s } { i j} 6∈ { t s} 6∈ (vid) ik (T ). Assume ik = Fτ for some facet τ K. Choose a vertex ∈ V 0 0 0 ∈ vj σ τ, and a vertex vs σ vj . Then we can let I be the same as∈ in (E.1\ ). ∈ \{ } (vie) ik (K). Choose a vertex vj σ ik , and a vertex vs σ vj, ik . Then∈ V we can also choose I to be∈ the\{ same} as in (E.1). ∈ \{ } It follows from the symmetry of (P ) that these cases exhaust all possibilities, E and then we ﬁnish the proof.

Appendix F. A generalization of Proposition 3.11

The following proposition is a generalization of Proposition 3.11, which is an essential ingredient in the second part of this paper. This result was also obtained by Erokhovets [30, Lemma 5.2] in the dual form. 52 F. FAN, J. MA & X. WANG

Γ0,1 Γ0,2 Γ0,3

Figure 5. Decomposition of Γ0

Proposition F.1. Let K be a simplicial 2-sphere from (see §6.1 for the deﬁni- Q tion), (K) = [m]. Suppose a missing face ω = i1, i2 MF (K) and a vertex i [mV] ω satisfy the following condition: { } ∈ s ∈ \ lk K is not a 4-circuit for any i (lk K) ω. i ∈ V is ∩ Then there exists a subset I [m] i with ω I such that K = S1 and ⊂ \{ s} ⊂ | I | ∼ He0(KJ ) = 0, where J = is I ω. 6 { } ∪ \

Proof. Since K is ﬂag, we can easily deduce that Γ := K[m]\{is} is a ﬂag triangula- 2 tion of D , and ∂Γ = KS , where = (lkis K). So by [34, Lemma 6.1] (see also [11, Lemma A.2]), there exists I S [mV] i with ω I such that K = S1. 0 s 0 I0 ∼ Note that K bounds a simplicial⊂ disk\{K } Γ. Let⊂J = i (|I |ω). If I0 0 ⊂ 0 { s} ∪ 0 \ He0(KJ ) = 0, then I = I0 is the desired subset. Otherwise, assign an orientation 0 6 to KI0 such that K0 is on our left when we are outside the sphere and go along + KI0 in the positive orientation. For the two components of KI0\ω, denote L0 (resp. − L0 ) the one from i1 to i2 (resp. from i2 to i1), the words ‘from’ and ‘to’ referring to the given orientation of KI0 . + − + Since He0(KJ0 ) = 0, (L0 ), (L0 ) = ∅. Taking a vertex l0 (L0 ), S ∩ V S ∩ V 6 ∈ S ∩ V the fact that ∂Γ = KS implies that Γ0 := Γ l0 = σ Γ: l0 σ is still a 2 − { } { ∈ 6∈ } triangulation of D , but ∂Γ0 may not be a full subcomplex of K. Let = σ Γ : dim σ = 1, σ ∂Γ , σ (∂Γ ) . A0 { ∈ 0 6∈ 0 ⊂ V 0 } Then we can uniquely write Γ0 as [ Γ0 = Γ0,i, Γ0,i = KV(Γ0,i), i 2 2 where each Γ0,i is a triangulation of D such that either Γ0,i = ∆ or ∂Γ0,i is a full subcomplex of K, and Γ0,j Γ0,k contains at most one edge from 0 for j = k (see Figure5). ∩ A 6 SOME RIGIDITY PROBLEMS IN TORIC TOPOLOGY: I 53

We claim that ω Γ for some i. Since otherwise there would be an edge ∈ 0,i j1, j2 0 (j1, j2 = i1, i2) separating Γ0 into two components, such that i1 and{ i }are ∈ A in different6 ones. By definition, j , j (∂Γ ), j , j / ∂Γ , so 2 1 2 ∈ V 0 { 1 2} ∈ 0 since ∂Γ0 = lkl0 Γ KS\{l0} and lkl0 Γ, KS\{l0} are full subcomplexes of K, we may assume j (lk ∪Γ) and j (st Γ). Hence K restricted to the subset 1 ∈ V l0 \S 2 ∈ S \ V l0 U := is, l0, j1, j2 is a 4-circuit, and then it is the link of a vertex. This vertex is either{ i or i ,} since i and i are in different components of Γ j , j , as 1 2 1 2 0 − { 1 2} well as components of K[m]\U . However this contradicts the assumption in the proposition. 2 Suppose ω Γ0,a . Clearly Γ0,a = ∆ , so ∂Γ0,a = KV(Γ ). Hence by applying ∈ 0 0 6 0 0,a0 [34, Lemma 6.1] again we can find a subset I1 (Γ0,a0 ) with ω I1 such that K = S1. Note that K bounds a simplicial⊂ V disk K Γ .⊂ Let J = | I1 | ∼ I1 1 ⊂ 0,a0 1 is (I1 ω). If He0(KJ1 ) = 0, then I = I1 is the desired subset. Otherwise, assign { }∪ \ 6 + − to KI1 the same orientation as KI0 , and let L1 (resp. L1 ) be the component of + KI1\ω from i1 to i2 (resp. from i2 to i1). As before, taking a vertex l1 (L1 ), we can show that Γ := Γ l is still a triangulation of D2,∈ and S ∩ weV can 1 0,a0 − { 1} uniquely write Γ1 as [ Γ1 = Γ1,i, Γ1,i = KV(Γ1,i), i 2 2 where each Γ1,i is a triangulation of D such that either Γ1,i = ∆ or ∂Γ1,i is a full subcomplex of K, and Γ1,j Γ1,k contains at most one edge from 1 = σ Γ1 : dim σ = 1, σ ∂Γ , σ (∩∂Γ ) for j = k. A { ∈ 6∈ 1 ⊂ V 1 } 6 If ω Γ1,a1 for some a1, then we can continue this process inductively. Actually, ∈ + − we will show that if we have defined I , K , L , L , l ,Γ = Γ − l , k k k k k k k 1,ak−1 − { k} Γk,i , there must exist Γk,ak such that ω Γk,ak . Then since in each step we eliminate{ } one vertex l from the finite set ∈, this process eventually terminates, k S say, at the nth step, and we can take I = In as the desired subset. The proof concludes by showing this fact. Suppose on the contrary that for some k > 0, there is an edge j , j Γ { 1 2} ∈ k separating Γk into two components, such that i1 and i2 are in the different ones. Let = and = (∂Γ − ) for k > 0. Then as before, we may assume S0 S Sk V k 1,ak−1 j1 (lklk K) k and j2 k (stlk K). Clearly, we have j1 . Moreover, consider∈ V the full\S subcomplex∈ S on \Vi , l , j , j , then the same argument6∈ S as in the { s k 1 2} third paragraph of the proof shows that j2 . We claim that j2 (lklj K) for some j < k. To see this, note that 6∈ S ∈ V

(∂Γ − ) = (lk Γ − ) − l − , Sk ⊂ V k 1 V lk−1 k 2,ak−2 ∪ Sk 1 \{ k 1} so j2 (lklk−1 K) or j2 k−1. The ﬁrst case is what we want, and for the second case we∈ V can continue this∈ S process by an inductive argument, and eventually get the desired l for some j < k, noting j = . j 2 6∈ S0 S 54 F. FAN, J. MA & X. WANG

+ Lj i1

j1 j2 lk lj

Figure 6. Γ

Since lj, lk = (∂Γ), j1, j2 , it is easy to see that the path consisting of three edges l∈, S j , Vj , j , j , l6∈ Sseparates Γ into two components such that i { k 1} { 1 2} { 2 j} 1 and i2 are in diﬀerent ones. Without loss of generality, we may assume i1 is in the upper components and i2 is in the lower one, and we ﬁrst consider the case lj is on the right side, just as shown in Figure6. From K = S1 it follows that | Ij | ∼ + − (L ) lk, j1, j2, lj , (L ) lk, j1, j2, lj = ∅. V j ∩ { } V j ∩ { } 6 This, together with the fact that l (L+) implies that (L+) l , j , j , l = j ∈ V j V j ∩ { k 1 2 j} lj or j2, lj . However in either case, from a geometric observation we can easily { } { } + + see that when we go from i1 to i2 along Lj , Kj would be on the right side of Lj , contradicting the given orientation. For the case that lk is on the right side, we + can apply the same argument to Lk .

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Feifei Fan, School of Mathematical Sciences, South China Normal University, Guangzhou, 510631, China. Email address: [email protected]

Jun Ma, College of Mathematics, Taiyuan University of Technology, Taiyuan 030024, China. Email address: [email protected]

Xiangjun Wang, School of Mathematical Sciences, Nankai University, Tianjin 300071, China. Email address: [email protected]