B -rigidity of the property to be an almost Pogorelov polytope
aa r X i v : . [ m a t h . A T ] M a y B -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOVPOLYTOPE NIKOLAI EROKHOVETS
Abstract.
Toric topology assigns to each n -dimensional combinatorial simple convex poly-tope P with m facets an ( m + n )-dimensional moment-angle manifold Z P with an action ofa compact torus T m such that Z P /T m is a convex polytope of combinatorial type P . Westudy the notion of B -rigidity. A property of a polytope P is called B -rigid, if any isomor-phism of graded rings H ∗ ( Z P , Z ) = H ∗ ( Z Q , Z ) for a simple n -polytope Q implies that it alsohas this property. We study families of 3-dimensional polytopes defined by their cyclic k -edge-connectivity. These families include flag polytopes and Pogorelov polytopes, that is polytopesrealizable as bounded right-angled polytopes in Lobachevsky space L . Pogorelov polytopesinclude fullerenes – simple polytopes with only pentagonal and hexagonal faces. It is knownthat the properties to be a flag 3-dimensional polytope or a Pogorelov polytope are B -rigid. Wefocus on almost Pogorelov polytopes, which are strongly cyclically 4-edge-connected polytopes.They correspond to right-angled polytopes of finite volume in L . There is a subfamily of idealalmost Pogorelov polytopes corresponding to ideal right-angled polytopes. We prove that theproperties to be an almost Pogorelov polytope and an ideal almost Pogorelov polytope are B -rigid. As a corollary we obtain that 3-dimensional associahedron As and permutohedron P e are B -rigid. We generalize methods known for Pogorelov polytopes. We obtain results on B -rigidity of subsets in H ∗ ( Z P , Z ) and prove an analog of the so-called separable circuit condition(SCC). As an example we consider the ring H ∗ ( Z As , Z ). Introduction
Consider a unit circle S ⊂ C and a unit disk D ⊂ C . Toric topology (see [BP15]) assignsto each simple convex n -polytope P with m faces F , . . . , F m an ( m + n )-dimensional moment-angle manifold Z P with an action of a compact m -dimensional torus T m = ( S ) m such that Z P /T m = P . One of the ways to define the space Z P is Z P = T m × P n / ∼ , where ( t , p ) ∼ ( t , p ) if and only if p = p , and t t − ∈ T G ( p ) , where T G ( p ) = { ( t , . . . , t m ) ∈ T m : t i = 1 for F i p } .It can be shown that topological type of the space Z P depends only on combinatorial typeof P and Z P has a smooth structure such that the action of T m obtained from its action onthe first factor, is smooth. Mathematics Subject Classification.
Key words and phrases.
Thee-dimensional polytope, toric topology, almost Pogorelov polytope, B -rigidity,cyclic k -edge-connectivity, fullerene, right-angled polytope.The research was supported by the RFBR grant No 18-51-50005. Definition 0.1.
A simple n -polytope P is called B -rigid , if for any simple n -polytope Q anyisomorphism of graded rings H ∗ ( Z P , Z ) ≃ H ∗ ( Z Q , Z ) implies that P and Q are combinatoriallyequivalent.We study families of combinatorial simple 3-polytopes defined by the condition of a cyclicedge k -connectivity ( ck -connectivity for short). For 3-polytopes by faces we mean facets. Twofaces are adjacent , if they have a common edge. Definition 0.2. A k -belt is a cyclic sequence of k faces with the property that faces are adjacentif and only if they follow each other, and no three faces have a common vertex. A k -belt is trivial , if it surrounds a face.A simple 3-polytope different from the simplex ∆ is ck -connected, if it has no l -belts for l < k , and is strongly ck -connected ( c ∗ k -connected) , if in addition any its k -belt is trivial. Bydefinition ∆ is c ∗ c P s ) are c P s ⊃ P aflag ⊃ P flag ⊃ P aP og ⊃ P P og ⊃ P
P og ∗ The family of c P flag of flag c ∗ c ∗ almost flag polytopes and denote P aflag .Results by A.V. Pogorelov [P67] and E.M. Andreev [A70a] imply that c P P og ) are exactly polytopes realizable in the Lobachevsky space L as boundedpolytopes with right dihedral angles. Moreover, the realization is unique up to isometries. Thesepolytopes were called Pogorelov polytopes , since the paper [P67] is devoted to exactly this classof polytopes. The paper [A70a] implies that flag polytopes are exactly polytopes realizable in L as polytopes with equal nonobtuse dihedral angles. Examples of Pogorelov polytopes are givenby k -barrels B k , k >
5, (
L¨obell polytopes in terminology of A.Yu. Vesnin [V17], or truncatedtrapezohedra ), see Fig. 1a). Results by T. Dˇosli´c [D98, D03] imply that the family P P og contains fullerenes , that is simple 3-polytopes with only pentagonal and hexagonal faces. Mathematicalfullerenes model spherical carbon atoms. In 1996 R. Curl, H. Kroto, and R. Smalley obtainedthe Nobel Prize in chemistry “for their discovery of fullerenes”. They synthesized
Buckmin-sterfullerene C (see Fig. 1b), which has the form of the truncated icosahedron (and also of asoccer ball). W.P. Thurston [T98] built a parametrisation of the fullerene family, which impliesthat the number of fullerenes with n carbon atoms grows like n when n tends to infinity.The family P aP og of c ∗ almost Pogorelov polytopes, and thefamily P P og ∗ of c ∗ strongly Pogorelov polytopes. In [FMW20, Section 6]the simplicial 3-polytopes dual to almost Pogorelov polytopes form the family Q . G.D. Birkhoff[B1913] reduced the 4-colour problem to the task to colour in 4 colours faces of polytopes ofthe family, which turned out to be the family P P og ∗ .T.E. Panov remarked that results by E. M. Andreev [A70a, A70b] should imply that almostPogorelov polytopes correspond to right-angled polytopes of finite volume in L . Such polytopes -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 3 Figure 1. a) k -barrel B k ; b) fullerene C may have 4-valent vertices on the absolute, while all proper vertices have valency 3. It was provedin [E19] that cutting of 4-valent vertices defines a bijection between classes of congruence ofright-angled polytopes of finite volume in L and almost Pogorelov polytopes different from thecube I and the pentagonal prism M × I .In the paper [V87] A.Yu. Vesnin introduced a construction of a 3-dimensional compact hy-perbolic manifold R ( P, Λ ) corresponding to a mapping Λ of the set of faces of a polytope P ∈ P P og to Z = ( Z / Z ) such that for any vertex the images of the faces containing it forma basis. The manifold is obtained by gluing 8 copies of the polytope along faces. The existenceof a mapping follows from the 4 colour theorem. In toric topology (see [BP15, DJ91]) suchmanifolds are obtained as small covers over right-angled polytopes. For a simple n -polytope P and a mapping Λ : { F , . . . , F m } → Z n such that for any vertex the images of facets containingit form a basis in Z n toric topology associates a 2 n -dimensional smooth quasitoric manifold M ( P, Λ) with an action of T n such that M ( P, Λ) /T n = P .In [FMW15] F. Fan, J. Ma and X. Wang proved that any Pogorelov polytope is B -rigid. In[B17] F. Bosio presented a construction of flag 3-polytopes, which are not B -rigid.In [BEMPP17] the following result was proved Theorem 0.3 ([BEMPP17]) . For polytopes
P, P ′ ∈ P P og with functions Λ , Λ , and Λ ′ , Λ ′ the manifolds M ( P, Λ) and M ( P ′ , Λ ′ ) are diffeomorphic if and only if their cohomology ringsover Z are isomorphic as graded rings, and if and only if there is a combinatorial equivalence ϕ between P and P ′ and a change of coordinates A ∈ Gl ( Z ) such that for any face F of P we have Λ ′ ( ϕ ( F )) = ± A Λ( F ) . The manifolds R ( P, Λ ) and R ( P ′ , Λ ′ ) are diffeomorphic if andonly if their cohomology rings over Z are isomorphic as graded rings, and if and only if thereis a combinatorial equivalence ϕ between P and P ′ and a change of coordinates A ∈ Gl ( Z ) such that for any face F of P we have Λ ′ ( ϕ ( F )) = ± A Λ ( F ) . In this paper we develop methods from [FMW15] (see also [BE17S]) to almost Pogorelovpolytopes.The cube and the pentagonal prism belong to P aP og . All the other polytopes from P aP og have no adjacent quadrangles. The simplest polytope of this type is the 3-dimensional Stasheffpolytope (or associahedron ) As (see Fig. 2 and Fig. 16), which is the cube with three pairwisedisjoint orthogonal edges cut. Result by D. Barnette [B74] imply that a simple polytope belongs N.YU. EROKHOVETS
Figure 2.
A 3-dimensional associahedron (Stasheff polytope) As as a truncated cube As Q Q Q Q Q (2,6) Figure 3.
Almost Pogorelov polytopes with at most 11 facesto P aP og \ { I , M × I } if and only if it can be obtained from As by a sequence of operationsof cutting off an edge not lying in quadrangles, and cutting off a pair of adjacent edges of atleast hexagonal face by one plane. On Fig. 3 we draw all possible ways to construct almostPogorelov polytopes with at most m = 11 faces. It should be mentioned that it follows from[K69] (see also [V15, BE15]) that a simple 3-polytope is flag if and only if it can be obtainedfrom the cube I by a sequence of operations of cutting off an edge and cutting off a pair ofadjacent edges of at least hexagonal face.A polytope in L is called ideal , if all its vertices lie on the absolute. An ideal polytopehas a finite volume. Its congruence class is uniquely defined by the combinatorial type (see[R96]). We will call almost Pogorelov polytopes corresponding to ideal right-angled polytopes -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 5 a) b) k-gon Figure 4. a) Medial graph of the k -gonal pyramid; b) The k -antiprism Figure 5.
An edge-twist ideal almost Pogorelov polytopes and denote their family P IP og . It is known (see more details in[E19]) that graphs of ideal right-angled 3-polytopes are exactly medial graphs of 3-dimensional(not necessarily simple) polytopes. A medial graph of the k -pyramid is known as k -antiprism,see Fig. 4 b). The boundaries of polytopes dual to the polytopes in P IP og appeared in [FMW20,Example 6.3(2)] as examples of flag triangulations such that any 4-circuit full subcomplex issimple.An operation of an edge-twist is drawn on Fig. 5. Two edges on the left lie in the same faceand are disjoint. Let us call an edge-twist restricted , if both edges are adjacent to an edge ofthe same face. It follows from [V17, BGGMTW05, E19] that a polytope is realizable as an idealright-angled polytope if and only if either it is a k -antiprism, k >
3, or it can be obtained fromthe 4-antiprism by a sequence of restricted edge-twists.Thus, the simplest ideal right-angled polytope is a 3-antiprism, which coincides with theoctahedron. The corresponding ideal almost Pogorelov polytope is known as 3-dimensionalpermutohedron
P e , see Fig. 6. This is a unique ideal almost Pogorelov polytope with minimalnumber m of faces (14 faces). It is easy to see that for ideal almost Pogorelov polytopes m =2( p + 1), where p is the number of quadrangles. It follows from the above construction thatfor m = 16 there are no polytopes in P IP og , and for m = 18 and 20 there are unique polytopes.First of the them corresponds to the 4-antiprism, and the second corresponds to a polytopeobtained from it by an edge-twist. Definition 0.4.
A property of an n -polytope P is called B -rigid , if any isomorphism of gradedrings H ∗ ( Z P , Z ) = H ∗ ( Z Q , Z ) for a simple n -polytope Q implies that it also has this property.It follows from [FMW15] that the properties to be flag, Pogorelov, and strongly Pogorelov3-polytope are B -rigid. N.YU. EROKHOVETS
Figure 6.
Three-dimensional permutohedron
P e Main result of our paper is the fact that the properties to be an almost Pogorelov polytopeand an ideal almost Pogorelov polytopes are B -rigid (Theorem 3.6 and Corollary 3.8). Thisimplies that polytopes of these families with m
11 and m
20 respectively, in particular As and P e , are B -rigid (Corollaries 3.7 and 3.9).Below in the plan of the paper.In Section 1 we give a brief description of the cohomology ring of the moment-angle manifoldof a simple 3-polytope.In Section 2 we give a scheme of the proof of B -rigidity of any Pogorelov polytope from[FMW15]. Also in Proposition 2.2 we simplify a criterion for a simple 3-polytope P = ∆ to beflag, namely rk H ( Z P ) = ( m − m − m − . In Section 3 we prove that the property to be an almost Pogorelov polytope and an idealalmost Pogorelov polytopes are B -rigid (Theorem 3.6 and Corollary 3.8). This implies thatpolytopes of these families with m
11 and m
20 respectively, in particular As and P e ,are B -rigid (Corollaries 3.7 and 3.9).In Section 4 we give another characterization of almost Pogorelov and ideal almost Pogorelovpolytopes in terms of the quotient ring of H ∗ ( Z P ).In Section 5 we prove an analog of the so-called separable circuit condition (SCC), which playsa crucial role in toric topology of Pogorelov polytopes. The same result in the dual setting forsimplicial polytopes was earlier proved in [FMW20, Proposition F.1].Denote by |B| the union of faces of the belt B . Lemma 0.5 (SCC, [FMW15]) . For any Pogorelov polytope P and any three pairwise differentfaces { F i , F j , F k } with F i ∩ F j = ∅ there exist l > and an l -belt B l such that F i , F j ∈ B l , F k / ∈ B l , and F k does not intersect at least one of the two connected components of |B l |\ ( F i ∪ F j ) . We prove (Lemma 5.2) that for an almost Pogorelov polytope P for three pairwise differentfaces { F i , F j , F k } such that F i ∩ F j = ∅ there exists an l -belt ( l > B l such that F i , F j ∈ B l , F k / ∈ B l , and F k does not intersect at least one of the two connected components of |B l |\ ( F i ∪ F j )if and only if F k does not intersect quadrangles among the faces F i and F j . Moreover, we prove(Proposition 5.3) that this condition is characteristic for a flag 3-polytope to be an almost -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 7 Pogorelov polytope or the polytope P obtained from the cube by cutting off two nonadjacentorthogonal edges.In Section 6 we prove (Lemma 6.1) an analog of the other crucial tool for Pogorelov polytopes– an annihilator lemma (which is not valid for almost Pogorelov polytopes, see Proposition 8.3),and analogs of its corollaries (Corollaries 6.7 and 6.8, see also Proposition 7.9), and study thearising notion of good and bad pairs of disjoint faces.In Section 7 we prove B -rigidity in the class P aP og \ { I , M × I } (see Definition 2.1) of • the sets of elements corresponding to 4-belts (Lemma 7.2); • the cosets corresponding to pairs of disjoint faces lying in 4-belts (Lemma 7.5); • the sets of elements corresponding to pairs of adjacent quadrangles (Proposition 7.9), • to (2 k )-belts containing k quadrangles (Lemma 7.12), • and to trivial belts of this type (Lemma 7.13).In Section 8 we consider the ring H ∗ ( Z P , Z ) for P = As . In particular, we prove (Proposition8.3) that the annihilator lemma is not valid for almost Pogorelov polytopes. Remark . In paper [FMW20] the authors generalize Theorem 0.3 ([FMW20, Theorem 5.1])and result on B -rigidity ([FMW20, Theorems 3.12 and 3.14]) for flag polytopes of dimension ofhigher than 3 satisfying a generalization of the SCC condition (see Lemma 0.5). In particular,products of Pogorelov polytopes satisfy these conditions. Namely, they prove that if P is aflag n -polytope satisfying generalized SCC condition, Q is a flag n -polytope, and there is anisomorphism of graded rings H ∗ ( Z P ) ≃ H ∗ ( Z Q ), then P and Q are combinatorially equivalent.Note that at this moment for n > n -polytope isdetermined by a graded ring H ∗ ( Z P ). Also it is announced without any details, that in the nextpaper [FMW20b] the authors will prove B -rigidity of any almost Pogorelov polytope and ananalog of Theorem 0.3 for them. But their Definition 2.16 of B -rigidity assumes isomorphismof bigraded rings, which is a more restrictive condition. Remark . Results of our paper imply (see Remark 6) that any ideal almost Pogorelov is B -rigid. This gives cohomologically rigid families of 3-dimensional and 6-dimensional manifoldscorresponding to a unique (up to a permutation of colours) colouring of faces of an ideal almostPogorelov polytope into 3 colours. They are ”pullbacks from the linear model” from [DJ91,Example 1.15(1)]. Details see in [E20b].1. Cohomology ring of a moment-angle manifold of a simple -polytope Details on cohomology of a moment-angle manifolds see in [BP15, BE17S]. If not specified,we study cohomology over Z and omit the coefficient ring.The ring H ∗ ( Z P ) has a multigraded structure: H ∗ ( Z P ) = M i > ,ω ⊂ [ m ] H − i, ω ( Z P ) , where H − i, ω ( Z P ) ⊂ H | ω |− i ( Z P ) , and [ m ] = { , . . . , m } . There is a canonical isomorphism H − i, ω ( Z P ) ≃ e H | ω |− i − ( P ω ), where P ω = S i ∈ ω F i , and e H − ( ∅ ) = Z . N.YU. EROKHOVETS
The multiplication between components is nonzero, only if ω ∩ ω = ∅ . The mapping H − i, ω ( Z P ) ⊗ H − j, ω ( Z P ) → H − ( i + j ) , ω ⊔ ω ) ( Z P )via the Poincare-Lefschets duality H i ( P ω ) ≃ H n − − i ( P ω , ∂P ω ) up to signs is induced by inter-section of faces of P . Denote by b H n − i − ( P ω , ∂P ω ) the subgroup corresponding to e H i ( P ω ).For any simple 3-polytope P and ω = ∅ the set P ω is a 2-dimensional manifold, perhapswith a boundary. Each connected component of P ω is a sphere with holes, and its bound-ary is a disjoint union of simple edge-cycles. Then e H k ( P ω ) is nonzero only for ( k, ω ) ∈{ ( − , ∅ ) , (0 , ∗ ) , (1 , ∗ ) , (2 , [ m ]) } . In particular, P has no torsion. There is a multigraded Poincareduality, which means that the bilinear form H − i, ω ( Z P ) × H − ( m − − i ) , m ] \ ω ) ( Z P ) → H − ( m − , m ] ( Z P ) = Z has in some basis matrix with determinant ±
1. In particular, for a basis { e α } in H − i, ω ( Z P )there is a dual basis { e ∗ β } in H − ( m − − i ) , m ] \ ω ) ( Z P ) such that e α · e ∗ β = δ α,β [ Z P ], where [ Z P ] isa fundamental class in cohomology. We have H ( Z P ) = e H − ( ∅ ) = Z = e H ( ∂P ) = H m +3 ( Z P ); H ( Z P ) = H ( Z P ) = 0 = H m +1 ( Z P ) = H m +2 ( Z P ); H k ( Z P ) = M | ω | = k − e H ( P ω ) ⊕ M | ω | = k − e H ( P ω ) , k m. Nontrivial multiplication occurs only for(1) e H − ( P ∅ ) ⊗ e H k ( P ω ) → e H k ( P ω ). This corresponds to multiplication by 1 in H ∗ ( Z P );(2) e H ( P ω ) ⊗ e H ( P [ m ] \ ω ) → e H ( ∂P ). This corresponds to the Poincare duality pairing.(3) e H ( P ω ) ⊗ e H ( P ω ) → e H ( P ω ⊔ ω ). This corresponds to the mapping b H ( P ω , ∂P ω ) ⊗ b H ( P ω , ∂P ω ) → H ( P ω ⊔ ω , ∂P ω ⊔ ω ) . Each group on the left has a basis with elements corresponding to consistently oriented con-nected components of P ω i with a unique relation that the sum of these elements is zero. Forconnected components in P ω and P ω the product up to a sign is the intersection of them,which is a sum of classes [ γ ], . . . , [ γ r ] of disjoint oriented edge paths on P ω ⊔ ω connectingcomponents of ∂P ω ⊔ ω (if the intersection is not a boundary cycle of each component), see Fig.7. Moreover, for the product to be nonzero at least one component of ∂P ω ⊔ ω should have onecommon point with some path γ i . For each connected component on the right a basis in thefirst homology group corresponds to any set of oriented edge paths connecting one boundarycycle to all the other. 2. B -rigidity of any Pogorelov polytope Definition 2.1.
Let P be some set of 3-polytopes.We call a number n ( P ), a set S P ⊂ H ∗ ( Z P ) or a collection of such sets defined for anypolytope P ∈ P B -rigid in the class P if for any isomorphism ϕ of graded rings H ∗ ( Z P ) ≃ -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 9 c.c. of P ω c.c. of P ω2 γ (cid:0) γ γ γ Figure 7.
Intersection of two connected components of P ω and P ω H ∗ ( Z Q ), P, Q ∈ P , we have n ( P ) = n ( Q ), ϕ ( S P ) = S Q , or each set from the collection for P is mapped bijectively to some set from the collection for Q respectively.To be short, B -rigidity in the class of all simple 3-polytopes we call B -rigidity .In [FW15] it was proved that the property to be a flag polytope is B -rigid. Namely, it isequivalent to the fact that the ring e H ∗ ( Z P ) / ([ Z P ]) is a (nonzero) indecomposable ring. Also itwas proved that for flag 3-polytopes e H ( P ω ) = M ω ⊔ ω e H ( P ω ) · e H ( P ω ) . This result was based on the fact that for any flag 3-polytope and any three pairwise differentfaces { F i , F j , F k } with F i ∩ F j = ∅ there exist l > and an l -belt B l such that F i , F j ∈ B l and F k / ∈ B l . This fact is characteristic for a polytope P = ∆ to be flag. Indeed, if P has a 3-belt,then for the faces F i and F j lying in different connected components of the complement to thisbelt there are no belts containing F i and F j .On the base of these facts in [BE17S] and [BEMPP17] a simple characterization of flagpolytopes was given: a simple 3-polytope P = ∆ is flag if and only if H m − ( Z P ) ⊂ ( e H ∗ ( Z P )) . Moreover, it turned out that there is another criterion when a 3-simple polytope is flag.
Proposition 2.2.
A simple -polytope P = ∆ with m faces is flag if and only if rk H ( Z P ) = ( m − m − m − . Proof.
Indeed, [BP15, Theorem 4.6.2] states that for a simple n -polytope(1 − t ) m − n ( h + h t + · · · + h n t n ) = X − i, j ( − i β − i, j t j , where h + h t + · · · + h n t n = ( t − n + f n − ( t − n − + · · · + f .In particular, for 3-polytopes we have(1 − t ) h (1 + ht + ht + t ) =1 − β − , t + h X j =3 ( − j − ( β − ( j − , j − β − ( j − , j ) t j +( − h − β − ( h − , h +1) t h +1) + ( − h t h +3) , where h = m − − ht + h ( h − t − h ( h − h − t + . . . )(1 + ht + ht + t ) =1 − β − , t + ( β − , − β − , ) t + . . . . For any simple 3-polytope P we have: • rk H ( Z P ) = β − , = | N ( P ) | = h ( h − ; • β − , = { } ; • β − , − β − , = ( h − h − .Then rk H ( Z P ) = β = β − , = β − , + ( h − h − { } + ( h − h − (cid:3) It is easy to see that a polytope has no 4-belts if and only if the multiplication H ( Z P ) ⊗ H ( Z P ) → H ( Z P )is trivial. Thus, the property to be a Pogorelov polytope is also B -rigid.In [FMW15] (details see also in [BE17S]) it was proved that any Pogorelov polytope is B -rigid.The proof consisted of several steps. Denote N ( P ) = {{ i, j } ⊂ [ m ] : F i ∩ F j = ∅ } . For any ω ∈ N ( P ) we have e H ( P ω ) = Z . Denote by e ω a generator of this group (it is definedup to a sign, we choose one of them arbitrarily). Then for any 3-polytope H ( Z P ) is a freeabelian group with the basis { e ω : ω ∈ N ( P ) } .First is was proved that the set of elements {± e ω : ω ∈ N ( P ) } ⊂ H ( Z P )corresponding to sets ω ∈ N ( P ) is B -rigid in the class of Pogorelov polytopes. The proof wasbased on the so-called separable circuit condition (SCC): for any Pogorelov polytope P and anythree different faces { F i , F j , F k } with F i ∩ F j = ∅ there exist l > and an l -belt B l such that F i , F j ∈ B l , F k / ∈ B l , and F k does not intersect at least one of the two connected components of -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 11 |B l | \ ( F i ∪ F j ). In fact, this condition is characteristic for a polytope P = ∆ to be Pogorelov.Indeed, if P has a 4-belt, then for the faces F i and F j lying in different connected components ofthe complement to this belt and F k lying on the belt any belt containing F i and F j intersects F k by both components. Later, in Lemma 5.2 we will prove an analog of SCC for almost Pogorelovpolytopes. It will be also characteristic to be an almost Pogorelov polytope with one exception,see Proposition 5.3.Now let us give some details of the proof, which we will need in what follows. Definition 2.3. An annihilator of an element r in a ring R is defined asAnn R ( r ) = { s ∈ R : rs = 0 } Since the group H ∗ ( Z P ) has no torsion, we have the isomorphism H ∗ ( Z P , Q ) ≃ H ∗ ( Z P ) ⊗ Q and the embedding H ∗ ( Z P ) ⊂ H ∗ ( Z P ) ⊗ Q . For polytopes P and Q the isomorphism H ∗ ( Z P ) ≃ H ∗ ( Z Q ) implies the isomorphism over Q . For the cohomology over Q all theoremsabout structure of H ∗ ( Z P , Q ) are still valid.It was proved that for a Pogorelov polytope P and an element α in H = H ∗ ( Z P , Q ): α = X ω ∈ N ( P ) r ω e ω with |{ ω : r ω = 0 }| > H ( α ) < dim Ann H ( e ω ) , if r ω = 0 . We call this result the annihilator lemma .For almost Pogorelov polytopes the annihilator lemma is not valid, see Proposition 8.3. InLemma 6.1 we will modify it to be valid for all simple 3-polytopes. Corollary 6.7 shows how thislemma can be applied to almost Pogorelov polytopes. Namely, under any isomorphism of gradedrings H ∗ ( Z P ) → H ∗ ( Z Q ) the element e ω for ω ∈ N ( P ) is mapped to ± e ω ′ for ω ′ ∈ N ( Q ), if ω = { p, q } , and both F p and F q are not adjacent to quadrangles. In Definition 6.3 for ω ∈ N ( P )we introduce the notion of good and bad elements in N ( P ), and in Corollary 6.5 we prove thatfor any simple 3-polytopes P and Q the element e ω, ω ∈ N ( P ) is mapped to a linear combinationof an element e ω ′ , ω ′ ∈ N ( Q ) and elements e ω ′′ , ω ′′ ∈ N ( Q ), where ω ′′ are bad for ω ′ .On the second step in [FMW15] it was proved that the set {± f B k : B k − a k -belt } ⊂ H k +2 ( Z P )is B -rigid in the class of Pogorelov polytopes. Here f B k is the element in H k +2 ( Z P ) correspondingto a generator in e H ( P ω ( B k ) ) ≃ Z , where ω ( B k ) = { i ∈ [ m ] : F i ∈ B k } . In Corollary 7.2 we willprove that the set of elements corresponding to 4-belts is B -rigid in the class of almost Pogorelovpolytopes. Also in Corollary 6.8 we will prove that if P ∈ P aP og \ { I , M × I } and all the facesof a k -belt B k do not have common points with quadrangles, then under any isomorphism ofgraded rings H ∗ ( Z P ) → H ∗ ( Z Q ) for a simple 3-polytope Q the element f B k is mapped to ± f B ′ k for some k -belt B ′ k of Q . Also from the arguments in [FMW15] it follows (see the detailed proof in [BE17S]) thatfor any simple 3-polytope P the subgroup B k ⊂ H k +2 ( Z P ), 4 k m −
2, with the basis { f B k : B k − a k -belt } is B -rigid. In particular, the number of k -belts, k >
4, is B -rigid.Then it was proved that the set of elements {± f B k : B k − a k -belt around a face } ⊂ H k +2 ( Z P )is B -rigid in the class of Pogorelov polytopes. Corollary 2.4.
The property to be a strongly Pogorelov polytope is B -rigid. This induces a bijection between the sets of faces of polytopes.Then it was proved that images of adjacent faces are adjacent. This finishes the proof of B -rigidity of any Pogorelov polytope.3. B -rigidity of the property to be an almost Pogorelov polytope Our aim is to prove that the property to be an almost Pogorelov polytope is B -rigid.First the cube and the pentagonal prism are the only flag polytopes with m = 6 and m = 7respectively. Hence, they are B -rigid.Now consider all the other almost Pogorelov polytopes. They have no adjacent quadrangles,since a pair of adjacent quadrangles should be surrounded by a nontrivial 4-belt.The image of the mapping H ( Z P ) ⊗ H ( Z P ) → H ( Z P ) is the subgroup B generated bythe elements f B corresponding to 4-belts.There is a B -rigid subgroup A ⊂ H ( Z P ): A = { x ∈ H ( Z P ) : x · y = 0 for all y ∈ H ( Z P ) } . It is generated by elements e ω such that the set ω ∈ N ( P ) does not belong to ω ( B ) for any4-belt B . Denote this set N ( P ).The subgroup A is a direct summand in H ( Z P ), therefore the quotient group H = H ( Z P ) /A is also a free abelian group. We have the mapping H ⊗ H → H ( Z P ). Also H ⊗ Q ≃ H ( Z P , Q ) / ( A ⊗ Q ) and there is an embedding H ⊂ H ⊗ Q . Lemma 3.1.
We have B > rk H , and the equality holds if and only if each set ω ∈ N ( P ) can be included in at most one set ω ( B ) .Proof. Each element f B can be represented uniquely up to a change of order as a product oftwo elements e ω · e ω , ω i ∈ N ( P ). On the other hand, an element e ω can be a factor of severalelements, corresponding to 4-belts, if the faces F p , F q can be included into several 4-belts, where ω = { p, q } . Let us calculate the number of pairs ( ω, B ), where ω ∈ N ( P ) and ω ⊂ ω ( B ). Onthe one hand, it is equal to 2 rk B , on the other hand, it is at least the number of elements in N ( P ) that can be included into a 4-belt, and the equality holds if and only if each this elementcan be included in exactly one set ω ( B ). (cid:3) Lemma 3.2.
Let P be a flag -polytope. If B = rk H , then P has no adjacent quadran-gles. -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 13 F i F j F k F l F r F i F j F j F i F i F j F k F l F r F k F k F l F l F r F r a) b) c) d) Figure 8. a) 4-belt ( F i , F j , F k , F l ) disjoint from F r ; b) 4-belt ( F i , F j , F k , F l ), and F r intersects only F j ; c) 4-belt ( F i , F j , F k , F l ) and F r intersects only F i and F j ;d) 5-belt ( F i , F j , F k , F l , F r ) Proof. If P is the cube I , then rk B = rk H = 3. Else if P has two adjacent quadrangles,then these quadrangles are surrounded by a 4-belt such that the two faces intersecting boththese quadrangles are included into two 4-belts surrounding the quadrangles. (cid:3) Lemma 3.3.
Let P be a flag -polytope.If e H ( P ω ) = 0 , then | ω | > . Moreover, if | ω | = 4 , then ω = ω ( B ) for some -belt.If B = rk H , and | ω | = 5 , then e H ( P ω ) = 0 if and only if P ω has one of the typesdrawn on Fig. 8.Proof. As we mentioned above, P ω is a disjoint union of spheres with holes. If e H ( P ω ) = 0,then one sphere P ω ′ , ω ′ ⊂ ω , has at least two holes. This is impossible for | ω ′ | = 1 or 2. Hence, | ω ′ | >
3. Consider the boundary component γ of P ω ′ . Walking along this simple edge-cyclewe obtain a cyclic sequence of faces L = ( F i , . . . , F i p ), where the index i l lies in Z p = Z /p Z .If p = 1, then P ω ′ consists of one face and is contractible. If p = 2, then P ω ′ consists of twoadjacent faces and also is contractible. If p = 3, then F i , F i and F i are different pairwiseadjacent faces. Since P has no 3-belts, they have a common vertex and P ω ′ consists of thesethree faces and is contractible. Thus, p > F i l = F i l +1 for any l . If F i l = F i l +2 for some l , then we have the configuration on Fig.9a), and P ω ′ is homeomorphic to P ω ′ \{ i l +1 } . If F i l ∩ F i l +1 is an edge, then we have the configurationon Fig. 9b), and P ω ′ is homeomorphic to P ω ′ \{ i l +1 } . If F i l = F i l +3 , then F i l , F i l +1 , F i l +2 are threedifferent pairwise adjacent faces. Since P has no 3-belts, the edge F i l +1 ∩ F i l +2 intersects F i l ,and we have the configuration on Fig. 9c). Then P ω ′ is homeomorphic to P ω ′ \{ i l +1 ,i l +2 } .If | ω ′ | = 3, then F i l = F i l +1 , F i l = F i l +2 , F i l = F i l +3 for all l by the previous argument. Acontradiction. Hence, | ω | > | ω ′ | > | ω ′ | = 4, then F i l = F i l +1 , F i l = F i l +2 , F i l ∩ F i l +2 = ∅ , F i l = F i l +3 for all l by the previousargument. If p = 4, then L is a 4-belt and ω = ω ′ = ω ( L ). Let p >
5. Then F i = F i = F i , inparticular, p >
6. But F i is separated from F i , F i , F i by F i = F i , which is a contradiction.Thus, p = 4. a) F il F i l+1 b) F il F il+1 F il+2 c) F il F il+1 F il+2 Figure 9. a) F i l = F i l +2 ; b) F i l ∩ F i l +2 is an edge, c) F i l = F i l +3 Let | ω | = 5 and 2 rk B = rk H . If ω ′ = ω , then ω ′ = ω ( B ) for some 4-belt, and P ω hasthe structure drawn on Fig. 8a). Now let ω ′ = ω . If F i l = F i l +2 for some l , then by the previousargument P ω has the structure drawn on Fig. 8b). If F i l ∩ F i l +2 is an edge for some l , then P ω hasthe structure drawn on Fig. 8c). If F i l = F i l +3 , then P ω is contractible, which is a contradiction.Now assume that F i l = F i l +1 , F i l = F i l +2 , F i l ∩ F i l +2 = ∅ , F i l = F i l +3 for all l . If F i l ∩ F i l +3 isan edge for some l , then B = ( F i l , F i l +1 , F i l +2 , F i l +3 ) is a 4-belt. Let i = ω \ { i l , i l +1 , i l +2 , i l +3 } .The face F i lies in one of the connected components of the complement to this belt in ∂P . If F i intersects two opposite faces of the belt, say F i l and F i l +2 , then it should also intersect F i l +1 and F i l +3 for otherwise the pair { F i l , F i l +2 } can be included into two 4-belts, which contradictsthe condition 2 rk B = rk H . Since P is flag, this means, that B is a belt around F i and P ω is contractible. Thus, F i intersects at most two faces and they should be successive in the belt.We obtain the cases b) and c) of Fig. 8 again. Now consider the case when F i l ∩ F i l +3 = ∅ forall l . Then p > F i l = F i l +4 and ω = { i l , i l +1 , i l +2 , i l +3 , i l +4 } for any l . If p = 5, then L is a5-belt (Fig. 8d). Else F i l = F i l +5 , in particular, p >
7. Then F i l +6 is separated from F i l +1 , F i l +2 , F i l +3 , F i l +4 by F i l = F i l +5 , which is a contradiction. Thus, p = 5. Now we have considered allthe possible cases and the proof is finished. (cid:3) Corollary 3.4.
Let P be a flag -polytope with B = rk H . Then the image I of themapping H ( Z P ) ⊗ H ( Z P ) → H ( Z P ) is a free abelian group with a basis corresponding to elements f B , where B is a -belt, andgenerators of the groups e H ( P ω ) ≃ Z for P ω drawn on Fig. 8 a)-c). Corollary 3.5.
Let P be a flag -polytope with B = rk H . rk I = { -belts } + ( m − { trivial -belts } + ( m − { nontrivial -belts } . Proof.
For any 4-belt B = ( F i , F j , F k , F l ) and any face F r / ∈ B if F r intersects opposite facesof the belt, say F i and F k , then F r ∩ F j = ∅ , for otherwise ( F i , F j , F k , F r ) is a 4-belt, and F r ∩ F l = ∅ for the same reason. Since P is flag, this implies that B surrounds F r . Otherwise, F r intersects at most two faces, and they should be successive. Then B ∪ F r has the form drawn -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 15 on Fig. 8a)-c). For a trivial belt we have m − F r , and for a nontrivial belt we have m − (cid:3) Theorem 3.6.
A flag simple -polytope P belongs to P aP og \ { I , M × I } if and only if B = rk H , and rk I = rk B + ( m − B , where rk B = { -belts } is a B -rigid number. In particular, the property to be an almostPogorelov polytope is B -rigid.Proof. Corollary 3.5 implies that if the conditions of the theorem hold, then every 4-belt of P is trivial, hence P is an almost Pogorelov polytope. Lemma 3.2 implies that P has no adjacentquadrangles. Hence, P is different from the cube and the pentagonal prism.Now let P ∈ P aP og \ { I , M × I } . It has no adjacent quadrangles, for otherwise thesequadrangles should be surrounded by a nontrivial 4-belt. Consider a pair of nonadjacent faces F i and F k that can be included into a 4-belt ( F i , F j , F k , F l ). The belt surrounds some quadrangle F r .Let F i and F k be included into the other 4-belt ( F i , F p , F k , F q ). This belt surrounds some otherquadrangle F t = F r . Then ( F i , F r , F k , F t ) is a 4-belt. It surrounds some quadrangle adjacent toboth F t and F r . A contradiction. Thus, any pair of nonadjacent faces of P can be included inat most one 4-belt, and 2 rk B = rk H by Lemma 3.1. Then rk I = rk B + ( m − B byCorollary 3.5. This finishes the proof. (cid:3) Corollary 3.7.
Any almost Pogorelov polytope with at most faces is B -rigid.Proof. Indeed, as we mentioned above for m = 6 the cube, and for m = 7 the pentagonal prismis a unique flag 3-polytope. For m = 8 there are no almost Pogorelov polytopes. For m = 9 theassociahedron As is a unique almost Pogorelov polytope. For m = 10 and 11 different almostPogorelov polytopes have different number of quadrangles, which is equal to rk B . (cid:3) Corollary 3.8.
A flag simple -polytope P is an ideal almost Pogorelov polytope if and only if B = rk H = m − , and rk I = rk B + ( m − m − . In particular, the property to be an ideal almost Pogorelov polytope is B -rigid.Proof. Indeed, an almost Pogorelov polytope is ideal if and only if any its vertex lies on aunique quadrangle, which is equivalent to the fact that the number of quadrangles p = rk B multiplied by 4 is equal to the number of vertices f , which is 2( m −
2) for any simple 3-polytope. (cid:3)
Corollary 3.9.
The -dimensional permutohedron P e is B -rigid. Moreover, any ideal almostPogorelov polytope with m is B -rigid.Proof. Indeed,
P e is a unique ideal almost Pogorelov polytope with minimal number of faces m = 14. Since m = 2 + 2 p , this number is even. For m = 16 there are no ideal almost Pogorelovpolytopes. For m = 18 and m = 20 there are unique ideal almost Pogorelov polytopes. (cid:3) The quotient ring of H ∗ ( Z P ) modulo the ideal generated by A . Now we will give equivalent characterizations of almost Pogorelov and ideal almost Pogorelovpolytopes in terms of the quotient ring of H ∗ ( Z P ) modulo the ideal generated by A . Lemma 4.1.
Let P be a flag -polytope.Then for k > any its k -belt B different from the beltsurrounding the base of a k -gonal prism, contains a pair of nonadjacent faces that can not beincluded into any -belt.Proof. Let B = ( F i , F i , . . . , F i k ) and C and C be components of its complement in ∂P .Let any pair of non-successive faces of B can be included into a 4-belt. Then there is a 4-belt( F i , F p , F i , F q ). At least one of the faces F p and F q , say F p is different from F i , hence liesin C or C , say C . There is a 4-belt ( F i , F p , F i , F q ). As before, F p lies in C or C . If F p liesin C , then F p = F p , since no other faces in C different from F p can intersect both F i and F i (this follows from the Jordan curve theorem, since the set F i ∩ C lies inside the disk boundedby a simple piecewise linear curve lying inside F i ∪ F i ∪ F i ∪ F p , and F i ∩ C lies outside thiscurve (see Fig. 10a)). Thus, F p is adjacent to F i , F i , F i , and F i . Since P is flag, we obtainconfiguration drawn on Fig. 10b). Then F q , F q ⊂ C . Thus, F q = F q by the same agrument,and we obtain the same picture as for F p . There is a 4-belt ( F i , F p , F i , F q ). The face F i intersects only faces F i , F i , F p , and F q . Since F i ∩ F i = ∅ , we have { F p , F q } = { F p , F q } ,and we can assume that F p = F p , and F q = F q . Repeating this argument, we obtain that F p i = F p , F q i = F q for all i , and B is the belt around the base of a k -gonal prism, which is acontradiction.Thus, F p ⊂ C , F q ⊂ F i ∪ C , and F q ⊂ F i ∪ C . Moreover, if F q = F i , then F p ∩ F i = ∅ .Else F p , F q ⊂ C , and at least one of these faces do not intersect F i . Without loss of generalitywe can assume that F p ∩ F i = ∅ . Similarly F p ∩ F i = ∅ . Repeating the argument, we obtain F p ⊂ C , F q ⊂ F i ∪ C , F p ∩ F i = ∅ , and so on. If k is odd, then F p k ⊂ C , F q k ⊂ F i k +1 ∪ C ,and F p ⊂ F i ∪ C , which is a contradiction. If k is even, then k >
6. Consider the 4-belt( F i , F s , F i , F t ). Both faces F s and F t do not belong to B , since they intersect both F i and F i .But in C there are no faces intersecting both F i and F i , because F p ∩ F i = ∅ . Similarly in C there are no faces intersecting both F i and F i , because F p k ∩ F i = ∅ . A contradiction.This finishes the proof. (cid:3) Corollary 4.2.
In the ring H ∗ ( Z P ) / h A i all the elements [ f B k ] corresponding to k -belts, k > ,are zero. Corollary 4.3.
Let P be a flag -polytope with B = rk H . Then the image A of themapping A ⊗ H ( Z P ) → H ( Z P ) is a free abelian group B with the basis corresponding to elements f B , where B is a -belt. -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 17 F i1 a(cid:1) F p C b)F (cid:2) F i2 F i (cid:3) F i (cid:4) F i F i1 F i2 F i (cid:5) F i (cid:6) F i (cid:7) F i1 c) F (cid:8)6 F i2 F i (cid:9) F i (cid:10) F i (cid:11) F i (cid:12) F (cid:13)(cid:14) Figure 10.
Configuration of faces near the belt
Proof.
Indeed, Lemma 4.1 implies, that B ⊂ A . Let x ∈ A . Then x = X ω ∈ N ( P ) λ ω e ω · X τ ⊂ [ m ] : | τ | =3 y τ = X η ⊂ [ m ] : | η | =5 z η , where each nonzero y τ corresponds to an element in b H ( P τ , ∂P τ ) = 0, and each nonzero z η corresponds to an element in b H ( P η , ∂P η ) = 0 such that for some ω ∈ N ( P ) we have ω ⊂ η and b H ( P η \ ω , ∂P η \ ω ) = 0. By Lemma 3.3 P η should have structure drawn on Fig. 8a)-d). Forcases a)-c) for any ω ∈ N ( P ), ω ⊂ η the set P η \ ω is contractible. Thus, each η has the form ω ( B ) for some 5-belt, and z η = µ η f B . Hence, A ⊂ B , and this finishes the proof. (cid:3) Denote I = I /A . Corollary 4.4.
Let P be a flag simple -polytope with B = rk H . Then group I is afree abelian group of rank ( m − { trivial -belts } + ( m − { nontrivial -belts > ( m − B . Corollary 4.5.
A flag simple -polytope P belongs to P aP og \ { I , M × I } if and only if B = rk H , and rk I = ( m − B . Corollary 4.6.
A flag simple -polytope P is an ideal almost Pogorelov polytope if and only if B = rk H = m − , and rk I = ( m − m − . Analog of the SCC for almost Pogorelov polytopes
In this section we generalize the SCC to almost Pogorelov polytopes.
Definition 5.1. A k -path , k >
1, (or a thick path ) is a sequence ( F i , . . . , F i k ) of faces suchthat successive faces are adjacent. A k -path is called simple , if all its faces are pairwise distinct.By definition a length of a k -path is equal to k . For any two faces in P there is a thick pathconnecting them. The shortest thick path between faces is simple. F k F i F j F i1 F i2 F i (cid:15) F i p-1 F ip F j1 F j2 F jq-1 F jq F w1 F w2 F wr-1 F wr F w3 F wr-2 Figure 11.
Case 1A 1 -loop is a face. By a k -loop , k >
2, we call a closed ( k + 1)-path ( F i , . . . , F i k +1 ): F i k +1 = F i . We denote the k -loop simply ( F i , . . . , F i k ). A k -loop is called simple , if all its faces aredistinct. For convenience we assume that the parameter j in the index i j of a k -loop belongsto Z k = Z /k Z , that is i k = i , i k +1 = i and so on.For a thick path, a k -loop or a k -belt L denote by |L| the union of its faces. The followingresult generalize the SCC, which is crucial for Pogorelov polytopes. Lemma 5.2.
Let P be an almost Pogorelov polytope. Then for three pairwise different faces { F i , F j , F k } such that F i ∩ F j = ∅ there exists an l -belt ( l > ) B l such that F i , F j ∈ B l , F k / ∈ B l ,and F k does not intersect at least one of the two connected components of |B l | \ ( F i ∪ F j ) if andonly if F k does not intersect quadrangles among the faces F i and F j .Proof. If F i is a quadrangle adjacent to F k , then any l -belt B l containing F i and not containing F k in the closure of each connected component of |B l | \ ( F i ∪ F j ) contains a face from the 4-beltaround F i adjacent to F k . Hence, F k intersects both connected components.Now let F k do not intersect quadrangles among the faces F i and F j .Since P is flag, there is an s -belt B = ( F i , F i , . . . , F i p , F j , F j , . . . , F j q ), with F k / ∈ B , s = p + q + 2, and p, q >
1. The complement ∂P \ |B | consists of two connected components C and C , both homeomorphic to disks. Without loss of generality assume that F k ⊂ C . Theneither C = F k , or ∂C ∩ F k consists of finite set of disjoint edge-segments γ , . . . , γ d .In the first case B surrounds F k , and F i and F j are adjacent to F k . Consider all faces { F w , . . . , F w r } in C adjacent to faces in { F i , . . . , F i p } (see Fig. 11), in the order we meetthem while walking round ∂C from C ∩ F i to C ∩ F j . Then F w a ∩ F j b = ∅ for any a, b ,for otherwise ( F k , F j b , F w a , F i c ) is a 4-belt for any i c with F i c ∩ F w a = ∅ , since F k ∩ F w a = ∅ and F j b ∩ F i c = ∅ . This belt should surround a quadrangle from B intersecting both F j b and F i c . This quadrangle can be only F i or F j , which is a contradiction. We have a thick path( F i , F w , . . . , F w r , F j ). Consider the shortest thick path of the form ( F i , F w s , . . . , F w st , F j ). If twofaces of this path intersect, then they are successive, else there is a shorter thick path. Thus wehave the belt B l = ( F i , F w s , . . . , F w st , F j , F j , . . . , F j q ) with faces of the segment ( F w s , . . . , F w st )not intersecting F k . -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 19 In the second case we can assume that F i ∩ F k = ∅ or F j ∩ F k = ∅ , say F i ∩ F k = ∅ , forotherwise we can take the belt B ′ around F k . Let γ a = ( F k ∩ F u a, , . . . , F k ∩ F u a,la ). Denote by U a the thick path ( F u a, , . . . , F u a,la ), and by S a the segment ( F s a, , . . . , F s a,ta ) of B between U a and U a +1 . Then B = ( U , S , U , . . . , U d , S d ).Consider the thick path Π a = ( F w a, , . . . , F w a,ra ) ⊂ C arising while walking round ∂C along edges ∂C ∩ F u a,j , j = 1 , . . . , l a (see Fig. 12). Then Π a ∩ Π b = ∅ for a = b , else( F w , F u a,j , F k , F u b,j ) is a 4-belt for any F w ∈ Π a ∩ Π b such that F w ∩ F u a,j = ∅ , F w ∩ F u b,j = ∅ .This belt surrounds a quadrangle F x adjacent to F k . If F x ⊂ C , then F x ∩ F w = ∅ , which is acontradiction. Otherwise, F x ∈ B , and the faces F u a,j and F u b,j are adjacent to it in the belt.Then a = b (since P is flag), which is a contradiction. Also F w a,j = F w a,j for j = j . This istrue for faces adjacent to the same face F u a,i . Let F w a,j = F w a,j . If the faces are adjacent tothe successive faces F u a,i and F u a,i +1 , then the flagness condition implies that j = j and F w a,j is the face in C intersecting F u a,i ∩ F u a,i +1 . If the faces are adjacent to non-successive faces F u a,i and F u a,j , then ( F w a,j , F u a,i , F k , F u a,j ) is a 4-belt around a quadrangle F x . If F x ⊂ C , then F x ∩ F w a,j = ∅ . Otherwise, F x ∈ B , and the faces F u a,i and F u a,j are adjacent to it in the belt.Without loss of generality we can assume that F x = F u a,i +1 , and j = i + 2. Then j = j , and F w a,j is a unique face in C intersecting F u a,i +1 . A contradiction.Now consider the thick path V b = ( F v b, , . . . , F v b,cb ) arising while walking round ∂C alongedges ∂C ∩ F s b,j (see Fig. 12). Then |V a | ∩ |V b | = ∅ for a = b , and | Π a | ∩ |V b | = ∅ for any a, b ,since interiors of the corresponding faces lie in different connected components of ∂P \ ( |B |∪ F k ).Now we will deform the segments I = ( F i , . . . , F i p ) and J = ( F j , . . . , F j q ) of the belt B to obtain a new belt ( F i , I ′ , F j , J ′ ) with I ′ not intersecting F k . First substitute the thickpath Π a for each segment U a ⊂ I and the thick path V b for each segment S b ⊂ J . Since F s a,ta ∩ F w a +1 , = ∅ , F w a,ra ∩ F s a, = ∅ , F v a,ca ∩ F u a +1 , = ∅ , and F u a,la ∩ F v a, = ∅ for any a and a + 1 considered mod d , we obtain a loop L = ( F i , I , F j , J ) instead of B .Since F i ∩ F k = ∅ , we have F i = F s ai,fi for some a i , f i . If F j = F s aj,fj for some a j , f j , thenwe can assume that a i = a j , else the faces in I or J already do not intersect F k , and B isa belt we need. If F j = F u aj,fj for some a j and some f j >
1, then substitute the thick path( F w aj, , . . . , F w aj,gj ), where g j – the first integer with F w aj,gj ∩ F j = ∅ (then F j ∩ F u aj,fj − ∩ F w aj,gj is a vertex), for the segment ( F u aj, , . . . , F u aj,fj − ) to obtain a loop L = ( F i , I , F j , J ) (elseset I = I and L = L ) with faces in I not intersecting F k . If 1 < f j < l a j , then F w aj,gj ∩ F u aj,fj +1 = ∅ , for otherwise ( F k , F u aj,fj − , F w aj,gj , F u aj,fj +1 ) is a 4-belt around the quadrangle F j .Then F w a,l ∩ F u aj,r = ∅ for any r ∈ { f j + 1 , . . . , l a j } and a, l , such that either a = a j , or a = a j ,and l ∈ { , . . . , g j } . Hence faces of the segment ( F u aj,fj +1 , . . . , F u aj,laj ) do not intersect faces in I .Now a face F i ′ a of I can intersect a face F j ′ b of J only if F i ′ a = F w c,h for some c, h , and F j ′ b = F s ai,l for l < f i , or F j ′ b = F s aj,l for F j = F s aj,fj and l > f j . In the first case take thesmallest l for all c, h , and the correspondent face F w c,h . Consider the face F u b,g = F i e ∈ I with F (cid:16) F u F (cid:17) F (cid:18) F u4,2 F w1,1 F w1,2 F w3,1 F w3,2 F w3,3 F w3,4 F w4,1 F w4,2 F u1,2 F u2,1 F u1,1 F v1,1 F v1,1 F v2,1 F v2,2 F v2,3 F v3,1 F v3,2 F v3,3 F v3,4 F v3,5 F s1,1 F s1,2 F s1,3 F s2,1 F s2,2 F s2,5 F s2,6 F s3,1 F s3,2 F s3,6 F s3,7 F s4,1 F s4,2 F s4,3 F s4,5 F s2,3 F s2,4 F s3,3 F s3,4 F s3,5 F j F i F v4,1 F v4,2 F v4,3 F z1 F w2,1 F w2,1 Figure 12.
Case 2 F u b,g ∩ F w c,h = ∅ . Then L ′ = ( F s ai,l , F s ai,l +1 , . . . , F i , F i , . . . , F i e , F w c,h ) is a simple loop. If f i < t a i ,then consider the thick path Z = ( F z , , . . . , F z ,y ) arising while walking in C along ∂C fromthe face F z , intersecting F i ∩ F i by the vertex, to the face F z ,y preceding F w ai +1 , . Considerthe thick path X = ( F v ai, , . . . , F v ai,x ) with x being the first integer with F v ai,x ∩ F i = ∅ .Consider the simple curve η ⊂ ∂P consisting of segments connecting the midpoints of thesuccessive edges of intersection of the successive faces of L ′ . It divides ∂P into two connectedcomponents E and E with J \ ( F s ai,l , . . . , F s ai,fi − ) lying in one connected component, say E ,and Z – in E ∪ F w c,h . Now substitute X for the segment ( F s ai, , . . . , F s ai,fi − ) of J . If f i < t a i substitute Z for the segment ( F s ai,fi +1 , . . . , F s ai,tai ) of I to obtain a new loop ( F i , I , F j , J )with faces in I not intersecting F k . A face F i ′′ a in I can intersect a face F j ′′ b in J only if F i ′′ a = F w c ′ ,h ′ for some c ′ , h ′ , F j = F s aj,fj , and F j ′′ b = F s aj,l for l > f j . The thick path Z lies in E ∪ F w c,h and the segment ( F j = F s aj,fj , . . . , F s aj,taj ) lies in E ; hence intersections of faces in I with faces in J are also intersections of the same faces in I and J , and F w c ′ ,h ′ is either F w c,h , or lies in E . We can apply the same argument for S a j as for S a i to obtain a new loop L = ( F i , I , F j , J ) with faces in I not intersecting F k and faces in J . Then take the shortestthick path from F i to F j in F i ∪ I ∪ F j and the shortest thick path from F j to F i in F j ∪ J ∪ F i to obtain a belt we need. (cid:3) -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 21 F (cid:19) F (cid:20) F u3,3 F u4,1 F u4,2 F w1,1 F w1,2 F w3,1 F w3,2 F w3,3 F w3,4 F w4,1 F w4,2 F u1,2 F u2,1 F u1,1 F v1,1 F v1,1 F v2,1 F v2,2 F v2,3 F v3,1 F v3,2 F v3,3 F v3,4 F v3,5 F s1,1 F s1,2 F s1,3 F s2,1 F s2,2 F s2,5 F s2,6 F s3,1 F s3,2 F s3,6 F s3,7 F s4,1 F s4,2 F s4,3 F s4,5 F s2,3 F s2,4 F s3,3 F s3,4 F s3,5 F j F i F v4,1 F v4,2 F v4,3 F z1 F w2,1 F w2,1 Figure 13.
Modified belta) b)
Figure 14. a) A polytope P ; b) its Schlegel diagramDenote by P a polytope obtained from the cube by cutting off two disjoint nonadjacentorthogonal edges (Fig. 14). It is easy to see, that the polytope P is flag. Remark . It is not difficult to show that there are only two flag 3-polytopes with m = 8: thepolytope P , and the hexagonal prism M × I . Moreover, P has five 4-belts (four trivial and F s F t F v F (cid:21) F i1 F i (cid:22) F (cid:23) F (cid:24)q F i (cid:25) Figure 15.
A polytope P assumed to satisfy the condition of Lemma 5.2one nontrivial), and M × I has nine 4-belts (six trivial and three nontrivial). Hence, both ofthem are B -rigid. Proposition 5.3.
A flag polytope P is an almost Pogorelov polytope or the polytope P if andonly if it satisfies the condition of Lemma 5.2.Proof. Indeed, let a flag polytope P satisfy the condition of Lemma 5.2. Assume that P has anontrivial 4-belt B = ( F s , F t , F u , F v ). If F s is not adjacent to some faces F i ∈ C and F j ∈ C ,where C and C are the connected components of ∂P \ |B | , then we can take F k = F s .Any belt B containing F i and F j and not containing F k should pass through F t and F v , inparticular, each component of |B| \ ( F i ∪ F j ) intersects F k . A contradiction. Thus, any face of B should be adjacent to all the faces either in C , or in C . Without loss of generality assumethat F s is adjacent to all the faces in C . Then C consists of the faces F i , . . . , F i p , where( F t , F i , . . . , F i p , F v ) is a part of the belt around F s . We have p >
2, since B is a nontrivial belt.In particular, F v does not intersect F i , . . . , F i p − , and F t does not intersect F i , . . . , F i p . Then F t and F v are adjacent to all the faces in C , and F u – in C . Since P is flag, F t ∩ F i ∩ F u , F i ∩ F i ∩ F u , . . . , F i p − ∩ F i p ∩ F u , F i p ∩ F v ∩ F u are vertices. Then all the faces F i , . . . , F i p are quadrangles. The same argument works for C . Then P has the structure drawn on Fig.15. Let p >
3. Consider the faces F i = F i , F j = F j , and F k = F i . Any belt B containing F i contains either the part ( F v , F i , F i ), or ( F s , F i , F u ). In the first case this belt also shouldcontain F k = F i . In the second case it intersects F k = F i by both connected components of |B| \ ( F i ∪ F j ). A contradiction. Hence, p = 2. For the same reason q = 2. Thus, P = P .At last, let us prove that P satisfies the condition of Lemma 5.2. Its group of combinatorialsymmetries acts transitively on quadrangles and pentagons. Denote by B the 4-belt of pen-tagons, and by C and C connected components of its complement in ∂P . Each component C and C consists of two quadrangles. -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 23 If F i is a quadrangle adjacent to F k , then any l -belt B l containing F i and not containing F k in the closure of each connected component of |B l | \ ( F i ∪ F j ) contains a face from the 4-beltaround F i adjacent to F k . Hence, F k intersects both connected components.Now let F k do not intersect quadrangles among the faces F i and F j .If F k is a quadrangle in C , then at least one of the faces F i and F j , say F i , should be apentagon. Since F i ∩ F j = ∅ , F j is either the pentagon opposite to F i in B , or a quadranglein C . In the first case, if F i intersects both quadrangles in C , then B is a belt we need. If F i intersects only one quadrangle in C , then the belt around a quadrangle in C is a belt we need.In the second case F i is adjacent to both quadrangles in C and the belt around the quadranglein C different from F j is a belt we need.If F k is a pentagon, then it is adjacent to three of four quadrangles of P . If both faces F i and F j are pentagons, then they are both adjacent to F k , and the belt around the fourth quadrangleis a belt we need. If F i is the fourth quadrangle, then it is adjacent to all the pentagons differentfrom F k . Then F j is a quadrangle, and F i and F j lie in different components C and C . Then F j is a quadrangle adjacent to F k , which is a contradiction. This finishes the proof. (cid:3) Remark . Thus, we have a characterization of the families of flag, almost Pogorelov andPogorelov polytopes it terms of the existence of the belt that passes through F i and F j , where F i ∩ F j = ∅ , and does not pass through F k . It should be mentioned, that there are othercharacterizations of these families in terms of belts. Namely, a simple 3-polytope is flag if andonly if any its face is surrounded by a belt [BE17I, BEMPP17]. It is almost Pogorelov orthe polytope P if and only if in addition the loop around any trivial belt is simple [E19]. Asimple 3-polytope is Pogorelov if and only if any pair of adjacent faces is surrounded by a belt[BE17I, BEMPP17]. Moreover, a simple 3-polytope is strongly Pogorelov if and only if any itsface is surrounded by two belts [B1913].6. Generalization of the annihilator lemma
In this section we generalise another crucial tool from [FMW15] – the annihilator lemma.
Lemma 6.1.
Let P be a simple -polytope and α ∈ Hα = X ω ∈ N ( P ) r ω e ω with |{ ω : r ω = 0 }| > . Then for any ω = { p, q } with r ω = 0 we have (1) dim Ann H ( α ) dim Ann H ( e ω ) ; (2) dim Ann H ( α ) < dim Ann H ( e ω ) , if there is ω ′ = { s, t } with r ω ′ = ∅ admitting an l -belt B l containing F s and F t such that either F p or F q does not belong to B l and does notintersect at least one of the two connected components of |B l | \ ( F s ∪ F t ) .Proof. Choose a complementary subspace C ω to Ann H ( e ω ) in H as a direct sum of complements C ω,τ to Ann H ( e ω ) ∩ b H ∗ ( P τ , ∂P τ , Q ) in b H ∗ ( P τ , ∂P τ , Q ) for all τ ⊂ [ m ] \ ω . Then for any β ∈ C ω \{ } we have β e ω = 0, which is equivalent to the fact that β = P β τ , β τ ∈ C ω,τ , τ ⊂ [ m ] \ ω , with β τ β e ω = 0 for some τ β ⊂ [ m ] \ ω . Moreover for any ω ′ = ω with r ω ′ = 0 and τ ⊂ [ m ] \ ω , τ = τ β , we have τ β ⊔ ω / ∈ { τ ∪ ω ′ , τ β ∪ ω ′ , τ ⊔ ω } ; hence ( β · α ) τ β ⊔ ω = r ω β τ β · e ω = 0, and βα = 0. Then C ω forms a direct sum with Ann( α ), and dim Ann H ( α ) dim Ann H ( e ω ).Now consider some ω ′ = ω , | ω ′ | = 2, r ω ′ = 0, satisfying additional condition of the lemma sayfor F q . Then there is an l -belt B l such that F s , F t ∈ B l , F q / ∈ B l , and F q does not intersect one ofthe two connected components B and B of |B l | \ ( F s ∪ F t ), say B . Take ξ = [ P i : F i ⊂ B F i ] ∈ b H ( P τ , ∂P τ , Q ) , τ = { i : F i ∈ B l \ { F s , F t }} , and [ F s ] ∈ b H ( P ω ′ , ∂P ω ′ , Q ). Then ξ · [ F s ] is agenerator in H ( B l , ∂ B l , Q ) ≃ Q . On the other hand, take [ F q ] ∈ b H ( P ω , ∂P ω , Q ). Then either F p ∈ B l \ { F s , F t } , and ξ · e ω = 0, since τ ∩ ω = ∅ , or F p / ∈ B l \ { F s , F t } , and ± ξ · e ω = ξ · [ F q ] = 0,since F q does not intersect B . In both cases ξ ∈ Ann( e ω ) and ξ · e ω ′ = 0. Then ξ · α = 0,since τ ⊔ ω ′ = τ ⊔ ω for ω = ω ′ . Consider any β = P τ ⊂ [ m ] \ ω β τ ∈ C ω \ { } . We have( β · α ) τ β ⊔ ω = 0. If ( ξ · α ) τ β ⊔ ω = 0, then since ξ is a homogeneous element, ( ξ · α ) τ β ⊔ ω = r ω ξ · f ω for ω = ( τ β ⊔ ω ) \ τ = { q, r } , r ∈ [ m ]. We have ξ · f ω = ± ξ · [ F q ] = 0, since F q does not intersect B . A contradiction. Thus, (( ξ + β ) · α ) τ β ⊔ ω = ( β · α ) τ β ⊔ ω = 0; hence ( ξ + β ) · α = 0, and thespace h ξ i ⊕ C ω forms a direct sum with Ann H ( α ). This finishes the proof. (cid:3) Remark . In Proposition 8.3 we show that for the 3-dimensional associahedron, which is thesimplest almost Pogorelov polytope without adjacent quadrangles, there is an example whendim Ann H ( α ) = dim Ann H ( e ω ). Corollary 6.2.
Ley P be a simple -polytope, and ω ∈ N ( P ) such that dim Ann H ( α ) < dim Ann H ( e ω ) for any α = λ ω ω + P ω ′ ∈ N ( P ) \{ ω } λ ω ′ e ω ′ with λ ω = 0 and λ ω ′ = 0 at least for one ω ′ , then forany isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) for a simple -polytope Q we have ϕ ( e ω ) = ± e ω ′ for some ω ′ ∈ N ( Q ) .Proof. Indeed, we have ϕ ( e ω ) = X ω ′′ ∈ N ( Q ) λ ω ′′ f ω ′′ . Then at least for one ω ′′ with λ ω ′′ = 0 we have ϕ − ( ω ′′ ) = µ ω e ω + X ω ′ ∈ N ( P ) \{ ω } µ ω ′ e ω ′ with µ ω = 0 . If µ ω ′ = 0, then by Lemma 6.1 we have in cohomology over Q :dim Ann H ( P ) ( e ω ) > dim Ann H ( P ) ( ϕ − ( f ω ′′ )) = dim Ann H ( Q ) ( f ω ′′ ) > dim Ann H ( Q ) ( ϕ ( e ω )) . A contradiction. Hence µ ω ′ = 0 for all ω ′ , and ϕ ( e ω ) = ± f ω ′′ . (cid:3) Definition 6.3.
Let P be a simple 3-polytope. We will call an element ω ′ = { s, t } ∈ N ( P ) good for an element ω = { p, q } ∈ N ( P ), if theres is an l -belt B l containing F s and F t such thateither F p or F q does not belong to B l and does not intersect at least one of the two connectedcomponents of B l \ { F s , F t } . Otherwise, we call ω ′ bad for ω . -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 25 Corollary 6.4.
Let P be a simple -polytope. If any ω ′ ∈ N ( P ) \ { ω } is good for an element ω ∈ N ( P ) , then for any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) for a simple -polytope Q we have ϕ ( e ω ) = ± e ω ′ for some ω ′ ∈ N ( Q ) . Corollary 6.5.
Let P be a simple -polytope and ω ∈ N ( P ) . Then for any isomorphism ofgraded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) for a simple -polytope Q we have ϕ ( e ω ) = P ω ′′ ∈ N ( Q ) λ ω ′′ f ω ′′ ,where there is ω ′′ ∈ N ( Q ) such that λ ω ′′ = 0 and all the other elements in N ( Q ) with nonzerocoefficients are bad for ω ′′ . Moreover, this is valid for any ω ′′ ∈ N ( Q ) such that λ ω ′′ = 0 and ϕ − ( f ω ′′ ) has a nonzero coefficient at e ω .Proof. Indeed, let ϕ ( e ω ) = X ω ′′ ∈ N ( Q ) λ ω ′′ f ω ′′ . Then at least for one ω ′′ with λ ω ′′ = 0 we have ϕ − ( ω ′′ ) = µ ω e ω + X ω ′ ∈ N ( P ) \{ ω } µ ω ′ e ω ′ with µ ω = 0 . If λ ω ′′ = 0, and ω ′′ is good for ω ′′ , then by Lemma 6.1 we have in cohomology over Q :dim Ann H ( P ) ( e ω ) > dim Ann H ( P ) ( ϕ − ( f ω ′′ )) = dim Ann H ( Q ) ( f ω ′′ ) > dim Ann H ( Q ) ( ϕ ( e ω )) . A contradiction. Hence λ ω ′′ = 0 for all elements ω ′′ ∈ N ( Q ), which are good for ω ′′ . Moreover,by a similar argument, µ ω ′ = 0 for all elements, which are good for ω . (cid:3) Corollary 6.6.
Let P be an almost Pogorelov polytope different from the cube and the pentag-onal prism, and ω = { p, q } ∈ N ( P ) . Then an element ω ′ = { s, t } ∈ N ( P ) \ { ω } is bad for ω if and only if at least one of the faces F s and F t is a quadrangle, and either F p and F q areopposite faces of the -belt around this face, or exactly one of the faces F p and F q belongs tothis belt, and the other face among F s and F t is either the other face among F p and F q , or is aquadrangle adjacent to it. In particular, (1) If F p and F q are not adjacent to quadrangles, then any element ω ′ ∈ N ( P ) \ { ω } is goodfor ω . In particular, this is valid, if both F p and F q are quadrangles. (2) If F p is a quadrangle and F q is adjacent to quadrangles, then ω ′ ∈ N ( P ) \ { ω ′ } is badfor ω if and only if ω ′ = { p, r } , where F r is a quadrangle adjacent to F q . (3) If each of the faces F p and F q is adjacent to quadrangles, and F p and F q do not belongto any -belt, then ω ′ = { s, t } ∈ N ( P ) \ { ω } is bad for ω if and only if either one ofthe faces F s and F t is F p or F q and the other face is a quadrangle adjacent to the otherface, or both F s and F t are quadrangles, and one of them is adjacent to exactly F p , andthe other – to exactly F q . (4) If the faces F p and F q belong to a -belt around the face F r , then ω ′ = { s, t } ∈ N ( P ) \{ ω } is bad for ω if and only if either ω ′ = { r, s } , or one of the faces F s and F t is F p or F q and the other face is a quadrangle adjacent to the other face, or both F s and F t arequadrangles, and one of them is adjacent to exactly F p , and the other – to exactly F q . Proof.
Without loss of generality assume that q / ∈ ω ′ . If ω ′ is bad for ω , then by Lemma 5.2one of the faces F s and F t , say F s , is a quadrangle adjacent to F q . Then F p = F s . If F p = F t ,then either F p is adjacent to F s , or F t is a quadrangle adjacent to F p .On the other hand, let F s be a quadrangle adjacent to F q . Then F p = F s . If either F p = F t ,or F p = F t and F p is adjacent to F s , or F t is a quadrangle adjacent to F p , then ω ′ is bad for ω . (cid:3) Corollary 6.7.
Let P be an almost Pogorelov polytope different from the cube and the pentag-onal prism, and ω = { p, q } ∈ N ( P ) such that F p and F q are not adjacent to quadrangles. Thenfor any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) for a simple -polytope Q we have ϕ ( e ω ) = ± e ω ′ for some ω ′ ∈ N ( Q ) Proof.
This follows directly from Corollary 6.4. (cid:3)
Corollary 6.8.
Let P be an almost Pogorelov polytope different from the cube and the pen-tagonal prism, and B k be a k -belt such that the set |B k | does not have common points withquadrangles. Then for any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) for a simple -polytope Q we have ϕ ( f B k ) = ± f B ′ k for some k -belt B ′ of Q .Proof. We know from Section 2 that ϕ ( f B k ) = P j µ j g B ′ k,j for k -belts B ′ k,j of Q . Lemma 6.9.
Let P be a simple -polytope and ω = { p, q } ∈ N ( P ) . Then an element x = X η α η + X τ β τ ∈ M η e H ( P η ) M τ e H ( P τ ) is divisible by e ω if and only if α η = 0 for all η , and each nonzero β τ is divisible by e ω (inparticular, ω ⊂ τ and e H ( P τ \ ω ) = 0 ).Proof. If x is divisible by e ω , then x = e ω P ζ ξ ζ for ξ ζ ∈ e H ( P ζ ). Then x = P ω ∩ ζ = ∅ e ωξ ζ , where e ωξ ζ ∈ e H ( P ω ⊔ ζ ), and ω ⊔ ζ = ω ⊔ ζ ′ for ζ = ζ ′ . Thus, each nonzero summand β τ has the form e ωξ ζ , in particular, ω ⊂ τ , and e H ( P τ \ ω ) = e H ( P ζ ) = 0. (cid:3) Corollary 6.10.
Let P be a simple -polytope and ω = { p, q } ∈ N ( P ) . Then an element x = P j µ j g B k,j ∈ B k is divisible by e ω if and only if each g B k,j with µ j = 0 is divisible by e ω , andif and only if each belt B k,j with µ j = 0 contains F p and F q .Proof. This follows directly from Lemma 6.9, since the element g B k,j is divisible by e ω if and onlyif ω ⊂ ω ( B k,j ), that is B k,j contains F p and F q . (cid:3) Take any ω = { p, q } ∈ N ( P ), ω ⊂ ω ( B k ). Since both faces F p and F q do not intersectquadrangles, Corollary 6.7 implies that ϕ ( e ω ) = ± e ω ′ for some ω ′ ∈ N ( Q ). Then ω ′ ⊂ ω ( B ′ k,j )for each µ j = 0. Thus, the isomorphism ϕ maps the set {± e ω : ω ⊂ ω ( B k ) } bijectively to thecorresponding set of any B ′ k,j with µ j = 0. But such a set defines uniquely the k -belt; hence wehave only one nonzero µ j , which should be equal to ±
1. This finishes the proof. (cid:3) -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 27 B -rigid subsets for almost Pogorelov polytopes In this section we study B -rigid subsets for almost Pogorelov polytopes. Lemma 7.1.
The set of subgroups G ( B ) ⊂ H of rank generated by elements [ e ω ] = e ω + A and [ e ω ′ ] = e ω ′ + A , where ω ⊔ ω ′ = ω ( B ) is B -rigid in the class of flag -polytopes with B = rk H .Proof. The elements { [ e ω ] } , where ω ∈ N ( P ) are subsets of ω ( B ) for 4-belts, form a basis in H . Take an element α = λ [ f ω ] + λ ′ [ f ω ′ ] + · · · + λ k [ f ω k ] + λ ′ k [ f ω ′ k ] , where B , ,. . . , B ,k are all 4-belts, and ω i ⊔ ω ′ i = ω ( B ,i ). We have [ e ω i ] · [ e ω ′ i ] = g B ,i .As mentioned above we have an embedding H ∗ ( Z P ) ⊂ H ∗ ( Z P ) ⊗ Q = H ∗ ( Z P , Q ). Images ofthese subgroups are 2-dimensional linear subspaces in H ⊗ Q . For the element α ∈ H ⊗ Q define a linear subspace A ( α ) = { β ∈ H ⊗ Q : α · β = 0 } We have β = µ [ f ω ] + µ ′ [ f ω ′ ] + · · · + µ k [ f ω k ] + µ ′ k [ f ω ′ k ] , and α · β = k X i =1 ( λ i µ ′ i − λ ′ i µ i ) g B ,i . Then β ∈ A ( α ) if and only if λ i µ ′ i − λ ′ i µ i = 0 for all i = 1 , . . . , k . Thus, dim A ( α ) = dim H ⊗ Q − l , where l is the number of i with ( λ i , λ ′ i ) = (0 , α ∈ h [ e ω i ] , [ e ω ′ i ] i for some i . Therefore, the collection of these subspaces is B -rigid over Q , andthe collection of the corresponding subgroups is B -rigid. (cid:3) Corollary 7.2.
The set of elements {± f B : B is a -belt } ⊂ H ( Z P ) is B -rigid in the class of flag -polytopes with B = rk H .Proof. Indeed, g B ,i is a generator of the image of the bilinear mapping h [ e ω i ] , [ e ω ′ i ] i × h [ e ω i ] , [ e ω ′ i ] i → H ( Z P ) . (cid:3) Corollary 7.3.
Let
P, Q ∈ P aP og \ { I , M × I } . Then any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) induces a bijection ϕ between the sets of quadrangles of P and Q by the rule ϕ ( F i ) = F ′ i ′ , where ϕ ( e B i ) = ± f B ′ i ′ for -belts B i and B ′ i ′ around F i and F ′ i ′ .Proof. Indeed, any 4-belt around a quadrangle does not surround a quadrangle on the otherside, for otherwise P = I . (cid:3) Notation 7.4.
In what follows we will use notations from Corollary 7.3, namely ϕ for theisomorphism between the sets of quadrangles of polytopes P, Q ∈ P aP og \ { I , M × I } inducedby an isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ), and B i and B ′ i ′ for the 4-belts aroundquadrangles F i and F ′ i ′ such that ϕ ( F i ) = F ′ i ′ . Lemma 7.5.
The set of elements: {± [ e ω ] : ω ∈ N ( P ) , ω ⊂ ω ( B ) for some -belt B } ⊂ H is B -rigid in the class of flag -polytopes with B = rk H .Proof. Consider the 4-belt B ,i corresponding to the pair ω i , ω ′ i ∈ N ( P ). Let ω i = { p, q } , ω ′ i = { s, t } . Consider the belt B l around the face F p . It contains both F s and F t . Let B and B be the connected components of B l \ { F s , F t } . They lie in different connected componentsof ∂P \ {B ,i } . Let F q intersect some face F a ∈ B and some face F b ∈ B . Then ( F p , F a , F q , F b )is a 4-belt. It is different from B ,i and contains F p and F q . A contradiction to Lemma 3.1.Now let ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) be an isomorphism of graded rings for flag polytopes P and Q with 2 rk B = rk H . By Lemma 7.1 ϕ ( e ω i ) = λ j e ω j + λ ′ j e ω ′ j + X e ω ∈ A ( Q ) µ ω e ω for some j ∈ { , . . . , k } . Since ϕ − ( A ( Q )) ⊂ A ( P ), for at least one of the elements e ω j and e ω ′ j with nonzero coefficient, say for e ω j , its preimage contains e ω i with a nonzero coefficient: ϕ − ( e ω j ) = τ i e ω i + τ ′ i e ω ′ i + X e ω ∈ A ( P ) η ω e ω, τ i = 0 . If τ ′ i = 0, then by Lemma 6.1 we have in cohomology over Q :dim Ann H ( P ) ( e ω i ) > dim Ann H ( P ) ( ϕ − ( e ω j )) = dim Ann H ( Q ) ( e ω j ) > dim Ann H ( Q ) ( ϕ ( e ω i )) . A contradiction. Hence τ ′ i = 0 and for the mapping b ϕ : H ( P ) → H ( Q ) we have: b ϕ − ([ e ω j ]) = τ i [ e ω i ]. Then τ i = ± b ϕ ([ e ω i ]) = ± [ e ω j ]. This finishes the proof. (cid:3) Lemma 7.6.
The sets of elements: {± [ e ω ] : ω ∈ N ( P ) , ω ⊂ ω ( B ) for a trivial -belt B } ⊂ H and {± f B : B is a trivial -belt } ⊂ H ( Z P ) are B -rigid in the class of flag -polytopes with B = rk H .Proof. Indeed, any generator of e H ( P ω ) = Z for P ω from Fig. 8 a)-c) is divisible by exactly e ω and e ω ′ for the corresponding belt B with e ω · e ω ′ = f B among all the elements in H ( Z P )corresponding to sets in N ( P ). Therefore, the set ω ∈ N ( P ) \ N ( P ) corresponds to a trivial 4-belt if and only if the free abelian subgroup [ e ω ] · H ( Z P ) ⊂ I has rank ( m − f B corresponds to a trivial belt if and only if it is divisible by such an element [ e ω ] ∈ H . (cid:3) -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 29 Corollary 7.7.
Let
P, Q ∈ P aP og \ { I , M × I } , and let B k be a k -belt passing through aquadrangle F i and its adjacent faces F p and F q . Then for any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) we have ϕ ( f B k ) = X j µ j g B ′ k,j for k -belts B ′ k,j of Q such that for any µ j = 0 the belt g B ′ k,j passes through the nonadjacent faces F ′ p ′ and F ′ q ′ of Q , where ϕ ( ^ { p, q } + A ( P )) = ± ^ { p ′ , q ′ } + A ( Q ) . Moreover, F ′ p ′ and F ′ q ′ areadjacent to the quadrangle F ′ i ′ = ϕ ( F i ) of Q (see Notation 7.4).Proof. We know from Section 2 that ϕ ( f B k ) = P j µ j g B ′ k,j for k -belts B ′ k,j of Q . Lemma 7.8.
Let P ∈ P aP og \ { I , M × I } , and let B k be a k -belt passing through a quadrangle F r and its adjacent faces F p an F q . Let x be a generator of e H ( P τ ) = Z for τ = ω ( B k ) (in thiscase x = ± f B k ), or τ = ω ( B k ) ⊔{ r } , where F r either is not adjacent to faces in B k , or is adjacentto exactly one face in B k . Then x is divisible by any element in the coset ^ { p, q } + A ( P ) .Proof. Indeed, consider the set P τ \{ p,q } . One of its connected components is F i , since F i can notbe adjacent to F r (for otherwise, F r is also adjacent to both F p and F q , which is a contradiction).Then for [ F i ] ∈ b H ( P τ \{ p,q } , ∂P τ \{ p,q } ) and [ F p ] ∈ b H ( P { p,q } , ∂P { p,q } ) we have [ F i ∩ F p ] is agenerator of H ( P τ , ∂P τ ) ≃ e H ( P τ ). On the other hand, for any { s, t } ∈ N ( P ) at least oneof the faces F s and F t does not belong to the belt around F i , say F s . Then either F s = F i ,or F s ∩ F i = ∅ . Therefore ξ · ] { s, t } = 0, where ξ is the element in H ∗ ( Z P ) corresponding to[ F i ] ∈ b H ( P τ \{ p,q } , ∂P τ \{ p,q } ). Thus, ± ξ · ( ^ { p, q } + P ω ∈ N ( P ) λ ω e ω ) = x. (cid:3) Then ϕ ( f B k ) is divisible by any element in the coset ± ^ { p ′ , q ′ } + A ( Q ), in particular, by ^ { p ′ , q ′ } .By Corollary 6.10 each belt B ′ k,j passes through F ′ p ′ and F ′ q ′ . The second part of the Corollaryfollows from the fact that f B ′ i ′ is also divisible by any element in the coset ± ^ { p ′ , q ′ } + A ( Q ) andpasses through F ′ p ′ and F ′ q ′ . (cid:3) Proposition 7.9.
Let
P, Q ∈ P aP og \ { I , M × I } , and let F p be a quadrangle of P not adjacentto a face F q . Assume that for an isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) we have ϕ ( ^ { p, q } ) = ± ] { s, t } for some { s, t } ∈ N ( Q ) . Then p ′ ∈ { s, t } for F ′ p ′ = ϕ ( F p ) (see Notation7.4). In particular, ϕ ( ^ { p, q } ) = ± ^ { p ′ , q ′ } for quadrangles F p and F q , and the set {± ^ { p, q } : F p and F q are quadrangles } ⊂ H ( Z P ) is B -rigid in the class P aP og \ { I , M × I } .Proof. We start with technical results.
Lemma 7.10.
Let F i and F j be two adjacent faces of a polytope P aP og \{ I , M × I } , and F u , F v be the faces intersecting F i ∩ F j by vertices. Then the loop L = ( F u , F i , . . . , F i k , F v , F j , . . . , F j r ) around F i and F j , where F i l ∩ F i are successive edges of F i , and F j l ∩ F i are successive edgesof F j is simple. It is a belt if and only if both F u and F v are not quadrangles. Moreover, if wedelete from L the quadrangles among F u and F v , we obtain a belt.Proof. Any face of a flag polytope is surrounded by a belt (see, for example, [BE17I, Proposition2.9.2.]). Hence, B = ( F u , F i , . . . , F i k , F v , F j ) and B = ( F v , F j , . . . , F j r , F u , F i ) are belts. Then L is not simple if and only if F i p = F j q for some p , q . But then ( F i , F j , F i p ) is a 3-belt, which isa contradiction. Thus, L is a simple loop. If it is not a belt, then two non-successive faces areadjacent. This can be possible only if they do not belong to B and B , that is one face is F i p ,and the other is F j q . Then ( F i , F j , F i p , F j q ) is a 4-belt. It should surround a quadrangle adjacentto both F i and F j . It can be only F u and F v . If both of these faces are not quadrangles, then L is a belt. Else delete these faces to obtain a new simple loop L ′ . If two non-successive faces ofthis loop are adjacent, they have the form F i p and F j q . By the previous argument these facesshould be adjacent to a quadrangle F u or F v . Then ( i p , j q ) ∈ { ( i k , j ) , ( i , j r ) } and F i p and F j q are successive in L ′ . A contradiction. (cid:3) Lemma 7.11.
Let P ∈ P aP og \ { I , M × I } . Then for any three pairwise different faces { F i , F j , F k } such that F i ∩ F j = ∅ , F k is a quadrangle, and at least one of the faces F i and F j is not adjacent to F k , there exists an l -belt ( l > ) B l such that F i , F j ∈ B l , F k / ∈ B l , and B l does not contain any of the two pairs of opposite faces of the -belt around F k .Proof. Let ( F i , F i , F i , F i ) be the 4-belt around F k , and F j , F j , F j , F j be the faces inter-secting the edges F i l ∩ F i l +1 by vertices different from F i l ∩ F i l +1 ∩ F k (for convenience we consider l ∈ Z = Z / Z for i l and j l ). By Lemma 7.10 for any l we have F j l = F j l +1 and F j l = F i l +2 ,since F i l +1 and F i l +2 are not quadrangles. Since there are at most one face among F i and F j inthe 4-belt around F k , there are at least two pairs ( F i l , F i l +1 ) not containing both F i and F j . Forthem there is at least one F j l / ∈ { F i , F j } . Thus, we obtain a triple { F i l , F i l +1 , F j l } not containingfaces in { F i , F j } . By Lemma 7.10 there is a belt corresponding to ( F i l , F i l +1 ). It contains F i l +2 , F i l +3 , and contains F j l if and only if it is not a quadrangle. Connect the midpoints of edges ofintersection of successive faces of this belt to obtain a piecewise linear closed cycle containing inone component of the complement F i l and F i l +1 . Substitute a disk for this component to obtaina spherical graph. By [BE17I, Lemma 5.1.1] (also [BE17S, Lemma 3.11]) this graph correspondsto a flag simple polytope. By the analog of SCC for flag polytopes (see Section 2) there is a beltin this polytope, which contains the faces arisen from F i and F j and does not contain the facecorresponding to the disk. In P this belt corresponds to a belt B l we need, since subdivision ofthe disk does not add or remove edges of intersection of faces of the belt. The belt B l contains F i and F j , but does not contain F i l and F i l +1 . This finishes the proof. (cid:3) Now we are ready to proof Proposition 7.9. Let p ′ / ∈ { s, t } . The faces F ′ s and F ′ t are notadjacent simultaneously to any quadrangle, for otherwise { s, t } / ∈ N ( Q ), { p, q } / ∈ N ( P ), and F p is adjacent to a quadrangle. In particular, they are not adjacent simultaneously to F ′ p ′ . ByLemma 7.11 there is a belt B ′ l in Q , which contains F ′ s and F ′ t , and does not contain any of the -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 31 pairs of opposite faces of the belt B p ′ around F ′ p ′ . Consider the preimage ϕ − ( e B ′ l ) = X s λ s f B l,s . By Corollary 6.10 any of the belts with λ s = 0 contains F p and F q . In particular, it passesthrough a pair of opposite faces in the belt B p around F p . By Corollary 7.7 e B ′ l = ϕ X s λ s f B l,s ! = X s λ s ϕ ( f B l,s ) = X s λ s X j µ s,j f B ′ l,j = X j X s λ s µ s,j ! f B ′ l,j where each belt B ′ l,j , such that λ s = 0 = µ s,j for some s , contains a pair of opposite faces in thebelt B ′ p . But B ′ l is equal to one of this belts, hence also contains a pair of opposite faces of thebelt B p ′ around F ′ p ′ . A contradiction. Thus, p ′ ∈ { s, t } .If F p and F q are both quadrangles, then Corollary 6.7 implies that ϕ ( ^ { p, q } ) = ± ] { s, t } forsome { s, t } ∈ N ( Q ). From the above argument q ′ ∈ { s, t } for F ′ q ′ = ϕ ( F q ). Thus, { s, t } = { p ′ , q ′ } . This finishes the proof. (cid:3) Lemma 7.12.
Let
P, Q ∈ P aP og \ { I , M × I } , and let B k be a (2 k ) -belt of P containing k quadrangles. Then for any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) we have ϕ ( f B k ) = ± f B ′ k for a (2 k ) -belt B ′ k in Q containing k quadrangles. In particular, the set ofelements {± f B k : B k − a (2 k ) -belt containing k quadrangles } ⊂ H k +2 ( Z P ) is B -rigid in the class P aP og \ { I , M × I } . Moreover, ϕ (see Notation 7.4) induces a bijectionbetween quadrangles of B k and quadrangles of B ′ k , and for any two faces of F p and F q of B k adjacent to a quadrangle F i in B k we have ϕ ( ^ { p, q } + A ( P )) = ± ^ { p ′ , q ′ } + A ( Q ) , where the faces F ′ p ′ and F ′ q ′ are adjacent to F ′ i ′ = ϕ ( F i ) .Proof. Since P has no adjacent quadrangles, B k = ( F i , F j , F i , F j , . . . , F i k , F j k ), where F i , . . . , F i k are quadrangles, and F j , . . . , F j k are not. Also k >
3, since each 4-belt surrounds aquadrangle and for this reason can not contain quadrangles.We have ϕ ( f B k ) = X s µ s ] B ′ k,s , where by Corollary 6.10 and Proposition 7.9 for each µ s = 0 the belt B ′ k,s contains all the pairsof quadrangles { F ′ i ′ a = ϕ ( F i a ) , F ′ i ′ b = ϕ ( F i b ) } , a = b . There are k ( k − such pairs. The belt B ′ k,s may contain at most this number of pairs of quadrangles, and it is achieved if and only if B ′ k,s contains exactly k quadrangles, and B ′ k,s = ( F ′ u , F ′ v , F ′ u , F ′ v , . . . , F ′ u k , F ′ v k ) for some quadrangles F ′ u , . . . , F ′ u k . Then ϕ induces a bijection between the sets { F i , . . . , F i k } and { F ′ u , . . . , F ′ u k } : F i a → F ′ u a ′ = ϕ ( F i a ). At this moment we do not know if this bijectionpreserves the cyclic order.For each pair { j a − , j a } (where we assume that a, a − ∈ Z k = Z /k Z ) we have ϕ ( ^ { j a − , j a } + A ( P )) = ± ^ { z, w } + A ( Q ) for faces F z and F w adjacent to F u a ′ . Moreover, by Corollary 7.7the belt B ′ k,s passes through F z and F w . Thus, { z, w } = { v a ′ − , v a ′ } . Since the belt B ′ k,s iscovered by segments ( F ′ v a ′− , F ′ u a ′ , F ′ v a ′ ), there is only one belt B ′ k,s with µ s = 0, and ϕ ( f B k ) = ± ] B ′ k,s . (cid:3) Lemma 7.13.
Let
P, Q ∈ P aP og \ { I , M × I } , and let B k be a trivial (2 k ) -belt of P containing k quadrangles. Then for any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) we have ϕ ( f B k ) = ± f B ′ k for a trivial (2 k ) -belt B ′ k in Q containing k quadrangles. In particular, the setof elements {± f B k : B k − a trivial (2 k ) -belt containing k quadrangles } ⊂ H k +2 ( Z P ) is B -rigid in the class P aP og \ { I , M × I } .Proof. As before, we see that k >
3, since each 4-belt surrounds a quadrangle and for thisreason can not contain quadrangles.By Lemma 7.13 we have ϕ ( f B k ) = ± f B ′ k for a (2 k )-belt B ′ k in Q containing k quadrangles.We will use notations from the proof of this lemma.Assume that the belt B k surrounds a face F i , and the belt B ′ k is not trivial.Consider a subgroup G ( B k ) ⊂ H k +3 ( Z P ) consisting of all the elements divisible by ^ { i a , i b } for all a = b , and by each member of a coset ^ { j a − , j a } + A ( P ) for all a . We have 2 k + 3 m ,since outside the belt there are at least two faces, for otherwise P should be a prism and haveadjacent quadrangles.Since ω ( B k ) = S a = b { i a , i b } S a { j a − , j a } Lemma 6.9 implies that any element in G ( B k ) hasthe form X τ ⊃ ω ( B l ) β τ , where β τ ∈ e H ( P τ ) , | τ | = l + 1 , and each nonzero β τ is divisible by all ^ { i a , i b } and ^ { j a − , j a } . Then τ = ω ( B k ) ⊔ { r } . We have F r = F i , for otherwise P τ is contractible. Let F r be adjacent to two non-successive faces of B k , say F p and F q . Then ( F r , F p , F i , F q ) is a 4-belt surrounding a quadrangle F i a in B k , and { F p , F q } = { F j a − , F j a } . There are k such quadrangles. If F r is adjacent to some face in B k different from F j a − , F i a , and F j a , then this face is not successive with F i a , and by the previousargument F i a belongs a 4-belt around a quadrangle, which is a contradiction. Thus, F r is notadjacent to other faces in B k . There are k faces of this type. For them P τ \{ i b } is connected forany i b = i a , hence β τ is not divisible by ^ { i a , i b } . If F r is adjacent to two successive faces of B k ,then one of this faces is a quadrangle, and F r is of previous type. If F r is adjacent to exactlyone face of B k , then this face is not a quadrangle, and P τ is homeomorphic to |B k | . If F r is not -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 33 adjacent to faces in B k , then P τ is a disjoint union of |B k | and F r . In both cases e H ( P τ ) = Z ,and its generator is divisible by all the elements ^ { i a , i b } and ^ { j a − , j a } . Moreover, by Lemma7.8 this generator is also divisible by any element in the coset ^ { j a − , j a } + A ( P ). Thus, G ( B k )is a free abelian group of rank m − k − − k = m − k − B ′ k . Since ϕ ( ^ { i a , i b } ) = ± ^ { u a ′ , u b ′ } for all a = b , and ϕ ( ^ { j a − , j a } + A ( P )) = ± ^ { v a ′ − , v a ′ } + A ( Q ) for all a , we have ϕ ( G ( B k )) = G ( B ′ k ). By the above argumenteach element in G ( B ′ k ) has the form X τ ⊃ ω ( B ′ k ) β ′ τ , where β ′ τ ∈ e H ( Q τ ) , | τ | = l + 1 , and each nonzero β ′ τ is divisible by all ^ { u a , u b } and ^ { v a − , v a } . Also τ = ω ( B ′ k ) ⊔ r . Consider aface F ′ r of Q . Let F ′ r be adjacent to two non-successive faces of B ′ k , say F ′ p and F ′ q . Since B ′ k is anontrivial belt and Q is a flag polytope, F ′ r is not adjacent to some face of B ′ k . We can assumethat for the segment T = ( F ′ p , F ′ w , . . . , F ′ w t , F ′ q ) of B ′ k the face F ′ r is not adjacent to F ′ w , . . . , F ′ w t . Then F ′ p and F ′ q are not quadrangles, for otherwise either F ′ w or F ′ w t is adjacent to F ′ r .Hence, B ′ t +3 = ( F ′ r , F ′ p , F ′ w , . . . , F ′ w t , F ′ q ) is a ( t + 3)-belt, where 4 t + 3 k , and t + 3 = 2 k only if F ′ p and F ′ q are faces adjacent to one face F ′ v in B ′ k . This face F ′ v is a quadrangle, since F ′ p and F ′ q are not quadrangles. Then F ′ v ∩ F ′ r = ∅ , for otherwise ( F ′ v , F ′ p , F ′ r , F ′ q ) is a 4-belt arounda quadrangle adjacent to the quadrangle F ′ v . Thus, if t + 3 = 2 k , then F ′ r is adjacent only tothe three faces F ′ v a − , F ′ u a , and F ′ v a in B ′ k for some a .Now assume that t + 3 < k . If t + 3 = 4, then B ′ t +3 surrounds a quadrangle adjacent tothree successive faces of B ′ k , which is a contradiction. Thus, t + 3 >
5. Moreover, t + 3 > F ′ p or F ′ q is a quadrangle. In particular, there are at leasttwo quadrangles in T . Consider the preimage ϕ − ( g B ′ t +3 ) = X s λ s ^ B t +3 ,s Each belt B t +3 ,s is divisible by any ^ { i a , i b } for { u a ′ , u b ′ } ⊂ ω ( B ′ t +3 ). Corollary 7.7 implies thatit is also divisible by any ^ { j a − , j a } with { v a ′ − , v a ′ } ⊂ ω ( B ′ t +3 ) (since in this case F u a ′ ∈ T ).In particular, it contains F j a − , F i a , F j a , F j b − , F i b , and F j b . The mapping ϕ − sends the set ofquadrangles in T injectively to some set of quadrangles in B t +3 ,s ∩ B k . Moreover, the image ofeach quadrangle in contained in B t +3 ,s ∩B k with two adjacent faces. Denote by M ⊂ B t +3 ,s ∩B k the union of the images of quadrangles and the faces in B t +3 ,s ∩ B k adjacent to this images.The set M is a disjoint union of segments of B k . Enumerate them in the order they appearin the belt. If we identify the last face of each segment with the first face of the next segment(both of them are not quadrangles), we obtain one segment consisting of | T | = t + 2 faces. If M consists of more than one connected component, then it has more than t + 2 faces. But M hasat most t + 3 faces, since M ⊂ B t +3 ,s . Therefore, M = B t +3 ,s ⊂ B k . This is impossible, since t + 3 < k . Thus, M is a segment of B k consisting of t + 2 faces. Then the ( t + 3)-th face in B t +3 ,s is some face F x / ∈ B k . Since B t +3 ,s is a belt, F x is different from F i . Let F j a and F j b be thefirst and the last faces of M (we know that these faces are not quadrangles, in particular, theyare not adjacent). Then ( F j a , F i , F j b , F x ) is a 4-belt surrounding some quadrangle F i c ∈ B k .Then { F j a , F j b } = { F j c − , F j c } , and F x is adjacent to F i c . Therefore M = B k \ { F i c } , and t + 2 = 2 k −
1, which is a contradiction. Hence, the case t + 3 < k is impossible.Thus, if the face F ′ r is adjacent in B k ′ to non-successive faces, then it is adjacent exactlyto some quadrangle F ′ u a and its adjacent faces F ′ v a − , F ′ v a . There are 2 k faces of this type. If F ′ r is adjacent to two successive faces, then one of them is a quadrangle, and we come to theprevious case. Also F ′ r can be adjacent to exactly one face F ′ v a , which is not a quadrangle, orto no faces in B ′ k . In each of these cases e H ( Q τ ) = Z . In the first case the generator is notdivisible by ^ { u a , u b } for any b = a , since Q τ \{ u a ,u b } is connected. In the second and the thirdcases the generator is divisible by all the elements ^ { u a , u b } and ^ { v a − , v a } . Moreover, by Lemma7.8 it is also divisible by any element in the coset ^ { v a − , v a } + A ( Q ). Thus, G ( B ′ k ) is a freeabelian group of rank m − k − k = m − k . Since m − k < m − k − k > B ′ k is a trivial belt. This finishes the proof. (cid:3) Lemma 7.14.
Let P ∈ P aP og \ { I , M × I } , F i be its quadrangle, and let B k be the (2 k ) -belt containing k quadrangles and surrounding a face F j . Then F i and F j are adjacent if andonly if the element f B k is divisible by any element in the coset ^ { p, q } + A ( P ) for one of thetwo pairs of opposite faces { F p , F q } of the -belt B i around F i . In particular, if F i and F j are adjacent, then for any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) for a polytope Q ∈ P aP og \ { I , M × I } we have ϕ ( f B k ) = ± f B ′ k for a (2 k ) -belt B ′ k containing k quadranglesand surrounding a face F ′ j ′ adjacent to F ′ i ′ = ϕ ( F i ) .Proof. If F j is adjacent to F i , then Lemma 7.8 implies that f B is divisible by any element inthe coset ^ { p, q } + A ( P ) for faces F p , F q adjacent to F i and lying in B k . On the other hand, ifthe element f B k is divisible by any element in the coset ^ { p, q } + A ( P ) for faces F p , F q adjacentto F i , then it is divisible by ^ { p, q } . Hence, B k contains F p and F q . If it does not contain F i ,then ( F i , F p , F j , F q ) is a 4-belt around some quadrangle adjacent to F i . A contradiction.By Lemma 7.13 we have ϕ ( f B k ) = ± f B ′ k for a belt B ′ k containing k quadrangles and surround-ing a unique face F ′ j ′ . The element f B ′ k is divisible by any element in the coset ^ { p ′ , q ′ } + A ( Q )for faces F ′ p ′ , F ′ q ′ adjacent to F ′ i ′ . Hence, the face F ′ j ′ is adjacent to F ′ i ′ . (cid:3) Lemma 7.15.
Let P ∈ P aP og \{ I , M × I } , and let B k be the (2 k ) -belt containing k quadranglesand surrounding a face F i , and B l be the (2 l ) -belt containing l quadrangles and surrounding aface F j . Then F i and F j are adjacent if and only if the elements f B k and f B l have exactly onecommon divisor among elements ^ { p, q } corresponding to pairs of quadrangles. In particular, if F i and F j are adjacent, then for any isomorphism of graded rings ϕ : H ∗ ( Z P ) → H ∗ ( Z Q ) fora polytope Q ∈ P aP og \ { I , M × I } we have ϕ ( f B k ) = ± f B ′ k for a (2 k ) -belt B ′ k containing -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 35 a) b) c) Figure 16. a) a 3-dimensional associahedron (Stasheff polytope) As as a trun-cated bipyramid; b) its Schlegel diagram based at a pentagon; c) its Schlegeldiagram based at a quadrangle k quadrangles and surrounding a face F ′ i ′ , and ϕ ( f B l ) = ± f B ′ l for a (2 l ) -belt B ′ l containing l quadrangles and surrounding a face F ′ j ′ adjacent to F ′ i ′ .Proof. We have k, l >
3, since a 4-belt surrounds a quadrangle and can not contain quadrangles.If F i and F j are adjacent, then, since P is flag, B ∩ B consists of two non-adjacent facesintersecting the edge F i ∩ F j by vertices, and both of them are quadrangles. On the other hand,let F i ∩ F j = ∅ , and { u, v } ⊂ ω ( B ) ∩ ω ( B ) with F u ∩ F v = ∅ . Then ( F u , F i , F v , F j ) is a 4-beltaround a quadrangle. If F u and F v are quadrangles, we have a contradiction. If F i = F j , then theelements f B k and f B l have k ( k − > (cid:3) Remark . Lemmas 7.13, 7.14, and 7.15 imply that any ideal almost Pogorelov polytope is B -rigid. The details see in [E20b].8. Cohomology ring H ∗ ( Z As )For the associahedron As we have the following picture. The group of symmetries of As acts transitively on quadrangles (3 quadrangles) and pentagons (6 pentagons).There are three equivalence classes of edges:(1) a common edge of a quadrangle and a pentagon (12 edges);(2) a common edge of pentagons connecting two vertices of quadrangles (3 edges);(3) a common edge of pentagons containing a common vertex of 3 pentagons (6 edges).There are two equivalence classed of vertices:(1) vertices of quadrangles (12 vertices);(2) common vertices of 3 pentagons (2 vertices). We have the following subsets P ω of faces with e H ( P ω ) = 0 (see Fig. 17). The complementarysubsets P [ m ] \ ω are exactly subsets with e H ( P [ m ] \ ω ) = 0. It should be | ω | > | ω | = 2 the sets ω = { i, j } correspond to pairs of nonadjacent faces F i , F j , F i ∩ F j = ∅ . We have 3 types of pairs encoded by numbers k , k , of edges of faces: 4 − − − Z = H ( Z P ). By the Poincare duality H ( Z P ) ≃ Z . The generators correspondto generators in e H ( P [ m ] \ ω ) for the complementary subsets.(2) For | ω | = 3 the sets ω = { i, j, k } correspond to either triples of disjoint faces or to pairsof disjoint sets: a face and a pair of adjacent faces; For the first type we have only oneset 4 − −
4. This corresponds to Z ⊂ H ( Z P ). For the second type we have cases:4 −
45 (for each quadrangle 4 pairs 45, in total 12 pairs, all equivalent); 4 −
55 (for eachquadrangle one pair, in total 3 pairs, all equivalent); 5 −
45 (for each pentagon two pairs,in total 12 pairs, all equivalent); 5 −
55 (for each pentagon one pair, in total 6 pairs,all equivalent). This corresponds to Z ⊂ H ( Z P ). In total, since no two faces can givenonzero contribution to H ( Z P ), we have H ( Z P ) = Z ⊕ Z = Z . By the Poincareduality H ( Z P ) ≃ Z ⊕ Z = Z . Here the summand Z corresponds to the complementto three disjoint quadrangles, which is a sphere with two holes. The generators in Z correspond to generators in e H ( P [ m ] \ ω ) for the complementary subsets.(3) Let | ω | = 4. Four can be expressed as a sum of positive integers in the following ways:4 = 1 + 1 + 1 + 1; 4 = 1 + 1 + 1 + 2; 4 = 1 + 3; 4 = 2 + 2. First let us mention that ifone of the connected components of P ω consists of one face, then either there are onlytwo components, or there are three components each consisting of a single quadrangle(it can be extracted from Fig. 16). Thus, we can have only 1 + 3 or 2 + 2. For thefirst case we may have 4 −
445 (two for each quadrangle, all equivalent, in total 6),4 −
455 (two for each quadrangle, all equivalent, in total 6), or 5 −
455 (one for eachpentagon, all equivalent, in total 6). For the second case we may have only 45 −
45 (allequivalent, in total 6). Since P has no 3-belts, we have H ( Z P ) ≃ Z . By the Poincareduality H ( Z P ) ≃ Z . The generators correspond to generators in e H ( P [ m ] \ ω ) for thecomplementary subsets.(4) For ω = 5 we have 5 = 1 + 4 = 2 + 3. For the first case we have 4 − Z ⊂ H ( Z P ). The complementary subsets are 4-belts and correspond togenerators in e H ( P [ m ] \ ω ), and we obtain a complementary summand Z ⊂ H ( Z P ) suchthat H ( Z P ) = Z ⊕ Z .Additive structure of H ∗ ( Z P ) is represented in Table 1.Choose in each group e H ( P ω ) = Z a generator β ω . We will call these generators canonicalgenerators . There is only one set P ω with rk e H ( P ω ) >
1. This is the union of three quadrangleswith e H ( P ω ) = Z . Choose in this group two generators β ω, and β ω, corresponding to twoquadrangles. We will also call these generators additional . Canonical and additional generators -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 37 (4-4-4) 12(4-45) 3(4-55) 6(5-55)12(5-45)6(4-445) 6(4-455) 6(5-455) 6(45-45)3(4-4455) Figure 17.
Collections P ω giving classes in e H ( P ω ) Table 1.
Additive structure of H ∗ ( Z As ) k H k ( Z P ) Z Z Z Z Z Z Z Z Z together form the set of multiplicative generators { β } of H ∗ ( Z As ), since As is a flag polytope.For short we will call them generators .To describe the multiplication in H ∗ ( Z P ) we need to describe how the product of each twogenerators can be expressed in terms of the elements { β ∗ } dual to generators, where β · β ∗ = [ Z P ]– a fundamental class in cohomology, and β · ( β ′ ) ∗ = 0 for β = β ′ .For each set P ω with e H ( P ω ) = 0 on Fig. 18 we describe all the ways how P [ m ] \ ω up tocombinatorial symmetries can be represented as a union of P ω ∪ P ω with [ m ] \ ω = ω ⊔ ω and e H ( P ω ) = 0, e H ( P ω ) = 0. Also we give the number of combinatorially symmetricdecompositions.On Fig. 19 we describe all the ways how the boundary ∂P of the polytope P up to com-binatorial symmetries can be decomposed as a union of three sets P ω , P ω , and P ω with[ m ] = ω ⊔ ω ⊔ ω , and e H ( P ω i ) = 0 for i = 1 , ,
3. We present the number of combinatorially
103 3 57102 2 5 51
Figure 18.
Decompositions of the sets P [ m ] \ ω with e H ( P ω ) = 0symmetric decompositions and for each set we give the number of decompositions where itarises.For generators β ∈ e H ( P ω ), β ∈ e H ( P ω ) their product β · β can be expressed uniquelyin terms of elements β ∗ ∈ e H ( P ω ⊔ ω ) dual to generators β ∈ e H ( P ω ), ω = [ m ] \ ( ω ⊔ ω ).If e H ( P ω ) = Z , then β · β is necessarily ± β ω . If e H ( P ω ) = Z , then P ω is the set of threedisjoint quadrangles, and due to Fig. 18 P ω and P ω both are disjoint unions of a pentagonand two adjacent pentagons. There are 3 such decompositions. The dual classes β ∗ ω , and β ∗ ω , in H ( P [ m ] \ ω , ∂P [ m ] \ ω ) are represented by the edges connecting a vertex of the quadranglecorresponding to the generator to a vertex of the third quadrangle. The edge connecting verticesof the two quadrangles corresponding to generators gives the element ± β ∗ ω , ± β ∗ ω , . Thus, thethree products β · β are equal to ± β ∗ ω , , ± β ∗ ω , , and ± β ∗ ω , ± β ∗ ω , .Up to signs this gives a full description of the ring H ∗ ( Z As , Z ). -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 39 = Figure 19.
Decompositions ∂P = P ω ∪ P ω ∪ P ω with [ m ] = ω ⊔ ω ⊔ ω , and e H ( P ω i ) = 0 for i = 1 , , e ω , ω = { F p , F q } , F p ∩ F q = ∅ , the complementary subspace in H ∗ ( Z P , Q ) toAnn H ( e ω ) is generated by the element e ω ∗ ∈ H m ( Z P , Q ) dual to e ω , and the canonical generatorsdividing e ω ∗ . In particular,(1) If F p and F q are both quadrangles, then codim Ann H ( e ω ) = 5;(2) If F p is a quadrangle, and F q is a pentagon, then codim Ann H ( e ω ) = 9;(3) If F p and F q are both pentagons, then codim Ann H ( e ω ) = 21.Also(1) If F p and F q are both quadrangles, then any element ω ′ ∈ N ( P ) \ { ω } is good for ω ;(2) If F p is a quadrangle, and F q is a pentagon, then an element ω ′ ∈ N ( P ) \ { ω } is badfor ω if and only if ω ′ = { p, r } , where F r is a quadrangle adjacent to F q . There are twosuch pairs;(3) If F p and F q are both pentagons, then they are both adjacent to a unique quadrangle F r .Then an element ω ′ = { s, t } ∈ N ( P ) \{ ω } is bad for ω if and only if either ω ′ = { F r , F u } ,where F u ∩ F r = ∅ , or ω ′ = { p, v } , where F v = F r is a quadrangle adjacent to F q , or ω ′ = { q, w } , where F w = F r is a quadrangle adjacent to F p , or ω ′ = { v, w } . There are 7bad elements in total. Corollary 8.1.
The set of elements {± e ω } corresponding to pairs of quadrangles in As ismapped to itself under each automorphism of a graded ring H ∗ ( Z P ) . Corollary 8.2.
Under each automorphism of a graded ring H ∗ ( Z P ) for P = As each element e ω , where ω = { p, q } , F p is a quadrangle, and F q is a pentagon, is mapped to the element ± e ω ′ + λ e ω ′ + λ e ω ′ , where ω ′ has the same type as ω , and ω ′ , ω ′ are bad for ω ′ . Proposition 8.3.
Let ω = { p, q } , where F p is a quadrangle, and F q is a pentagon, and let F u and F v be quadrangles different from F p . Then for ω = { q, u } , ω = { q, v } and any λ , λ wehave dim Ann H ( e ω ) = dim Ann H ( e ω + λ e ω + λ e ω ) . Proof.
Denote 4-5 = e ω , 4-4 = f ω , 4-4 = f ω , α = 4-5 + λ + λ . Then due to Fig. 18(1) 4-4 ∗ is divided by 5-55 , 5-55 , 5-455 , and 5-455 . Moreover, choosing appropriatesigns of generators, we can assume that4-4 ∗ = 5-55 · = 5-55 · . (2) 4-4 ∗ is divided by 5-55 , 5-55 , 5-455 , and 5-455 . Here all the elements listed aboveare different. Moreover, we can assume that4-4 ∗ = 5-55 · = 5-55 · . (3) 4-5 ∗ is divided by 5-45 , 5-45 , 5-455 , 5-455 , 5-55 , 5-55 , 45-45 and 45-45 . More-over, choosing appropriate signs of generators, we have4-5 ∗ = 5-45 · = 5-45 · = 5-55 · = 5-55 · . An element belongs to Ann H ( e ω ) if and only if in its expression in terms of the canonicalgenerators and dual elements coefficients at 4-5 ∗ and 5-45 , 5-45 , 5-455 , 5-455 , 5-55 ,5-55 , 45-45 and 45-45 are zero.Let x = P β ( ϕ ( β ) β + ϕ ( β ∗ ) β ∗ ) be the expression of an element x ∈ H ∗ ( Z P , Q ) in terms ofgenerators and dual elements. We have x = x + ϕ (4-4 ∗ )4-4 ∗ + ϕ (5-55 )5-55 + ϕ (5-55 )5-55 + ϕ (5-455 )5-455 + ϕ (5-455 )5-455 ++ ϕ (4-4 ∗ )4-4 ∗ + ϕ (5-55 )5-55 + ϕ (5-55 )5-55 + ϕ (5-455 )5-455 + ϕ (5-455 )5-455 ++ ϕ (4-5 ∗ )4-5 ∗ + ϕ (5-45 )5-45 + ϕ (5-45 )5-45 + ϕ (45-45 )45-45 + ϕ (45-45 )45-45 , where x is a linear combination of generators and dual elements having zero product with 4-4 ,4-4 , and 4-5. Also α = 4-5 + λ + λ . We have5-455 · · − ∗ ;5-55 · · − ∗ ;5-455 · − ∗ ;5-55 · − ∗ . -RIGIDITY OF THE PROPERTY TO BE AN ALMOST POGORELOV POLYTOPE 41 Therefore, − x · α = [ ϕ (4-5 ∗ ) + ϕ (4-4 ∗ ) λ + ϕ (4-4 ∗ ) λ ] [ Z P ]++ ϕ (5-455 ) λ ∗ + [ ϕ (5-455 ) λ + ϕ (45-45 )] 5-55 ∗ ++ [ ϕ (5-55 ) λ + ϕ (5-45 )] 5-455 ∗ + ϕ (5-55 ) λ ∗ ++ ϕ (5-455 ) λ ∗ + [ ϕ (5-455 ) λ + ϕ (45-45 )] 5-55 ∗ ++ [ ϕ (5-55 ) λ + ϕ (5-45 )] 5-455 ∗ + ϕ (5-55 ) λ ∗ ++ ϕ (5-455 )5-45 ∗ + ϕ (5-455 )5-45 ∗ + ϕ (5-55 )45-45 ∗ + ϕ (5-55 )45-45 ∗ . Thus, x · α = 0 if and only if0 = ϕ (5-455 ) = ϕ (5-455 ) = ϕ (5-55 ) = ϕ (5-55 ) == ϕ (5-55 ) λ = ϕ (5-455 ) λ = ϕ (5-55 ) λ = ϕ (5-455 ) λ == ϕ (5-455 ) λ + ϕ (45-45 ) = ϕ (5-55 ) λ + ϕ (5-45 ) == ϕ (5-455 ) λ + ϕ (45-45 ) = ϕ (5-55 ) λ + ϕ (5-45 ) == ϕ (4-5 ∗ ) + ϕ (4-4 ∗ ) λ + ϕ (4-4 ∗ ) λ . This is equivalent to 9 linearly independent equations: ϕ (5-455 ) = 0; ϕ (5-455 ) = 0; ϕ (5-55 ) = 0; ϕ (5-55 ) = 0; ϕ (45-45 ) = − ϕ (5-455 ) λ ; ϕ (5-45 ) = − ϕ (5-55 ) λ ; ϕ (45-45 ) = − ϕ (5-455 ) λ ; ϕ (5-45 ) = − ϕ (5-55 ) λ ; ϕ (4-5 ∗ ) + ϕ (4-4 ∗ ) λ + ϕ (4-4 ∗ ) λ = 0 . Thus, codim Ann H ( α ) = 9 = codim Ann H ( e ω ). (cid:3) Acknowledgements
The author is grateful to Victor Buchstaber for his encouraging support and attention tothis work, and Donald Stanley for a fruitful discussion. He is grateful to the Fields Institutefor Research in Mathematical Sciences (Canada) for providing excellent research conditionsand support while working on this paper at the Thematic Program on Toric Topology andPolyhedral Products.
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