An infinite family of tight triangulations of manifolds
aa r X i v : . [ m a t h . G T ] J un An infinite family of tight triangulations of manifolds
Basudeb Datta and Nitin Singh
Department of Mathematics, Indian Institute of Science, Bangalore 560 012, India. Revised: June 11, 2013
Abstract
We give explicit construction of vertex-transitive tight triangulations of d -manifoldsfor d ≥
2. More explicitly, for each d ≥
2, we construct two ( d + 5 d + 5)-vertexneighborly triangulated d -manifolds whose vertex-links are stacked spheres. The onlyother non-trivial series of such tight triangulated manifolds currently known is the seriesof non-simply connected triangulated d -manifolds with 2 d + 3 vertices constructed byK¨uhnel. The manifolds we construct are strongly minimal. For d ≥
3, they are alsotight neighborly as defined by Lutz, Sulanke and Swartz. Like K¨uhnel’s complexes, ourmanifolds are orientable in even dimensions and non-orientable in odd dimensions.
MSC 2000 :
Keywords:
Stacked sphere; Tight triangulation; Strongly minimal triangulation.
In [20], Walkup introduced the class K ( d ), d ≥
2, of simplicial complexes whose vertex-links are stacked ( d − K ( d ) is a triangulated d -manifold for d ≥ K (2). The followingresult by Kalai [11] shows that the members of this class triangulate a very natural class ofmanifolds obtained by handle additions on a sphere. Proposition 1.1 (Kalai) . For d ≥ , a connected simplicial complex X is in K ( d ) if andonly if X is obtained from a stacked d -sphere by β ( X ) combinatorial handle additions. Inconsequence, any such X triangulates either ( S d − × S ) β or ( S d − × − S ) β according towhether X is orientable or not. ( Here β = β ( X ) = β ( X ; Z ) . )Walkup’s class K ( d ) has also been a major source of examples of tight triangulations.Recall that, for a field F , a d -dimensional simplicial complex X is called tight with respectto F (or F -tight ) if (i) X is connected, and (ii) for all induced sub-complexes Y of X and forall 0 ≤ j ≤ d , the morphism H j ( Y ; F ) → H j ( X ; F ) induced by the inclusion map Y ֒ → X isinjective [13, 4]. In this paper, by tight we mean tight with respect to the field Z .Very few examples of tight triangulations are known. Apart from the trivial ( d + 2)-vertex triangulation S dd +2 of the d -sphere S d , the only non-trivial series of such triangulationscurrently known is the (2 d + 3)-vertex non-simply connected triangulated manifolds K d d +3 constructed by K¨uhnel [12]. The complex K d d +3 triangulates an S d − -bundle over S .Not surprisingly, K¨uhnel’s triangulations are members of K ( d ). Walkup’s class also relatesto one of the few combinatorial criteria for tightness that are known (for more generalcombinatorial criteria see [4, Theorem 3.10]). For example, Effenberger [9] showed that: E-mail addresses: [email protected] (B. Datta), [email protected] (N. Singh). roposition 1.2 (Effenberger) . For d = 3 , the neighborly members of K ( d ) are tight. Analogous to Walkup’s class K ( d ), let K ( d ) be the class of all simplicial complexes whosevertex-links are stacked ( d − K ( d ) is a triangulated d -manifoldwith boundary. Recently, Bagchi and Datta [5] proved Proposition 1.3 (Bagchi and Datta) . If M is a neighborly member of K (3) then thefollowing are equivalent. (i) M is tight, (ii) M is the boundary of a neighborly member of K (4) , and (iii) β ( M ; Z ) = ( f ( M ) − f ( M ) − / . Walkup’s class is also closely related to the notion of tight neighborly triangulation asintroduced by Lutz, Sulanke and Swartz in [14]. In particular, we have the following:
Proposition 1.4 (Novik and Swartz) . Let X be a connected triangulated d -manifold. ( a ) If d ≥ then (cid:0) d +22 (cid:1) β ( X ; Z ) ≤ f ( X ) − ( d + 1) f ( X ) + (cid:0) d +22 (cid:1) . ( b ) Further, if (cid:0) d +22 (cid:1) β ( X ; Z ) = f ( X ) − ( d + 1) f ( X ) + (cid:0) d +22 (cid:1) and d ≥ then X ∈ K ( d ) . From Propositions 1.4 and 1.1, one can deduce the following :
Corollary 1.5 (Lutz, Sulanke and Swartz) . Let X be a connected triangulated d -manifold.If d ≥ , then (cid:18) d + 22 (cid:19) β ( X ; Z ) ≤ (cid:18) f ( X ) − d − (cid:19) . (1) Moreover for d ≥ , the equality holds if and only if X is a neighborly member of K ( d ) . For d ≥
3, a triangulated d -manifold is called tight neighborly if it satisfies (1) withequality.In this paper, we present the second infinite series of neighborly members of K ( d ) afterK¨uhnel’s series K d d +3 . Like K¨uhnel’s complexes, our manifolds also exhibit vertex-transitiveautomorphism groups. They are orientable in even dimensions, non-orientable in odd di-mensions. In view of the above results, it follows that the triangulated d -manifolds weconstruct are tight for d ≥ d ≥
3. Our examples are alsostrongly minimal. More explicitly we have
Theorem 1.6.
For d ≥ and n = d +5 d +5 , there exist n -vertex non-isomorphic members M dn and N dn of K ( d ) with the following properties. (a) M dn and N dn are neighborly for all d . (b) M dn and N dn are tight for all d . (c) M dn and N dn are tight neighborly for d ≥ . (d) β ( M dn ; Z ) = β ( N dn ; Z ) = (cid:0) n − d − (cid:1) / (cid:0) d +22 (cid:1) = d + 5 d + 6 for d ≥ . (e) If d ≥ is even then M dn and N dn triangulate ( S d − × S ) β and if d ≥ is odd then M dn and N dn triangulate ( S d − × − S ) β , where β = d + 5 d + 6 . (f) M dn and N dn are strongly minimal for all d . (g) Z n acts vertex-transitively on M dn and N dn , respectively for all d . d ≥
3, apart from the ( d + 2)-vertex standard spheres S dd +2 , K¨uhnel’s complexes K d d +3 and a few sporadic examples, our examples M dd +5 d +5 and N dd +5 d +5 are the onlyknown tight neighborly triangulated manifolds (cf. Table 1 in Section 6). For d ≥ K d d +3 is the unique (2 d +3)-vertex triangulated manifold with β = 0 [1, 6]. We pose the following. Conjecture 1.7.
For d ≥ , if X is a ( d + 5 d + 5) -vertex triangulated d -manifold with β ( X ; Z ) = d + 5 d + 6 then X is isomorphic to M dd +5 d +5 or N dd +5 d +5 . The remainder of the paper is organized as follows. In Section 2, we review some basicdefinitions and results. Explicit description of the manifolds in Theorem 1.6 appears inSection 3. In Section 4, we present a purely combinatorial way of constructing neighborlymembers of K ( d ) and use it to construct the families M dd +5 d +5 and N dd +5 d +5 . In Section5, we prove properties of the aforementioned manifolds mentioned in Theorem 1.6. All simplicial complexes considered here are finite and abstract. We identify two complexesif they are isomorphic. By a triangulated manifold, sphere or ball, we mean a simplicialcomplex whose geometric carrier is a topological manifold, sphere or ball, respectively.A d -dimensional simplicial complex is called pure if all its maximal faces (called facets )are d -dimensional. A d -dimensional pure simplicial complex is said to be a weak pseudo-manifold if each of its ( d − d -dimensional weakpseudomanifold X , the boundary ∂X of X is the pure subcomplex of X whose facets arethose ( d − X which are contained in unique facets of X . The dualgraph Λ( X ) of a pure simplicial complex X is the graph whose vertices are the facets of X , where two facets are adjacent in Λ( X ) if they intersect in a face of codimension one.A pseudomanifold is a weak pseudomanifold with a connected dual graph. All connectedtriangulated manifolds are necessarily pseudomanifolds.If X is a d -dimensional simplicial complex then, for 0 ≤ j ≤ d , the number of its j -facesis denoted by f j = f j ( X ). The vector ( f , . . . , f d ) is called the face vector of X and thenumber χ ( X ) := P di =0 ( − i f i is called the Euler characteristic of X . As is well known, χ ( X ) is a topological invariant, i.e., it depends only on the homeomorphic type of | X | and,for any field F , χ ( X ) = P di =0 ( − i β i ( X ; F ), where β i ( X ; F ) = dim F ( H i ( X ; F )) is the i -thBetti number of X with respect to the field F . A simplicial complex X is said to be l -neighbourly if any l vertices of X form a face of X . By a neighborly complex, we shall meana 2-neighborly complex.Let X be a weak pseudomanifold with disjoint facets γ , δ and let ψ : γ → δ be a bijection.Let X ψ denote the weak pseudomanifold obtained from X \ { γ, δ } by identifying x with ψ ( x ) for each x ∈ γ . Then X ψ is said to be obtained from X by a combinatorial handleaddition . If u and ψ ( u ) have no common neighbor in X for each u ∈ γ (such a ψ is calledan admissible map) and X is in K ( d ) then X ψ is also in K ( d ) (see [2]).A standard d -ball is a pure d -dimensional simplicial complex with one facet. The stan-dard ball with facet σ is denoted by σ . A standard d -sphere is a simplicial complex isomor-phic to the boundary complex of a standard ( d + 1)-ball. The standard d -sphere on thevertex-set V is denoted by S dd +2 ( V ) (or simply by S dd +2 ). A simplicial complex X is calleda stacked d -ball if there exists a sequence B , . . . , B m of simplicial complexes such that B is a standard d -ball, B m = X and, for 2 ≤ i ≤ m , B i = B i − ∪ σ i and B i − ∩ σ i = τ i , where σ i is a d -face of B i and τ i is a ( d − σ i . Clearly, a stacked ball is a pseudomanifold.A simplicial complex is called a stacked d -sphere if it is (isomorphic to) the boundary of3 stacked ( d + 1)-ball. A trivial induction on m shows that a stacked d -ball actually tri-angulates a topological d -ball, and hence a stacked d -sphere is a triangulated d -sphere. If X is a stacked ball then clearly Λ( X ) is a tree. So, the dual graph of a stacked ball is atree. But, the converse is not true (e.g., the 7-vertex 3-pseudomanifold P whose facets are1234 , , , , P ) is a tree but P is not a triangulated ball). We have ([7]) Lemma 2.1.
Let X be a pure simplicial complex of dimension d . (i) If the dual graph Λ( X ) is a tree then f ( X ) ≤ f d ( X ) + d . (ii) The graph Λ( X ) is a tree and f ( X ) = f d ( X ) + d if and only if X is a stacked ball.Proof. Let f d ( X ) = m and f ( X ) = n . So, Λ( X ) is a graph with m vertices. We prove (i)by induction on m . If m = 1 then the result is true with equality. So, assume that m > m . Since Λ( X ) is a tree, it has a vertex σ ofdegree one (leaf) and hence Λ( X ) − σ is again a tree. Let Y be the pure simplicial complex(of dimension d ) whose facets are those of X other than σ . Since σ has a ( d − Y , it follows that f ( Y ) ≥ n −
1. Since f d ( Y ) = m −
1, the result is true for Y and hence f ( Y ) ≤ ( m −
1) + d . Therefore, n ≤ f ( Y ) + 1 ≤ m −
1) + d = m + d . This proves (i).If X is a stacked d -ball with m facets then X is a pseudomanifold and by the definition(since at each of the m − n = ( d + 1) +( m −
1) = m + d . Conversely, let Λ( X ) be a tree and n = f ( X ) = m + d . Let Y , σ beas above. Since f ( Y ) ≥ n −
1, it follows that f ( Y ) = n or n −
1. If f ( Y ) = n then f ( Y ) = n > ( m −
1) + d = f d ( Y ) + m , a contradiction to part (i). So, f ( Y ) = n − Y ∩ σ is a ( d − σ . Since f d ( Y ) = m −
1, by induction hypothesis, Y is astacked d -ball and hence X = Y ∪ σ is a stacked d -ball. This proves (ii). Corollary 2.2.
Let X be a pure d -dimensional simplicial complex and let CX denote a cone over X . Then CX is a stacked ( d + 1) -ball if and only if X is a stacked d -ball.Proof. Notice that f d +1 ( CX ) = f d ( X ) and f ( CX ) = f ( X ) + 1. Also Λ( CX ) is naturallyisomorphic to Λ( X ). The proof now follows from Lemma 2.1.Clearly, if N ∈ K ( d ) then N is a triangulated manifold with boundary and satisfiesskel d − ( N ) = skel d − ( ∂N ) . (2)Here skel j ( N ) := { α ∈ N : dim( α ) ≤ j } is the j -skeleton of N . From [5, Remark 2.20], itfollows : Proposition 2.3 (Bagchi and Datta) . For d ≥ , the map M ∂M is a bijection between K ( d + 1) and K ( d ) . The following corollary follows from Proposition 2.3 (cf. [7]).
Corollary 2.4.
For d ≥ , if M ∈ K ( d + 1) then Aut( M ) = Aut( ∂M ) . Note that any automorphism ϕ of a pure simplicial complex X induces an automorphism¯ ϕ of the dual graph Λ( X ) given by σ ϕ ( σ ) for any facet σ of X . Here we have : Lemma 2.5.
Let X be a pseudomanifold which is not a cone ( i.e., not all the facets arethrough a single vertex ) . Then, ϕ ¯ ϕ is an injective group homomorphism from Aut( X ) into Aut(Λ( X )) . Thus, Aut( X ) is naturally isomorphic to a subgroup of Aut(Λ( X )) . roof. Clearly, ϕ ¯ ϕ is a group homomorphism. Let ϕ be such that ¯ ϕ is identity onΛ( X ). Thus ϕ ( σ ) = σ for each facet σ in X . Let x ∈ V ( X ) be arbitrary. Choose facets α, β such that x ∈ α and x β . As Λ( X ) is connected, there is a path α α · · · α k in Λ( X )with α = α and α k = β . Since x ∈ α and x α k , there exists l < k such that x is in α , α , . . . , α l and x α l +1 . Hence α l \ α l +1 = { x } . Now ϕ ( α l ) = α l and ϕ ( α l +1 ) = α l +1 imply ϕ ( x ) = x . Since x was arbitrary, we see that ϕ is identity on X .A d -dimensional simplicial complex X is called minimal if f ( X ) ≤ f ( Y ) for everytriangulation Y of the geometric carrier | X | of X . We say that X is strongly minimal if f i ( X ) ≤ f i ( Y ), 0 ≤ i ≤ d , for all such Y . In [4], Bagchi and Datta have shown the following. Proposition 2.6 (Bagchi and Datta) . For any field F , each F -tight member of K ( d ) isstrongly minimal. In this section, we present our examples of neighborly members of K ( d ) for every d ≥ Example 3.1.
Let d ≥ n = d + 5 d + 5. Consider the ( d + 1)-dimensional puresimplicial complex M d +1 n on the vertex set { a , a , . . . , a n − } whose ( d + 2) n facets are σ i = { a i − j : 0 ≤ j ≤ d + 1 } , µ i = { a i } ∪ { a i + j ( d +3) − : 1 ≤ j ≤ d + 1 } ,α k,i = { a i } ∪ { a i − j : 2 ≤ j ≤ d + 2 − k } ∪ { a i + j ( d +3) − : 1 ≤ j ≤ k } , (3)0 ≤ i ≤ n −
1, 1 ≤ k ≤ d . The subscripts (except the first subscript on α ) are to be takenmodulo n . For all d ≥ M d +1 n is a neighborly member of K ( d + 1) (see Lemma 4.6). Wefurther define M dn := ∂ M d +1 n . (4)Since M d +1 n ∈ K ( d + 1), we have M dn ∈ K ( d ). By (2), skel d − ( M dn ) = skel d − ( M d +1 n ). Thisimplies that f ( M dn ) = f ( M d +1 n ) = n and, since d ≥ M dn is neighborly. Example 3.2.
Let d ≥ n = d + 5 d + 5. Consider the ( d + 1)-dimensional puresimplicial complex N d +1 n on the vertex set { a , a , . . . , a n − } whose ( d + 2) n facets are σ i = { a i − j : 0 ≤ j ≤ d + 1 } , µ i = { a i − j ( d +3) : 0 ≤ j ≤ d + 1 } ,α k,i = { a i } ∪ { a i − j : 2 ≤ j ≤ d + 2 − k } ∪ { a i − j ( d +3) : 2 ≤ j ≤ k + 1 } , (5)0 ≤ i ≤ n −
1, 1 ≤ k ≤ d . The subscripts (except the first subscript on α ) are to be takenmodulo n . For all d ≥ N d +1 n is a neighborly member of K ( d + 1) (see Lemma 4.7). Wefurther define N dn := ∂ N d +1 n . (6)Since N d +1 n ∈ K ( d + 1), we have N dn ∈ K ( d ). By the similar arguments as in the case of M dn , N dn has n vertices and is neighborly.From the definition of M d +1 n (resp., N d +1 n ), the permutation ψ := ( a , a , . . . , a n − ) (7)is an automorphism of M d +1 n (resp., N d +1 n ). Since any automorphism of M d +1 n is anautomorphism of ∂ M d +1 n , it follows that ψ ∈ Aut( M dn ). Similarly, ψ ∈ Aut( N d +1 n ) ⊆ Aut( N dn ). Since the order of ψ is n , we get 5 emma 3.3. Z n acts vertex-transitively on M d +1 n , N d +1 n , M dn and N dn , respectively. Observe that the induced automorphism ¯ ψ of Λ( M d +1 n ) (resp., Λ( N d +1 n )) is given by¯ ψ = ( σ , . . . , σ n − )( µ , . . . , µ n − ) d Y k =1 ( α k, , . . . , α k,n − ) . We remark that triangulations of surfaces with cyclic automorphism group were alsoconstructed by Ringel and Youngs as part of their proof of the Map Color Theorem [16,Chap 2, Sec 2.3]. As part of a series of neighborly triangulations on 12 s + 7 vertices,they obtained a neighborly triangulation of an orientable surface on 19 vertices with Z action. For d = 2, the 19-vertex triangulated 2-manifold M (resp., N ) is obtained as theboundary of the triangulated 3-manifold M (resp., N ). Our examples also exhibit Z action and are different (non-isomorphic) from the one obtained in [16]. In the terminologyof [16, Chap 2, Sec 2.3], the triangulations R (Ringel), M and N are described byfollowing cyclic permutations as “row 0” (we identify the vertex a i with i ). R : 1 11 14 13 15 3 8 9 7 4 17 10 18 5 16 12 2 6 ,M : 1 7 3 2 11 6 18 16 4 14 8 10 15 12 13 5 9 17 ,N : 1 12 3 2 6 11 18 16 9 5 13 15 10 7 8 14 4 17 . Consider the 3-dimensional example M . Figure 1 shows the link lk M ( a ) of thevertex a in M . Clearly, lk M ( a ) is a 28-vertex triangulation of the 2-sphere S . Byconstruction, we know that Z acts vertex-transitively on M . These imply that the linkof each vertex is a triangulated 2-sphere and hence M is a triangulated 3-manifold. Herewe prove • • • • • • • • • • • • • • • • • • •• •••• • •• • ••• •• ••
17 27 710 5122221 6 1319 2 91420 1615 125
Figure 1: Link of vertex a in M ( i stands for a i ) Lemma 3.4.
The full automorphism group of M ( resp., N ) is isomorphic to Z . roof. We present a proof for M . Similar arguments work for N .Since Z acts vertex-transitively, it is sufficient to show that the stabilizer of the vertex a in Aut( M ) is the trivial subgroup.Let β ∈ Aut( M ) and β ( a ) = a . So, β ∈ Aut(lk M ( a )). For 1 ≤ i ≤
28, letdeg( a i ) denote the number of edges through a i in lk M ( a ). Clearly, deg( β ( a i )) = deg( a i ).Since deg( a i ) = 7 for i = 12 and 17, β ( { a , a } ) = { a , a } and hence β ( { a , a } ) = { a , a } . This implies that β ( a ) = a and hence β (lk lk M ( a ) ( a )) = lk lk M ( a ) ( a ). Sincelk lk M ( a ) ( a ) is the 9-cycle a a a a a a a a a a , deg( a ) = 5 and deg( a ) = 6,it follows that β is identity on lk lk M ( a ) ( a ). Then β ( σ ) = σ for all simplices σ in lk M ( a )containing the vertex a . Since lk M ( a ) is a pseudomanifold, this implies that β is theidentity on lk M ( a ). This proves that Aut( M ) is isomorphic to Z .For the geometric carriers of M and N , we have Lemma 3.5.
The simplicial complex M ( resp., N ) is obtained from a stacked -sphereby combinatorial handle additions.Proof. We present a proof for M . Similar arguments work for N .Consider the pure 4-dimensional simplicial complexes B and B on the vertex sets V = { a j : − ≤ j ≤ } and V = { b , b , . . . , b , b , b , b , b } ∪ { u j : − ≤ j ≤ } ∪ { v j : − ≤ j ≤ } ∪ { w j : − ≤ j ≤ } , respectively, given by B = { e σ i : 0 ≤ i ≤ } , B = { e µ i : 0 ≤ i ≤ } ∪ { e α k,i : 0 ≤ i ≤ , ≤ k ≤ } , (8)where e σ i := { a i − j : 0 ≤ j ≤ } , e µ i := { b i +5+6 j : 0 ≤ j ≤ } , e α ,i := { w i − , b i , b i +5 , b i +11 , b i +17 } , e α ,i := { v i − , w i − , b i , b i +5 , b i +11 } , e α ,i := { u i − , v i − , w i − , b i , b i +5 } , for m ≥
29 and m = 34 , , ,
52 we have b m := b m − . Clearly, B is a stacked 4-ball whose dual graph is apath. The dual graph of B is a tree (a comb with 29 teeth) with 116 vertices. Since B has120 vertices, by Lemma 2.1, B is a stacked 4-ball. Let B be the simplicial complex obtainedfrom B ⊔ B by identifying the 3-simplices α = a − a − a − a and β = u − v − w − b bythe map ψ : u − a − , v − a − , w − a − , b a . Again by Lemma 2.1, B is astacked 4-ball. Let S := ∂B . Then S is a stacked 3-sphere with 149 vertices.Now consider the sixty 3-simplices α i = { u i − , v i − , w i − , b i } , β i = { a i − , a i − , a i − , a i } ,1 ≤ i ≤ α = { a − , a − , a − , a − } , β = { a , a , a , a } , α = { b , b , b , b } , β = { b , b , b , b } and the 30 maps ψ i : α i → β i given by ψ i ( u i − ) = a i − , ψ i ( v i − ) = a i − , ψ i ( w i − ) = a i − , ψ i ( b i ) = a i , for 1 ≤ i ≤ ψ ( a j ) = a j +29 , ψ ( b j ) = b j − .For 1 ≤ i ≤
30, let M i = M ψ i i − , where M = S . For 0 ≤ j ≤
30, let N j ( x ) denotethe set of neighbors of x in M j . Then N i − ( a i ) \ β i = { a i + k : − ≤ k ≤ , k = 0 } , N i − ( b i ) \ α i = { b i +5+6 k : 0 ≤ k ≤ , k = 4 } . Therefore, N i − ( a i ) ∩ N i − ( b i ) = ∅ . Similarly, N i − ( a i − ) ∩ N i − ( w i − ) = N i − ( a i − ) ∩ N i − ( v i − ) = N i − ( a i − ) ∩ N i − ( u i − ) = ∅ . Thus ψ i is admissible for 1 ≤ i ≤
28. Similarly, we can show that ψ and ψ are admissible.Since M = S ∈ K (3), inductively it follows that M ∈ K (3). It is now easy to see that M is isomorphic to M . This completes the proof.If ∼ is the equivalence relation generated by x ∼ ψ i ( x ) for x ∈ α i , 1 ≤ i ≤
30, then thequotient complex B/ ∼ is isomorphic to M , where B and ψ i are as in the above proof.In Lemma 5.2, we show that M is non-orientable.7 Construction in K ( d ) In this section, we present constructions of neighborly members of K ( d + 1). In particular,we construct manifolds in K ( d + 1) whose boundaries are M dn and N dn , respectively. Ourconstructions are based on Lemma 4.1 below ([7]). Given a graph G and a family T = { T i } i ∈I of induced subtrees of G , we say that u ∈ V ( G ) defines the subset ˆ u := { i ∈ I : u ∈ V ( T i ) } of I . Lemma 4.1.
Let G be a finite graph and T = { T i } ni =1 be a family of ( n − d ) -vertex inducedsubtrees of G . Suppose that (i) any two of the T i ’s intersect, (ii) each vertex of G is inexactly d + 1 members of T and (iii) for any two vertices u = v of G , u and v are togetherin exactly d members of T if and only if uv is an edge of G . Then the pure simplicialcomplex M whose facets are ˆ u , where u ∈ V ( G ) , is an n -vertex neighborly member of K ( d ) ,with Λ( M ) ∼ = G .Proof. First we prove that ˆ u = ˆ v for u = v in V ( G ). Assume that ˆ u = ˆ v for some u = v .Let P be a u - v path in G . Let w be the neighbor of u on P . Then uw is an edge of G andhence d = u ∩ ˆ w ) = v ∩ ˆ w ). Therefore wv is an edge in G . Let i ∈ ˆ u \ ˆ w = ˆ v \ ˆ w . Let Q be the u - v path in the tree T i . Let z be the neighbor of u on Q . Since i ˆ w , z = w .As before, we have d = u ∩ ˆ z ) = v ∩ ˆ z ). Therefore, zv is an edge in G . Since d ≥ u ∩ ˆ w ∩ ˆ z = ∅ . Let j ∈ ˆ u ∩ ˆ w ∩ ˆ z . Since ˆ u = ˆ v , it follows that j is in ˆ u , ˆ v ,ˆ w and ˆ z . Then T j contains u, v, w and z . Since T j is an induced subgraph it contains thecycle uwvzu , a contradiction to the fact that T j is a tree.Let S ⊆ { , . . . , n } be of size d . We show that at most two facets of M contain S . Ifpossible, let ˆ u , ˆ v and ˆ w be three facets of M that contain S . Then by assumption, uv , uw and vw are edges in G . Let i ∈ S . Then u , v and w are vertices of T i . Since T i isinduced subgraph, we conclude that uv , uw , vw are edges of T i , which is a contradiction tothe fact that T i is a tree. Thus M is a d -dimensional weak pseudomanifold. Clearly u ˆ u is an isomorphism between G and Λ( M ). Further the conditions on ( G, T ) imply that G is connected. Thus M is a d -pseudomanifold. Since any two members of T intersect, itfollows that M is neighborly. Let S i = st M ( i ) be the star of the vertex i in M . Then byconstruction Λ( S i ) ∼ = T i and thus f d ( S i ) = V ( T i )) = n − d . Also from the neighborlinessof M , f ( S i ) = n . Thus f ( S i ) = f d ( S i ) + d and hence, by Lemma 2.1, S i is a stacked d -ball.Therefore, by Corollary 2.2, lk M ( i ) is a stacked ( d − M is a member of K ( d ).We consider two examples of intersecting families of induced subtrees of a graph whichwe use to show that M d +1 n and N d +1 n are in K ( d + 1) (cf. Lemmata 4.6 and 4.7). Example 4.2.
Let d ≥ n = d + 5 d + 5. Consider the graph G d on n ( d + 2) verticesconsisting of two n -cycles C , C and n disjoint paths P i , 0 ≤ i ≤ n −
1, given by C = σ σ · · · σ n − σ , C = µ µ d +3 µ d +3) · · · µ ( n − d +3) µ , P i = σ i α ,i α ,i · · · α d,i µ i , (9)where the subscripts (except the first subscript on α ) are to be taken modulo n . Let T = { T i } n − i =0 be the family of induced trees where the vertex-set V ( T i ) of T i is given by V ( T i ) = { σ i + j : 0 ≤ j ≤ d + 1 } ∪ { µ i + j ( d +3) : 0 ≤ j ≤ d + 1 } ∪ { α j,i : 1 ≤ j ≤ d }∪ ( d +1 [ k =2 { α j,i + k : 1 ≤ j ≤ d + 2 − k } ) ∪ ( d +1 [ k =2 { α j,i + k ( d +3) : d + 2 − k ≤ j ≤ d } ) , (10)see Figure 2. Figure 4 shows the graph G with the tree T in black.8 • • • •••• • • • •••• •••• ••• • ••• •••••• µ i µ i + d +3 µ i +2( d +3) µ i + d ( d +3) µ i +( d +1)( d +3) α d,i α ,i σ i σ i +1 σ i +2 σ i +3 σ i + d σ i + d +1 α d,i +2 α d,i +2( d +3) α d − ,i +3 α ,i + d α ,i +( d +1)( d +3) Figure 2: Schematic representation of the tree T i in T . Example 4.3.
Let d ≥ n = d + 5 d + 5. Let G d be the graph as defined in Example4.2. Further let, T = { T i } ni =1 be the family of induced subtrees of G d , where the vertex-set V ( T i ), of the tree T i is given by (see Figure 3): V ( T i ) = { σ i + j : 0 ≤ j ≤ d + 1 } ∪ { µ i + j ( d +3) : 0 ≤ j ≤ d + 1 } ∪ { α j,i : 1 ≤ j ≤ d }∪ ( d +1 [ k =2 { α j,i + k : 1 ≤ j ≤ d + 2 − k } ) ∪ ( d +1 [ k =2 { α j,i + k ( d +3) : k − ≤ j ≤ d } ) . (11) • • • • •••• • • • •••• •••• ••••• µ i µ i + d +3 µ i +2( d +3) µ i + d ( d +3) µ i +( d +1)( d +3) α d,i α ,i σ i σ i +1 σ i +2 σ i + d σ i + d +1 α d,i +2 α d,i +( d +1)( d +3) α ,i + d +1 α ,i + d α d − ,i + d ( d +3) α ,i +2( d +3) Figure 3: Schematic representation of the tree T i in T . Lemma 4.4.
Let d ≥ and n = d + 5 d + 5 . Let the graph G d and the family of inducedsubtrees T = { T i } n − i =0 be as in Example . . Then T i ∩ T j = ∅ for all ≤ i, j ≤ n − .Proof. Let ϕ be a bijection on V ( G d ) given by ϕ = ( σ , . . . , σ n − )( µ , . . . , µ n − ) d Y j =1 ( α j, , . . . , α j,n − ) . It is easily seen that ϕ is an automorphism of G d and further we have T i +1 = ϕ ( T i ). Thus T i = ϕ i ( T ). Thus to show that T i ∩ T j = ∅ for 0 ≤ i, j ≤ n −
1, it is sufficient to show that T i ∩ T = ∅ for 0 ≤ i ≤ n −
1. 9 laim :
For 0 ≤ i ≤ n −
1, if there exist integers l, k with 2 ≤ l ≤ k ≤ d + 1 which satisfyeither (i) i + k ( d + 3) = n + l or (ii) i + l = k ( d + 3) then T i intersects T .Suppose i + k ( d + 3) = n + l for some integers l, k satisfying 2 ≤ l ≤ k ≤ d + 1. Thus i + k ( d + 3) ≡ l (mod n ). Then from (10), we see that { α j,l : d + 2 − k ≤ j ≤ d } ⊆ V ( T i ).Also from (10), { α j,l : 1 ≤ j ≤ d + 2 − l } ⊆ V ( T ). For l ≤ k , we see that the intersectionof the above two sets is { α j,l : d + 2 − k ≤ j ≤ d + 2 − l } ⊆ V ( P l ). Next supposethat i + l = k ( d + 3) for some integers 2 ≤ l ≤ k ≤ d + 1. Again from (10), we have { α j,i + l : 1 ≤ j ≤ d + 2 − l } ⊆ V ( T i ) and { α j,i + l : d + 2 − k ≤ j ≤ d } ⊆ V ( T ). Thus for l ≤ k , the two sets intersect, and hence T and T i intersect. This proves the claim.Clearly, we have the following six cases.(a) 0 ≤ i ≤ d + 1: In this case, T i intersects T in σ i .(b) i > ( d + 1)( d + 3): It is easy to see that T i contains σ and hence T i ∩ T = ∅ .(c) i = k ( d + 3), 1 ≤ k ≤ d + 1: In this case, T i intersects T in µ i .(d) i = k ( d + 3) −
1, 1 ≤ k ≤ d + 1: Then i + l ( d + 3) ≡ n ≡ n ), where l = d + 2 − k ≤ d + 1. This implies T i contains µ ∈ V ( T ). So, T i ∩ T = ∅ .(e) j ( d + 3) < i < ( j + 1)( d + 2), 1 ≤ j ≤ d : Let i = j ( d + 3) + t , where 1 ≤ t < d + 2 − j .Let k = d + 2 − j . Then i + k ( d + 3) = n + l where l = t + 1. Since 1 ≤ j ≤ d , wehave 2 ≤ l ≤ k ≤ d + 1. Hence, by the claim, T i intersects T .(f) k ( d + 2) ≤ i < k ( d + 3) −
1, 2 ≤ k ≤ d + 1: Let i = k ( d + 2) + t where 0 ≤ t < k − l = k − t . Then i + l = k ( d + 3) and 2 ≤ l ≤ k ≤ d + 1. Therefore, by the claim, T i intersects T .This completes the proof of the lemma. Lemma 4.5.
Let d ≥ and n = d + 5 d + 5 . Let the graph G d and the family of inducedsubtrees T = { T i } n − i =0 be as in Example . . Then (a) T i is a tree on n − d − vertices. (b) For all v ∈ V ( G d ) , ˆ v is a ( d + 2) -element set. (c) For u, v ∈ V ( G d ) , ˆ u ∩ ˆ v is a ( d + 1) -element set if and only if uv is an edge in G d .Proof. From (10) we have V ( T i ) = ( d + 2) + ( d + 2) + d + d +1 X k =2 ( d + 2 − k ) + d +1 X k =2 ( k −
1) = d + 4 d + 4 = n − d − . This proves (a). Recall that for v ∈ V ( G d ), ˆ v = { i : v ∈ T i } . Then from (10), we see thatˆ α l,m = { m } ∪ { m − k : 2 ≤ k ≤ d + 2 − l } ∪ { m − ( d + 3) j : d + 2 − l ≤ j ≤ d + 1 } , ˆ σ m = { m − k : 0 ≤ k ≤ d + 1 } , ˆ µ m = { m − k ( d + 3) : 0 ≤ k ≤ d + 1 } , (12)for 1 ≤ l ≤ d and 0 ≤ m ≤ n −
1. (Here ˆ σ m , ˆ µ m , ˆ α l,m ⊆ Z n .) Clearly ˆ σ l , ˆ µ l are sets ofsize d + 2. Further, we have: for 2 ≤ k, j ≤ d + 1, k ( d + 3) j (mod n ), and hence α l,m ) = 1 + ( d + 1 − l ) + l = d + 2. This proves (b).Let us define a metric ∆ on the set V ( G d ) as ∆( u, v ) := u \ ˆ v ) = v \ ˆ u ). It is easyto see that ∆ indeed defines a metric on V ( G d ), the proof of which will be omitted here.Clearly, σ i ∩ ˆ µ j ) < d + 1 and σ i µ j is not an edge of G d for 0 ≤ i, j ≤ n −
1. Thus, toprove (c), we need to show the following: 10 • ◦ ◦ ◦ ◦ •• ◦ ◦ ◦ ◦•• ◦ ◦ ◦ ◦•• ◦ ◦ ◦ ◦•• ◦ ◦ ◦ ◦••◦◦◦◦••◦◦◦◦••◦◦◦◦••◦◦◦◦••◦◦◦◦••◦◦◦◦••◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦• •◦◦◦◦ • •◦◦◦◦ • •◦◦◦◦ • •◦◦◦◦ • •◦◦◦◦ ••◦◦◦◦ ••◦◦◦◦ ••◦◦◦◦ ••◦◦◦◦ ••◦◦◦◦ ••◦◦◦◦ ••◦◦◦◦ •• ◦ ◦ ◦ ◦ •• ◦ ◦ ◦ ◦ •• ◦ ◦ ◦ ◦ •• ◦ ◦ ◦ ◦ •◦◦◦◦• ••◦◦◦◦ •◦◦◦ •◦◦•◦••◦•◦◦ •◦◦◦ •◦◦◦◦
Figure 4: Graph G and the tree T ( ∈ T ) in black(i) ∆( σ i , σ j ) = 1 ⇔ i − j ≡ ± n ).(ii) ∆( µ i , µ j ) = 1 ⇔ i − j ≡ ± ( d + 3) (mod n ).(iii) ∆( α l,m , α r,s ) = 1 ⇔ m = s , l − r = ± σ i , α l,m ) = 1 ⇔ i = m, l = 1.(v) ∆( µ i , α l,m ) = 1 ⇔ i = m, l = d .In all the above cases, the reverse implications follow from the definitions of the setsin (12). Before we proceed with the proofs of the forward implications, we introduce somenotation. For integers i, j , let | i − j | n denote the smallest non-negative integer k such thateither i + k ≡ j (mod n ), or j + k ≡ i (mod n ). If we think of Z n as the vertex set of the n -cycle C n whose edges are { i, i + 1 } then | i − j | n is the distance between vertices i and j in C n . Thus, | · | n is a metric on Z n and | i − j | n ≤ n/ i, j . For integers i ≤ j , let[ i, j ] n := { z ∈ Z : z ≡ k (mod n ), for some k ∈ { i, i + 1 , . . . , j }} . Claim 1: ∆( σ i , σ j ) ≥ min {| i − j | n , d + 2 } .If | i − j | n = 0 then there is nothing to prove. So, assume that t := | i − j | n > j ≡ i + t (mod n ). Let T = { j − k : 0 ≤ k ≤ t − } . Weclaim that T ∩ ˆ σ i = ∅ . Assume that T ∩ ˆ σ i = ∅ . Then there exist integers k, k ′ , where0 ≤ k ≤ t − ≤ k ′ ≤ d + 1, such that j − k ≡ i − k ′ (mod n ). So, j ≡ i + ( k − k ′ )(mod n ). Since k − k ′ ≤ t −
1, this implies (by the definition of | i − j | n ) that k − k ′ < t − ( k − k ′ ) > t − ( k − k ′ ) ≡ n ) (since t = j − i ≡ k − k ′ (mod n )). So, t − ( k − k ′ ) = pn for some positive integer p . Then n ≤ pn ≤ t + k ′ ≤ n/ d + 1). Thisimplies n ≤ d + 2, a contradiction. Thus, T ∩ ˆ σ i = ∅ .Now, if t ≤ d + 1, then T ⊆ { j − k : 0 ≤ k ≤ d + 1 } = ˆ σ j and hence ∆( σ i , σ j ) ≥ T ) = t .On the other hand, if t ≥ d + 2, then T ⊇ { j − k : 0 ≤ k ≤ d + 1 } = ˆ σ j , and hence ˆ σ i ∩ ˆ σ j = ∅ .Therefore ∆( σ i , σ j ) = d + 2. This proves Claim 1.11or 1 ≤ i ≤ d , 0 ≤ j ≤ n −
1, let A i,j := { j } ∪ { j − k : 2 ≤ k ≤ d + 2 − i } , B i,j := { j − k ( d + 3) : d + 2 − i ≤ k ≤ d + 1 } , C i,j := { j } ∪ { j − k ( d + 3) : d + 2 − i ≤ k ≤ d + 1 } and D i,j := { j − k : 2 ≤ k ≤ d + 2 − i } . So, ˆ α i,j = A i,j ⊔ B i,j = C i,j ⊔ D i,j . Claim 2: (a) If | m − s | n > d + 1 then A l,m ∩ A r,s = ∅ and A l,m ∩ B r,s ) ≤ ≤ l, r ≤ d .(b) If 0 < | m − s | n ≤ d + 1 then C l,m ∩ C r,s = ∅ and C l,m ∩ D r,s ) ≤ ≤ l, r ≤ d .Suppose | m − s | n > d + 1. Assume that z ∈ A l,m ∩ A r,s . Then there exist integers k, k ′ with 0 ≤ k ≤ d + 2 − l ≤ d + 1 and 0 ≤ k ′ ≤ d + 2 − r ≤ d + 1 such that m − k ≡ z ≡ s − k ′ (mod n ). Then | m − s | n ≤ d + 1, a contradiction. Thus A l,m ∩ A r,s = ∅ .Assume that z, x ∈ A l,m ∩ B r,s , where z = x . Since z, x ∈ A l,m , there exist a, b ∈{ , . . . , d + 1 } such that z = m − a and x = m − b . Then z − x = b − a ∈ [ − ( d + 1) , ( d + 1)] n .Since z, x ∈ B r,s , there exist k, k ′ ∈ { , . . . , d + 1 } such that z ≡ s − k ( d + 3) (mod n ) and x ≡ s − k ′ ( d + 3) (mod n ). So, z − x ≡ ( k ′ − k )( d + 3) (mod n ). Assume without lossthat k ′ > k . Then 1 ≤ k ′ − k ≤ d − d + 1 < ( k ′ − k )( d + 3) < n − ( d + 1).This implies that z − x = ( k ′ − k )( d + 3) [ − ( d + 1) , ( d + 1)] n , a contradiction. Therefore, A l,m ∩ B r,s ) ≤
1. This proves part (a). By similar arguments, part (b) of Claim 2 follows.
Claim 3:
If ∆( α l,m , α r,s ) = 1 then m = s .Assume that ∆( α l,m , α r,s ) = 1. Then α l,m ∩ ˆ α r,s ) = d + 1. Assume that m = s . Then | m − s | n >
0. We have the following two cases.
Case 1. | m − s | n > d + 1. Then, by Claim 2 (a), we have A l,m ∩ A r,s = ∅ , A l,m ∩ ˆ α r,s ) ≤ A r,s ∩ ˆ α l,m ) ≤
1. Also, B l,m ) = l ≤ d , B r,s ) = r ≤ d and hence B l,m ∩ B r,s ) ≤ d . Since α l,m ∩ ˆ α r,s ) = d + 1, these imply B l,m ∩ B r,s ) = d . This implies B l,m = B r,s and B l,m ) = d = B r,s ). Therefore l = d = r . In particular, B d,m = B d,s . Thenthere exist integers 2 ≤ k, k ′ ≤ d + 1 such that m − d + 3) ≡ s − k ( d + 3) (mod n ), and m − k ′ ( d + 3) ≡ s − d + 3) (mod n ). Subtracting we get ( k ′ − d + 3) ≡ (2 − k )( d + 3)(mod n ). Multiplying by d + 2, we get k ′ − ≡ − k (mod n ) and hence k + k ′ ≡ n ).Since 4 ≤ k + k ′ ≤ d + 2 < n , it follows that k = k ′ = 2. Thus m − d + 3) ≡ s − d + 3)(mod n ) and hence m ≡ s (mod n ). This is not possible since 0 ≤ m, s ≤ n − m = s . Case 2. < | m − s | n ≤ d +1. Then, by Claim 2 (b), we have C l,m ∩ C r,s = ∅ , C l,m ∩ ˆ α r,s ) ≤ C r,s ∩ ˆ α l,m ) ≤
1. Since α l,m ∩ ˆ α r,s ) = d + 1, we must have D l,m ∩ D r,s ) = d .This implies (as in Case 1) D l,m = D r,s and D l,m ) = d = D r,s ). Then, from thedefinition of D l,m (resp., D r,s ), l = r = 1. So, D ,m = D ,s . As in Case 1, we get m ≡ s (mod n ). Again this is not possible.Thus, we get contradictions in both cases. Therefore, m = s . This proves Claim 3.If ∆( σ i , σ j ) = 1 then, by Claim 1, | i − j | n ≤ i − j ≡ ± n ). Thisproves (i).Since ( d + 2)( d + 3) ≡ n ), we see that the map π : Z n → Z n given by i ( d + 2) i is a bijection, with the inverse map π − given by i ( d + 3) i . From the definitions ofˆ σ i and ˆ µ i , we see that π (ˆ µ i ) = ˆ σ i ( d +2) . Thus ∆( µ i , µ j ) = 1 ⇔ ∆( σ i ( d +2) , σ j ( d +2) ) = 1 ⇔| i ( d + 2) − j ( d + 2) | n = 1. Thus i ( d + 2) − j ( d + 2) ≡ ± n ), where multiplying by d + 3 gives i − j ≡ ± ( d + 3) (mod n ). This proves (ii).Now, assume ∆( α l,m , α r,s ) = 1. By Claim 3, m = s . So, ∆( α l,m , α r,m ) = 1. Fromthe reverse implication, we have ∆( u, v ) = 1 whenever uv is an edge in G d . Notice thatˆ σ m ∩ ˆ µ m = { m } . Thus ∆( σ m , µ m ) = d + 1. Assume that l = r ±
1. Then we canassume, without loss, that r > l and r − l ≥
2. Then by the triangle inequality we have d + 1 = ∆( σ m , µ m ) ≤ ∆( σ m , α l,m ) + ∆( α l,m , α r,m ) + ∆( α r,m , µ m ) ≤ l + 1 + ( d + 1 − r ) < d + 1,a contradiction. Therefore, l = r ±
1. This completes the proof of (iii).12y again using the triangle inequality for ∆ along the path P m = σ m α ,m · · · α d,m µ m ,we can prove (iv) and (v). This completes the proof of the lemma. Lemma 4.6.
For d ≥ and n = d + 5 d + 5 , the simplicial complex M d +1 n defined inExample . is a neighborly member of K ( d + 1) .Proof. Let ( G d , T ) be as in Example 4.2. By Lemma 4.5, ˆ v = { i : v ∈ T i } is a set of d + 2elements for each v ∈ V ( G d ). Consider the ( d + 1)-dimensional simplicial complex M ( G d )consisting of facets ˆ v , v ∈ V ( G d ). From Lemmata 4.4 and 4.5, we see that ( G d , T ) satisfythe hypothesis of Lemma 4.1 and hence, by Lemma 4.1, M ( G d ) is a neighborly member of K ( d + 1). Since { i + j ( d + 3) − ≤ j ≤ k } = { i − j ( d + 3) : d + 2 − k ≤ j ≤ d + 1 } assubsets of Z n for 0 ≤ i ≤ n − ≤ k ≤ d + 1, by (12), i a i defines an isomorphismfrom M ( G d ) to M d +1 n . This proves the lemma. Lemma 4.7.
For d ≥ and n = d + 5 d + 5 , the simplicial complex N d +1 n defined inExample . is a neighborly member of K ( d + 1) .Proof. Let ( G d , T ) be as in Example 4.3. It can be shown (as in Lemmata 4.4 and 4.5), that( G d , T ) satisfies the hypothesis of Lemma 4.1. Therefore, the simplicial complex N ( G d )consisting of facets ˆ v , v ∈ V ( G d ), is a neighborly member of K ( d + 1). Again, i a i definesan isomorphism from N ( G d ) to N d +1 n . This proves the lemma.We know that Z n acts vertex-transitively on each of M d +1 n , N d +1 n , M dn and N dn , respec-tively (see Lemma 3.3). Here we prove Lemma 4.8.
Let ψ be the map given in (7) . Then Aut( M dn ) = Aut( M d +1 n ) = h ψ i =Aut( N d +1 n ) = Aut( N dn ) ∼ = Z n .Proof. Since ψ is an automorphism of M d +1 n (resp., N d +1 n ), it follows that Aut( M dn ) ⊇ Aut( M d +1 n ) ⊇ h ψ i ⊆ Aut( N d +1 n ) ⊆ Aut( N dn ).Let β ∈ Aut( M d +1 n ) and let ¯ β ∈ Aut(Λ( M d +1 n )) be the induced automorphism. If β ( a ) = a then β (lk M d +1 n ( a )) = lk M d +1 n ( a ) and hence ¯ β ( T ) = T and ¯ β is an auto-morphism of the tree T . Then ¯ β ( σ i ) = σ i , ¯ β ( µ i ( d +3) ) = µ i ( d +3) and ¯ β ( α j, ) = α j, for0 ≤ i ≤ d + 1, 1 ≤ j ≤ d . These imply ¯ β | C = Id, ¯ β | C = Id and this in term implies that¯ β is the identity of Aut(Λ( M d +1 n )). Then, by Lemma 2.5, β is the identity of Aut( M d +1 n ).Thus the only automorphism of M d +1 n which fixes a is the identity. Since h α i is transitiveon V ( M d +1 n ), this implies that h α i = Aut( M d +1 n ).Let β ∈ Aut( N d +1 n ). Let ¯ β ∈ Aut(Λ( N d +1 n )) be the induced automorphism. Then¯ β ( T i ) = T j , where β ( a i ) = a j , for all i . If β ( a ) = a then β (lk N d +1 n ( a )) = lk N d +1 n ( a )and hence ¯ β ( T ) = T and ¯ β is an automorphism of the tree T . This implies that¯ β ( { σ , . . . , σ d +1 , µ d +3 , . . . , µ ( d +1)( d +3) } ) = { σ , . . . , σ d +1 , µ d +3 , . . . , µ ( d +1)( d +3) } . Since T and T are the only trees which contain { σ , . . . , σ d +1 , µ d +3 , . . . , µ ( d +1)( d +3) } , it followsthat ¯ β ( T ) = T . Inductively, we get ¯ β ( T i ) = T i for all i . Since ¯ β is an automorphismof Λ( N d +1 n ), ¯ β is an automorphism of T i for all i . This implies that ¯ β ( α ⌊ ( d +1) / ⌋ ,i ) = α ⌊ ( d +1) / ⌋ ,i or α ⌈ ( d +1) / ⌉ ,i for all i . Now, either ¯ β is identity on T or ¯ β ( σ j ) = µ j ( d +3) for0 ≤ j ≤ d + 1. In the second case, ¯ β ( α ⌊ ( d +1) / ⌋ , ) = α ⌈ ( d +1) / ⌉ , d +3) . This is not possiblesince ¯ β ( α ⌊ ( d +1) / ⌋ , ) = α ⌊ ( d +1) / ⌋ , or α ⌈ ( d +1) / ⌉ , . Therefore, ¯ β | T is the identity. Now, bythe similar argument as in the case M d +1 n it follows that β is the identity in Aut( N d +1 n )and Aut( N d +1 n ) = h ψ i . Thus, Aut( M dn ) ⊇ Aut( M d +1 n ) = h ψ i = Aut( N d +1 n ) ⊆ Aut( N dn ).If d ≥ d = 3, the result follows fromLemma 3.4. So, assume that d = 2. Using simpcomp [10], we found that Aut( M ) ∼ = Z ∼ =Aut( N ). The result follows from this for d = 2.13 emark 4.9. Observe that a tree in T (in Example 4.2) is non-isomorphic to a tree in T (in Example 4.3). Since a tree in T (resp., T ) is isomorphic to the dual graph of the link ofthe corresponding vertex of M d +1 n (resp., N d +1 n ), it follows that M d +1 n is non-isomorphic to N d +1 n for all d ≥
2. However, the dual graphs of M d +1 n and N d +1 n are isomorphic (both areisomorphic to the graph G d ). Since M d +1 n and N d +1 n are non-isomorphic, by Proposition2.3, M dn and N dn are non-isomorphic for d ≥
4. Notice that M has edges (namely, a i a i +12 )which are in seven facets. But, N does not have any such edge. Thus, M and N arenon-isomorphic. Using simpcomp [10], we found that M and N are non-isomorphic. For d ≥ m ≥
2, let D d +1 m + d +1 be the stacked ( d + 1)-ball with vertex-set { , , . . . , m + d + 1 } and facets { k, k + 1 , . . . , k + d + 1 } , 1 ≤ k ≤ m . Let M = ∂D d +1 m + d +1 , A be the d -simplex { , , . . . , d + 1 } and B be the d -simplex { m + 1 , m + 2 , . . . , m + d + 1 } . Then M ′ := M \ { A, B } triangulates I × S d − . Recall that, a bijection ψ : A → B is called admissible if for each vertex u ∈ A there does not exist v ∈ V ( M ) such that both { u, v } and { ψ ( u ) , v } are edges in M (see [1]). If m ≤ d +2 then { d +1 , d +1+ ⌈ m ⌉} and { d +1+ ⌈ m ⌉ , j } are edges in M for d + 1 + ⌈ m ⌉ 6 = j ∈ B . Thus, existence of an admissible map impliesthat m ≥ d + 3. On the other hand, if m ≥ d + 3 then there is no common neighbourof i and j in M for i ∈ A , j ∈ B and hence any bijection ψ : A → B is admissible. Let σ be a permutation on the set { , . . . , d + 1 } (i.e., σ ∈ Sym( d + 1)). Consider the bijection ϕ σ : A → B given by ϕ σ ( i ) = m + σ ( i ). Consider the quotient complexes Y := D d +1 m + d +1 /ϕ σ and X dm ( σ ) := M ′ /ϕ σ . Then ∂Y = X dm ( σ ). If ϕ σ is admissible then X dm ( σ ) ∈ K ( d ) andtriangulates an S d − -bundle over S . The case when md is even of the following lemmawas proved in Lemma 3.3 of [1]. Lemma 5.1.
For d ≥ , let X dm ( σ ) be as above, where ϕ σ is admissible. Then X dm ( σ ) is orientable if and only if either md is even and σ is an even permutation or md is oddand σ is an odd permutation. ( In particular, X dm (Id) is orientable when md is even andnon-orientable when md is odd. ) Proof.
For 1 ≤ k ≤ m , 1 ≤ l ≤ d , let δ k,l denote the d -simplex { k, k +1 , . . . , k + d +1 }\{ k + l } of M . Since | M ′ | is homeomorphic to [0 , × | ∂B | , M ′ is orientable. Observe that thefollowing defines an orientation on | M ′ | . (Here ∂B = S d − d +1 ( B ).)+ δ k,l = ( − kd + l +1 h k, k + 1 , . . . , k + l − , k + l + 1 , . . . , k + d + 1 i . (13)(To check that (13) defines a coherent orientation, one can take any orientation on ( d − M ′ . In particular, one can take positively oriented ( d − | ∂B | so that the orientation on | M ′ | as the product[0 , × | ∂B | is the same as the orientation given in (13). This also induces an orientation on | ∂A | . Let S B (resp., S A ) denote the oriented sphere | ∂B | (resp., | ∂A | ) with this orientation.Now, as the boundary of an oriented manifold, ∂ | M ′ | = S A ∪ ( − S B ) (cf. [8, pages 371–372]).Therefore, | M ′ /ϕ σ | = | M ′ | / | ϕ σ | is orientable if and only if | ϕ σ | : S A → S B is orientationpreserving (cf. [19, pages 134–135]). (Here, | ϕ σ | : | ∂A | → | ∂B | is the homeomorphisminduced by ϕ σ .)Note that ( d − M are δ k,i,j = { k, k + 1 , . . . , k + d + 1 } \ { k + i, k + j } ,0 ≤ i < j ≤ d + 1, ( i, j ) = (0 , d + 1), 1 ≤ k ≤ m . Consider the orientation on the14 d − M ′ as :+ δ k,i,j = ( − kd + i + j h k, . . . , k + i − , k + i + 1 , . . . , k + j − , k + j + 1 , . . . i . (14)Then [ δ m,i +1 , δ m, ,i +1 ] = − δ ,i , δ ,i,d +1 ] = 1) for 0 ≤ i ≤ d . This implies that | ∂B | (resp., | ∂A | ) with orientation given in (14) is S B (resp., S A ). (For β = { v , v , . . . , v d } ∈ M ′ and α = { v , . . . , v d } ∈ ∂B , if [ β, α ] = − α = h v , . . . , v d i with the orientationgiven in (14) ⇐⇒ + β = h v , v , v , . . . , v d i with the orientation given in (13) ⇐⇒ ( −→ v v , −→ v v , . . . , −→ v v d ) is the orientation of | M ′ | ⇐⇒ ( −→ v v , . . . , −→ v v d ) is the orientation of | ∂B | ⇐⇒h v , v , . . . , v d i is positive in S B .)For 1 ≤ i ≤ d + 1, consider the ( d − δ ,i,d +1 = { , . . . , d + 1 } \ { i + 1 } of ∂A . Then ϕ Id ( δ ,i,d +1 ) = { m + 1 , . . . , m + d + 1 } \ { m + i + 1 } = δ m, ,i +1 . Therefore,from (14), ϕ Id (+ δ ,i,d +1 ) = ( − md δ d, ,i +1 . Thus, | ϕ Id | : S A → S B is orientation preserving(resp., reversing) if md is even (resp., odd). Also | σ | : S A → S A is orientation preserving(resp., reversing) if σ is an even (resp., odd) permutation. Since ϕ σ = ϕ Id ◦ σ , it followsthat | ϕ σ | : S A → S B is orientation preserving if and only if md is even and σ is an evenpermutation or if md is odd and σ is an odd permutation. The lemma now follows. Lemma 5.2.
For d ≥ and n = d + 5 d + 5 , let M dn and N dn be as in Examples . and . , respectively. Then M dn , N dn are orientable if d is even and are non-orientable if d isodd.Proof. We present a proof for M dn . Similar arguments work for N dn . Let M d +1 n be as inExample 3.1. Let E (resp., E ) be the pure ( d + 1)-dimensional subcomplex of M d +1 n whose facets are σ . . . , σ n − (resp., µ . . . , µ n − ). So, Λ( E i ) = C i , 1 ≤ i ≤ E is isomorphic to the pseudomanifold D d +1 n + d +1 /ϕ Id , where D d +1 n + d +1 is thestacked ( d + 1)-ball defined at the beginning of this section. Thus, ∂E is isomorphicto X dn (Id). Therefore, ∂E triangulates an S d − -bundle over S and, by Lemma 5.1, isorientable if and only if dn is even. Thus (since n is odd), ∂E is orientable if and only if d is even. So, if d is odd then ∂E is non-orientable and hence (since | M dn | can be obtainedfrom | ∂E | by attaching handles) M dn is non-orientable.Again, the bijection f : Z n → V ( E ) given by f ( i ) = a ( d +3) i defines an isomorphism be-tween D d +1 n + d +1 /ϕ Id and E . Thus, ∂E is isomorphic to X dn (Id). Therefore, ∂E triangulatesan S d − -bundle over S and, by Lemma 5.1, orientable if and only if d is even.For 0 ≤ i ≤ n −
1, let F i be the stacked ( d + 1)-ball whose facets are α ,i , . . . , α d,i . Then, M d +1 n = E ∪ E ∪ ( ∪ n − i =0 F i ) and M dn is obtained from ∂E ∪ ∂E by attaching handles ∂F i \ { A i , B i } , where A i = a i − d − · · · a i − a i and B i = a i + d +2 a i +2( d +3) − · · · a i + d ( d +3) − a i ,0 ≤ i ≤ n − n ).Now, assume that d is even. So, ∂E , ∂E , ∂F i , 0 ≤ i ≤ n −
1, are orientable. Considerthe orientation on ∂E , ∂E and ∂F i , 0 ≤ i ≤ n −
1, given by :+ σ i,l = ( − l h a i − d − , . . . , a i − d − l , a i − d + l , . . . a i i , (15a)+ µ i,l = ( − l +1 h a i +( d +2) , . . . , a i + l ( d +3) − , a i +( l +2)( d +3) − , . . . , a i +( d +1)( d +3) − , a i i , (15b)+ α k,i,l = ( − l h b k,i, , . . . , b k,i,l b k,i,l +2 , · · · b k,i,d +2 i , (15c)where ( b k,i, , b k,i, , . . . , b k,i,d +2 ) = ( a i − − d + k , . . . , a i − a i a i +( d +2) a i +2( d +3) − , . . . , a i + k ( d +3) − ),for 1 ≤ k ≤ d , 0 ≤ l ≤ d + 1. From the proof of Lemma 5.1, (15a) (resp., (15b)) defines anorientation on ∂E (resp., ∂E ). Also, (15c) defines an orientation on ∂F i , 0 ≤ i ≤ n − A i = σ i,d = α ,i,d +1 and + σ i,d = − α ,i,d +1 . Also, B i = µ i,d = α d,i, and+ µ i,d = − α d,i, . Now, let γ be a ( d − A i . Let γ E (resp., γ F i ) be the d -face of ∂E (resp., ∂F i ) other than A i which contains γ . Then (with any orientation of γ )[ γ E , γ ] = − [ σ i,d , γ ] = [ α ,i,d +1 , γ ] = − [ γ F i , γ ] . (16)Similarly, if β is a ( d − B i and β E (resp., β F i ) is the d -face of ∂E (resp., ∂F i )other than B i which contains β . Then (with any orientation of β )[ β E , β ] = − [ µ i,d , β ] = [ α d,i, , β ] = − [ β F i , β ] . (17)Since M dn = ( ∂E \ { A , . . . , A n − } ) ∪ ( ∂E \ { B , . . . , B n − } ) ∪ ( n − [ i =0 ( ∂F i \ { A i , B i } )) , it follows from (16) and (17) that the orientations defined by (15a), (15b) and (15c) give acoherent orientation on M dn . Thus, M dn is orientable. This completes the proof.We now in a position to prove our main result. Proof of Theorem 1.6.
Let d ≥ n = d + 5 d + 5.Part (a): From Lemmata 4.6 and 4.7, M d +1 n and N d +1 n are neighborly members of K ( d + 1).This implies (by Proposition 2.3 and (2)) that M dn and N dn are neighborly membersof K ( d ). Also, M dn and N dn are non-isomorphic (see Remark 4.9).Part (b): If d = 3 then tightness follows from Proposition 1.2. Since M = ∂ M and N = ∂ N , tightness follows from Proposition 1.3 for d = 3.Part (d): If d ≥ d = 3 then the result followsfrom Proposition 1.3.Part (c): The result follows from Part (d).Part (e): If d ≥ M and N are orientable (by Lemma 5.2) and neighborly, β ( M ; Z ) = β ( N ; Z ) =2 − (19 − (cid:0) (cid:1) + (cid:0) (cid:1) ) = 40 and hence M and N both triangulate ( S × S ) .This proves the result for d = 2. If d = 3 then the result follows from Lemmata 3.5and 5.2.Part (f): By Part (b) and Proposition 2.6, M dn and N dn are strongly minimal for d ≥ K ( d ) Any triangulated 2-manifold is a member of K (2). In Table 1, we summarize the knownand some open cases for neighborly members of Walkup’s class K ( d ) for d ≥ Acknowledgement:
This work is supported in part by UGC Centre for Advanced Studies.The second author thanks IISc Mathematics Initiative for support. The authors thankBhaskar Bagchi and the anonymous referees for many useful comments.16 ( K ) d n K | K | References0 d d + 2 S dd +2 S d d even 2 d + 3 K d d +3 S d − × S [12]1 d odd 2 d + 3 K d d +3 S d − × − S [12]2 d ≥ − Not possible [17]3 4 15 M ( S × − S ) [3]3 4 15 N ( S × S ) [18]5 5 21 ? [9]7 4 20 ? [13]8 4 21 M ( S × S ) [7]8 4 21 N ( S × − S ) [7]14 4 26 N ( S × − S ) [7] d + 5 d + 6 d even d + 5 d + 5 M dd +5 d +5 ( S d − × S ) β this paper d + 5 d + 6 d odd d + 5 d + 5 M dd +5 d +5 ( S d − × − S ) β this paper Table 1: Known and some open cases for neighborly members of K ( d ), d ≥ References [1] B. Bagchi, B. Datta, Minimal triangulations of sphere bundles over the circle,
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