Minimal triangulations of (S^3\times S^1)^{#3} and (S^3 \(twisted product) S^1)^{#3}
aa r X i v : . [ m a t h . G T ] J un Minimal triangulations of ( S × S ) and ( S × − S ) Nitin Singh
Department of Mathematics, Indian Institute of Science, Bangalore-560012, India.email: [email protected]
September 25, 2018
Abstract
A triangulated d -manifold K , satisfies the inequality (cid:0) f ( K ) − d − (cid:1) ≥ (cid:0) d +22 (cid:1) β ( K ; Z ) for d ≥ d -manifolds that meet the bound with equality are called tight neighborly . Inthis paper, we present tight neighborly triangulations of 4-manifolds on 15 vertices with Z asautomorphism group. One such example was constructed by Bagchi and Datta in 2011. We showthat there are exactly 12 such triangulations up to isomorphism, 10 of which are orientable. MSC 2000 :
Keywords:
Stacked sphere; Tight neighborly triangulation; Minimal triangulation.
Minimal triangulations play an important role in combinatorial topology. In this regard, lowerbounds on the number of vertices needed in a triangulation of a topological space, in terms ofit’s topological invariants are particulary important. For compact surfaces, Heawood’s inequalityis one such lower bound, which states that a surface with Euler characteristic χ , requires at least ⌈ (7 + √ − χ ) ⌉ vertices for it’s triangulation. A similar analogoue for higher dimensions wasproved by Novik and Swartz in [10]. They prove that a triangulation of any manifold K , of dimension d ≥ (cid:18) f ( K ) − d − (cid:19) ≥ (cid:18) d + 22 (cid:19) β ( K ; Z ) . (1)The triangulations which satisfy (1) with equality were called tight neighborly by Lutz et al. in [9].It must be noted that tight neighborly triangulations, when they exist, are vertex minimal. Infact,for d ≥
4, they have compontent-wise minimal f -vector. A trivial example of a tight neighborlytriangulation is the ( d + 2)-vertex sphere S dd +2 for d ≥
3. In 1986, K¨uhnel constructed 2 d + 3 vertextriangulations of S d − × S for even d , and S d − × − S for odd d . K¨uhnel’s triangulations are tightneighborly with β = 1. Until recently, very few examples of tight neighborly triangulations apartfrom K¨uhnel’s series were known. The triangulation of a 4-manifold constructed by Bagchi andDatta in [1], was a first sporadic example of a tight neighborly triangulation. Very recently Dattaand Singh [4], have given another infinite family of tight neighborly triangulations, exploiting someunique combinatorial properties of such triangulations. These techniques were further extended in[11] to show the non-existence of tight neighborly triangulations for ( S d − × S ) and ( S d − × − S ) .In this paper, our aim is to employ the recently developed combinatorial criteria in [4] and [11] toenumerate all possible tight neighborly triangulations of ( S × − S ) and ( S × S ) , which have anon-trival Z action as the Bagchi-Datta example. As a result, we obtain one more tight neighborlytriangulation of ( S × − S ) and 10 new tight neighborly triangulations of ( S × S ) . Preliminaries
All simplicial complexes considered here are finite and abstract. Members of a simplicial complex arecalled faces . The empty set is a face of every simplicial complex. A simplicial complex is called pure if all it’s maximal faces (called facets ) have the same dimension. A pure d -dimensional simplicialcomplex is called a weak pseudomanifold without boundary (resp., with boundary), if all it’s d − X , we associate the graph Λ( X ), whose vertices are facets of X , and two facets are adjacent inΛ( X ) if they intersect in a face of co-dimension one. A weak pseudomanifold X of dimension d iscalled a d -pseudomanifold, if Λ( X ) is a connected graph. Triangulations of connected manifolds arenaturally pseudomanifolds.If X is a d -dimensional simplicial complex then, for 0 ≤ j ≤ d , the number of its j -faces aredenoted by f j = f j ( X ). The vector ( f , . . . , f d ) is called the face vector of X . For simplicialcomplexes X and Y , we define the join of X and Y , denoted by X ∗ Y as, X ∗ Y := { α ⊔ β : α ∈ X, β ∈ Y } . If X and Y are pseudomanifolds of dimension m and n respectively then, X ∗ Y is a pseudomanifoldof dimension m + n + 1. When X consists of a single vertex x , we denote the join as x ∗ Y , and callthe join as cone over Y .For a vertex x ∈ X , we define the subcomplex lk X ( x ), called the link of x in X as,lk X ( x ) := { α ∈ X : α ∪ { x } ∈ X, x α } . The cone x ∗ lk X ( x ) is called the star of x in X and is denoted by st X ( x ). By the k -skeleton of asimplicial complex X , denoted by skel k ( X ), we shall mean the subcomplex of X consisting of facesof dimension at most k . We call the 1-skeleton of a simplicial complex as it’s edge graph . When theedge graph is a complete graph on the vertex set, we will call the complex to be neighborly . K ( d ) A very natural class of triangulations of a topological ball is the class of stacked balls. A standard d -ball is the pure d -dimensional simplicial complex with one facet. A simplicial complex X is called a stacked d -ball if it is obtained from standard d -ball by successively pasting d -simplices along ( d − stacked d -sphere is defined as the boundary of a stacked ( d + 1)-ball.Clearly a stacked d -ball and a stacked d -sphere triangulate the d -ball and d -sphere, respectively.Following result from [4], gives a combinatorial characterization of a stacked ball. Proposition 2.1.
Let X be a pure d -dimensional simplicial complex. (a) If Λ( X ) is a tree then f ( X ) ≤ f d ( X ) + d . (b) Λ( X ) is a tree and f ( X ) = f d ( X ) + d if and only if X is a stacked d -ball. In [12], Walkup introduced the class K ( d ) of simplicial complexes, all whose vertex-links are stacked spheres . Clearly members of K ( d ) are triangulated d -manifolds. The following result ofNovik and Swartz shows that for dimensions four and above, tight neighborly triangulations liewithin the class K ( d ). Proposition 2.2 (Novik & Swartz [10]) . For d ≥ , if M is tight neighborly, then M is a neighborlymember of K ( d ) . Further, the following result by Kalai specifies the topological space determined by members of K ( d ). Proposition 2.3 (Kalai [7]) . Let X ∈ K ( d ) and let β = β ( X ; Z ) . If d ≥ , then X triangulates ( S d − × S ) β if it is orientable, and it triangulates ( S d − × − S ) β if it is non-orientable. he 4-manifold constructed by Bagchi-Datta in [1], is tight neighborly with parameters ( f , β ) =(15 ,
3) and is a member of K (4). Analogous to the class K ( d ), we define the class K ( d ) of triangulatedmanifolds, where the link of each vertex is a stacked ( d − K ∗ ( d )and K ∗ ( d ) to denote neighborly members of class K ( d ) and K ( d ) respectively. From the results in[2], we have the following correspondence: Proposition 2.4 (Bagchi & Datta) . If d ≥ , then M ∂M is a bijection from K ( d + 1) to K ( d ) . Proposition 2.4 immediately suggests that one can look for members of K ( d ) as boundaries ofmembers of K ( d + 1). In particular, we can obtain members of K (4) as boundaries of members of K (5). This is exactly the approach we take in this paper. Another easy consequence of the abovecorrespondence is the following: Corollary 2.5.
For d ≥ , let M, N ∈ K ( d + 1) . Then ϕ : V ( M ) → V ( N ) is an isomorphism from M to N , if and only if it is an isomorphism from ∂M to ∂N . In particular, for M = N , we get Aut( M ) = Aut( ∂M ) for M ∈ K ( d ). Example 3.1.
Let V = S i =1 { a i , b i , c i } be a 15-element set and let ϕ : V → V be the permutationdefined by ϕ := ( a , b , c )( a , b , c )( a , b , c )( a , b , c )( a , b , c ). For 1 ≤ i ≤
12, let N i denotethe simplicial complex with vertex set V and the set of facets { z i } ∪ { u i, , . . . , u i, } ∪ { v i, , . . . , v i, } ∪ { w i, , . . . , w i, } , (2)where z i = a b c a b c , v i,k = ϕ ( u i,k ) and w i,k = ϕ ( u i,k ) for 1 ≤ i ≤ , ≤ k ≤
8. The facetsmodulo the automorphism ϕ are given in Table 1.The following are the main results of this paper. Theorem 3.2.
Let N ∈ K ∗ (5) with f ( N ) = 15 and Aut( N ) ⊇ Z . Then N ∼ = N i for some i ∈ { , , . . . , } . Theorem 3.3.
Let M ∈ K ∗ (4) with f ( M ) = 15 and Aut( M ) ⊇ Z . Then M ∼ = ∂N i for some i ∈ { , , . . . , } . By Theorem 3.2 and Proposition 2.3, the complexes ∂N i , 1 ≤ i ≤
12, triangulate ( S × S ) or ( S × − S ) . Using a combinatorial topology software such as simpcomp [6], one can checkthat complexes ∂N and ∂N are non-orientable, while the complexes ∂N i for i ∈ { , , . . . , } are orientable. Thus, ∂N i triangulates ( S × − S ) for i = 1 , S × S ) for3 ≤ i ≤
12. We also point out that the example N obtained here is isomorphic to the triangulation N obtained by Bagchi and Datta in [1]. Consequently, the triangulation M of ( S × − S ) in [1]is isomorphic to ∂N . In this section, we give a broad outline of the enumeration strategy. By Proposition 2.4 and Corollary2.5, to obtain neighborly members of K (4), with Z automorphism, we can instead look for neighborlymembers of K (5) with Z automorphism. This has the advantage that all vertex-links are stackedballs, and by Proposition 2.1, we have a succint combinatorial description for it’s dual graph; thatit is a tree. Moreover, the dual graph of a vertex x , in a pseudomanifold X , is isomorphic to theinduced subgraph Λ( X )[ V x ], where V x is the set of facets containing x . For M ∈ K ( d ), let T x denotethe subtree of Λ( M ) induced by facets containing x . Then from [11], we have the following: , = a b c a b a , u , = a b a b a a , u , = a a b a a a , u , = a a a a b a , u , = a a a b a b , u , = a b a b a b , u , = c a b b a b , u , = b c b b a b ; u , = a b c a b a , u , = a b a b a a , u , = a a b a a a , u , = a a a a b a , u , = a a a b a b , u , = a a b b a b , u , = a b b c a b , u , = b b b c a b ; u , = a b c a b a , u , = a b a b a a , u , = a b a a a a , u , = a a a a b a , u , = a a a b a b , u , = a a b b a b , u , = a b b c a b , u , = b b b c a b ; u , = a b c a b a , u , = a b a b a a , u , = a b a a a a , u , = b a a a a c , u , = a a b a a c , u , = a a b a b a , u , = a b a b a b , u , = c b a b a b ; u , = a b c a b a , u , = a b c a a a , u , = a b a a a a , u , = a a a a a b , u , = a a a b a b , u , = a a b b a b , u , = a a b b c b , u , = a b b c b c ; u , = a b c a b a , u , = a b c a a a , u , = a b a a a a , u , = a a a a a b , u , = a a a b a b , u , = a a b a b b , u , = a a b b b c , u , = a b b c b c ; u , = a b c a b a , u , = a b c a a a , u , = a b a a a a , u , = b a a a a c , u , = a a a b a c , u , = a a b b a c , u , = a a b b c a , u , = a b b c a b ; u , = a b c a b a , u , = a b c a a a , u , = a b a a a a , u , = b a a a a c , u , = a a a b a c , u , = a a b a b a , u , = a a b b a b , u , = a b b c a b ; u , = a b c a b a , u , = a b c a a a , u , = a b a a a a , u , = a a a a a b , u , = a a a c a b , u , = a a b a c b , u , = a a b a b c , u , = a b a b b c ; u , = a b c a b a , u , = a b c a a a , u , = a b a a a a , u , = b a a a a c , u , = a a a c a c , u , = a a b a c a , u , = a a b a a b , u , = a b a b a b ; u , = a b c a b a , u , = a b c a a a , u , = a b a a a a , u , = a a a a a b , u , = a a b a a b , u , = a b a b a b , u , = b b a b a b , u , = b b c a b b ; u , = a b c a b a , u , = a b c a a a , u , = a b a a a a , u , = b a a a a c , u , = a a b a a c , u , = a b a b a c , u , = b b a b a c , u , = b b c a b a ; Table 1:
Facets of complexes modulo automorphism ϕ Proposition 4.1.
For M ∈ K ∗ ( d ) , let T x for x ∈ V ( M ) be as defined above. Then (a) Λ( M ) is a two connected graph. (b) Λ( M ) contains n ( n − d ) / ( d + 1) vertices and n ( n − d − /d edges where n = f ( M ) . (c) T x contains n − d vertices for each x ∈ V ( M ) . In [11], a set of facets S of M ∈ K ∗ ( d ) was defined to be critical in M if each of the connectedcomponents of Λ( M ) − S contained fewer than f ( M ) − d vertices. The following observations werealso made there. Proposition 4.2.
Let S be a critical set of facets of M ∈ K ∗ ( d ) . Then the facets in S togethercontain all the vertices of M . Proposition 4.3.
Let M ∈ K ∗ ( d ) with f ( M ) > d + 1 . Then the set of facets with degree three ormore in Λ( M ) together contain all the vertices of M . Identifying critical set of facets helps us reduce the possibilities for members of K ∗ (5). Wealready know from Proposition 4.1, that the dual graph Λ( M ) for M ∈ K ∗ ( d ) is two connected.When working with complexes with non-trivial automorphism groups, we can further narrow downthe admissible dual graphs, due to the following observation. Proposition 4.4 (cf. [4]) . Let M ∈ K ( d ) , then Aut( M ) is a subgroup of Aut(Λ( M )) . We now summarize what the above propositions imply for M ∈ K ∗ (5), with f ( M ) = 15 andAut( M ) ⊇ Z . We have,(a) Λ( M ) is a two connected graph,(b) Λ( M ) contains 25 vertices and 27 edges,(c) Z is a subgroup of Aut(Λ( M )), andd) for each vertex x of M , x appears in 10 facets of M , which induce a tree on Λ( M ).The program we carry out in the next section, leading to the classification is the following. Firstwe classify all two connected graphs on 25 vertices, with 27 edges which exhibit Z symmetry. Thenfor each of these graphs, we consider all possible members of K ∗ (5) which will have the graph astheir dual graph. Example 5.1.
For r, s ≥
1, let G r,s be the graph on 3 r + 3 s − V = { z } ∪ ( S r + s − i =1 { u i , v i , w i } ) and consisting of six edge disjoint paths, namely (see Figure 1(a)), p zu := z u · · · u r , p zv := z v · · · v r , p zw := z w · · · w r , p uv := u r u r +1 · · · u r + s − v r , p vw := v r v r +1 · · · v r + s − w r , p wu := w r w r +1 · · · w r + s − u r . Example 5.2.
For r, s ≥
1, let T r,s be the graph on 3 r + s vertices with vertex set V = { x , . . . , x s }∪ ( S ri =1 { u i , v i , w i } ) and consisting of four edge disjoint paths, namely (see Figure 1(b)), p := x x · · · x s , p := x u · · · u r x s , p := x v · · · v r x s , p := x w · · · w r x s . z u v w u v w u v w u v w u v w u v w u v w u v w (a) G , x x x x x x x v u w v u w v u w v u w v u w v u w (b) T , Figure 1:
Graphs G , and T , Lemma 5.3.
Let G be a -connected graph on vertices with edges. If Aut( G ) ⊇ Z , then G ∼ = G r,s for some r, s > with r + s = 9 or G ∼ = T r,s for some r, s > with r + s = 25 . We will defer the proof of Lemma 5.3 to the Appendix A. For M ∈ K ∗ (5) with f ( M ) = 15,by Proposition 4.3, we see that the set of facets with degree three or more in Λ( M ) together mustcontain all the vertices of M . But the graphs T r,s contain only two vertices of degree three ormore, and hence can contain at most 2 × M , as f ( M ) = 15. Thus, we have a further constraint on the graph Λ( M ). In particular from Proposition4.4 and Lemma 5.3 we have, Lemma 5.4.
Let M ∈ K ∗ (5) with f ( M ) = 15 and Aut( M ) ⊇ Z . Then Λ( M ) ∼ = G r,s for some r, s > with r + s = 9 . The following result from [11] further restricts the structure of Λ( M ), for M ∈ K ∗ ( d ). Proposition 5.5.
Let M ∈ K ∗ ( d ) with f ( M ) > d + 1 . Let u u · · · u r be a path in Λ( M ) whereall the internal vertices u i , ≤ i ≤ r − , have degree two in Λ( M ) . Let x i be the unique element in u i − \ u i for ≤ i ≤ r . Then we have, (a) x , . . . , x r are distinct, (b) x i ∈ u for ≤ i ≤ r , c) r ≤ d + 1 . By Part (c) of the above proposition, Λ( M ) contains induced paths of length at most d + 1.For M ∈ K ∗ (5), therefore, Λ( M ) contains induced paths of length at most 6. By Lemma 5.4, weknow that Λ( M ) ∼ = G r,s for some r + s = 9. Since G r,s contains the induced path of length at leastmax( r, s ), we must that max( r, s ) ≤
6. Together with the constraint r + s = 9, we conclude Lemma 5.6.
Let M ∈ K ∗ (5) with f ( M ) = 15 . If Aut( M ) ⊇ Z , then Λ( M ) ∼ = G r, − r for some r ∈ { , , , } . Let M ∈ K ( d ) and for each x ∈ V ( M ), let T x denote the subtree of Λ( M ) induced by the facetsof M containing x . We note the following: Lemma 5.7.
For M ∈ K ( d ) , let T x for x ∈ V ( M ) be as defined. Then, (a) T x = T y for x = y , (b) If σ ∈ V (Λ( M )) is a leaf of some tree T x , then d Λ( M ) ( σ ) < .Proof. We first prove (a). Suppose T x = T y for some x = y . Since T x = Λ( M ), and Λ( M ) isconnected graph, there exists an edge uv in Λ( M ) such that u ∈ V ( T x ) and v = V ( T x ). Now since u ∈ V ( T x ) = V ( T y ), we have { x, y } ⊆ u . Since v = T x , T y , we have { x, y }∩ v = ∅ . Thus { x, y } ⊆ u \ v ,which is a contradiction as uv is an edge in Λ( M ).To prove (b), suppose σ is a leaf of tree T x and d Λ( M ) ( σ ) ≥
3. Since at most one neighbor of σ is on T x , we have at least two neighbors of σ , say α and β which are not on T x . Thus x α, β . Butthen σ \{ x } ⊆ α, β . This is a contradiction as σ \{ x } is a face of co-dimension one which is containedin three facets, namely σ, α and β . This completes the proof.Let M ∈ K ( d ) and let { T x : x ∈ V ( M ) } be the collection of trees as before. Clearly the trees T x and T y intersect if and only if x ∈ st M ( y ). Thus the number of trees that a tree T x intersects(counting itself), is same as the number of vertices in it’s star. From Proposition 2.1 we have (sincest M ( x ) is a stacked d -ball), f (st M ( x )) = f d (st M ( x )) + d. However, f d (st M ( x )) is the number of vertices in the tree T x , and hence we can write the above as,Number of trees intersected by T x = | V ( T x ) | + d. (3)We can however, count the number of trees intersected by T x in the following way. Designate a fixedvertex r ∈ V ( T x ) to be the root . Next we orient each edge uv of T x as −→ uv where v is the vertex nearerto r (see Figure 2). To each oriented edge −→ uv of T x , we associate a label l ( −→ uv ) = y where y is theunique element of u \ v . Now there are d + 1 trees that intersect T x (including itself), at vertex r . Forany tree T y which intersects T x , but does not contain r , there must be an oriented edge e in T x suchthat l ( e ) = y . Thus the number of trees that T x intersects is at most ( d +1)+ | V ( T x ) |− | V ( T x ) | + d .For the equality to hold, which it should for M ∈ K ( d ), all the edge labels must be distinct, anddifferent from vertices of the facet r . We note this observation as the following lemma. Lemma 5.8.
For M ∈ K ( d ) and x ∈ V ( M ) , let T x be an oriented tree with root r , as describedbefore. Then the labels on the oriented edges are distinct, and different from vertices of the facet r . Lemma 5.9.
Let M ∈ K ( d ) and let u u · · · u r be a path in Λ( M ) with r < d + 1 . Let x i be theunique element in u i − \ u i , and y i be the unique element in u i \ u i − for ≤ i ≤ r . Then we have, (a) { x , x , . . . , x r } ⊆ u \ u r , and (b) { y , y , . . . , y r } ⊆ u r \ u .Proof. Since r < d + 1, there exists z ∈ M such that { u , u , . . . , u r } ⊆ V ( T z ). Part (a) follows byorienting T z with the u r as root and applying Lemma 5.8, and Part (b) follows similarly by orienting T z with u as root. Figure 2:
Oriented tree with root r Let M ∈ K ∗ (5) with f ( M ) = 15 and Aut( M ) ⊇ Z . Recall that Λ( M ) ∼ = G r, − r for some3 ≤ r ≤ M with vertices of G r, − r . Let ϕ be anautomorphism of M . Then ϕ induces an automorphism ¯ ϕ : u ϕ ( u ) of Λ( M ). Following lemmahas been proved in [4]. Lemma 5.10.
Let M ∈ K ( d ) . For ϕ ∈ Aut( M ) , let ¯ ϕ be an induced automorphism of Λ( M ) . Then ϕ ¯ ϕ is an injective homomorphism from Aut( M ) to Aut(Λ( M )) . Now let ϕ be an order three automorphism of M . Then, it induces an order three automorphism¯ ϕ of Λ( M ). Since Λ( M ) ∼ = G r, − r , ¯ ϕ is an order three automorphism of G r, − r . Without loss ofgenerality, we may assume ϕ is such that, ¯ ϕ = Q i =1 ( u i , v i , w i ). We show that ϕ does not have anyfixed points. Lemma 5.11.
Let M ∈ K ∗ (5) with f ( M ) = 15 and Aut( M ) ⊇ Z . If ϕ is an order threeautomorphism of M , then ϕ has no fixed points.Proof. By Lemma 5.6, Λ( M ) ∼ = G r, − r for some 3 ≤ r ≤
6. We identify facets of M with verticesof G r, − r . Without loss of generality assume ϕ induces the automorphism ¯ ϕ = Q i =1 ( u i , v i , w i ) of G r, − r . Now the orbits of ϕ are either singleton or contain 3 elements. Let x be a fixed point of ϕ .Let V x be the facets of M containing x . By definition of ¯ ϕ , we have ¯ ϕ ( V x ) = V x . Thus V x is unionof orbits of ¯ ϕ = Q i =1 ( u i , v i , w i ). Since | V x | = 10 and each orbit of ¯ ϕ has cardinality three or one, V x must contain a fixed point of ¯ ϕ . Since z is the only fixed point of ¯ ϕ , we conclude z ∈ V x . Also,observe that we must have, | V x ∩ { u , . . . , u }| = | V x ∩ { v , . . . , v }| = | V x ∩ { w , . . . , w }| . Thus we must have | V x ∩ { u , . . . , u }| = 3. Since V x induces a connected subgraph of G r, − r and r ≥
3, we have V x ∩ { u , . . . , u } = { u , u , u } . Similarly, V x ∩ { v , . . . , v } = { v , v , v } and V x ∩ { w , . . . , w } = { w , w , w } . Thus V x = { z } ∪ ( ∪ i =1 { u i , v i , w i } ). However if ϕ has one fixedpoint, it has at least three fixed points. Let y = x be another fixed point of ϕ . Then we will have V y = { z } ∪ ( ∪ i =1 { u i , v i , w i } ) = V x , and thus T x = T y , which contradicts Lemma 5.7. This provesthe lemma. Definition 5.12 (Class C of complexes) . Let V := S i =1 { a i , b i , c i } be a 15-vertex set. Let V beordered as a < b < c < a < b < · · · < b < c . Let Φ := Q i =1 ( a i , b i , c i ). We will denote theorbit { a i , b i , c i } of Φ as Φ i . Let us define the class C of complexes as C := { M ∈ K ∗ (5) : V ( M ) = V, Φ ∈ Aut( M ) } . (4)We notice that any M ∈ K ∗ (5) with f ( M ) = 15 and Aut( M ) ⊇ Z is isomorphic to a memberof the collection C . Therefore it suffices to enumerate C up to isomorphism. Before we proceed withenumeration, we consider an efficient string representation of a member of the class C .f M ∈ C , then we know that Λ( M ) ∼ = G r, − r for some 3 ≤ r ≤
6. We identify the facets of M with vertices of G r, − r . We may assume that Φ induces the automorphism Q i =1 ( u i , v i , w i ) of G r, − r . To each facet u of M , we associate a string [ u ] = a a . . . a k where a < a < · · · < a k are vertices of u . With the complex M , we associate the string representation, which we denote bystr( M ) as, str( M ) := [ z ] + [ u ] + [ u ] + · · · + [ u ] , (5)where + denotes the concatenation of strings. Note that the above representation uniquely specifiesa complex in C as remaining facets may be obtained by applying the automorphism Φ. Clearly anyother orbit such as { z , v , . . . , v } or { z , w , . . . , w } could have been used. However, we assumethat { z , u , . . . , u } is the one that yeilds lexicographically least representation, i.e, we assume[ u ] < min([ v ] , [ w ]). Throughout the remainder of the paper, let V, C and Φ be as in Definition 5.12. We order C withthe ordering M ≤ M for M , M ∈ C if str( M ) ≤ str( M ), where the ordering on strings islexicographic. We say M ∈ C is minimal , if M ≤ N for all N ∈ C , such that N is isomorphic to M .Thus each isomorphism class contains exactly one minimal element. We will look for such minimalmembers of C . Lemma 5.13.
Let M ∈ C . If Γ is a permutation of V such that ΦΓ = ΓΦ i for some i ∈ { , } , then Γ( M ) ∈ C .Proof. Clearly Γ( M ) ∈ K ∗ (5) for M ∈ K ∗ (5). We need to show that Φ is an automorphism ofΓ( M ). Suppose ΦΓ = ΓΦ i , where i ∈ { , } . Let u be a facet of Γ( M ). Then u = Γ( v ) for somefacet v of M . Now Φ( u ) = ΦΓ( v ) = ΓΦ i ( v ) = Γ( w ) where w = Φ i ( v ) is a facet of M , and henceΓ( w ) is a facet of Γ( M ). This Φ maps facets to facets in Γ( M ), and hence is an automorphism ofΓ( M ). The lemma follows.Lemma 5.13 implies that for a mimimal complex M ∈ C and for Γ, a permutation of V satisfyingthe conditions in the lemma, we must have M ≤ Γ( M ). Next we define some permutations of V ,which satisfy the conditions in Lemma 5.13. • π i := ( a i , b i , c i ) for 1 ≤ i ≤ • π i,j := ( a i , a j )( b i , b j )( c i , c j ) for 1 ≤ i < j ≤ • γ α,β := ( α , β )( α , β )( α , β )( α , β )( α , β ), where { α i , β i } ⊆ { a i , b i , c i } for 1 ≤ i ≤ π i as shifting the elements cyclically within an orbit. The permuata-tions π i,j “interchange” the orbits i and j , while the permutations γ α,β interchange an adjacent pairin each of the orbits. We will need the above permutations in pruning our candidates for minimalelement of C . Lemma 5.14.
Let M ∈ C be minimal. Then [ z ] = a b c a b c .Proof. Since Φ( z ) = z , z must be a union of Φ-orbits. Since | z | = 6, it contains exactly twoorbits. For M to be minimal, [ z ] should be minimal for M , among all complexes in it’s isomorphismclass. Thus z should be union of orbits Φ and Φ , for otherwise using one of the permutations π i,j we can get the complex π i,j ( M ) with lexicographically smaller z . Thus z = Φ ∪ Φ , or[ z ] = a b c a b c .We introduce a succint representation of complexes in C . Recall that with an oriented edge −→ uv in Λ( M ), we associated a label l ( −→ uv ) as the unique element of u \ v . To a complex M ∈ C , with( M ) ∼ = G r, − r , we associate tuples X = ( x , x , . . . , x ) and Y = ( y , y , . . . , y ), where (see Figure3) x i = l ( −−→ z u ) if i = 1 ,l ( −−−−→ u i − u i ) if 2 ≤ i ≤ ,l ( −−→ u v r ) if i = 9 , and y i = l ( −−→ u z ) if i = 1 ,l ( −−−−→ u i u i − ) if 2 ≤ i ≤ ,l ( −−→ v r u ) if i = 9 . (6)Clearly, the triple ( z , X, Y ) uniquely specifies a complex in C . Further, by Lemma 5.14, for minimalcomplexes, z is constant, and hence the pair ( X, Y ) uniquely specifies a minimal complex in C . Wewill frequently make use of the following lemma. In the remainder of the paper, we shall alwaysassume that for a minimal complex M ∈ C , X = ( x , . . . , x ) and Y = ( y , . . . , y ) are the tuples asdefined in (6). z u v w u v w u v w u v w u v w u v w u v w u v w x y x y x y x y x y x y x y x y x y Figure 3:
Succint representation of a minimal complex in C Lemma 5.15.
Let M ∈ C be minimal. For i ∈ { , , } , define m i = min { j : y j ∈ Φ i } . For i ∈ { , } , define n i = min { j : x j ∈ Φ i } . Then we have, (a) m < m < m . (b) n < n . (c) y m i = a i for i ∈ { , , } . (d) x n i = c i for i ∈ { , } .Proof. To prove part (a), we prove m < m and m < m . Assume that m > m . Let π = π , .Consider the complex M ′ = π ( M ). Since π ( z ) , π ( u ) , . . . , π ( u ) is an orbit of M ′ under Φ, we havestr( M ′ ) ≤ [ π ( z )] + [ π ( u )] + · · · + [ π ( u )] . Note that for i < m < m , we have π ( u i ) = u i as none of the u i ’s contain elements from Φ or Φ . However, for i = m , we have u i = S ∪ { y } where S ∩ Φ = S ∩ Φ = ∅ and y ∈ Φ .Thus π ( u i ) = π ( S ) ∪ { π ( y ) } = S ∪ { π ( y ) } . However, π ( y ) ∈ Φ as y ∈ Φ , and hence π ( y ) < y ,or [ π ( u i )] < [ u i ]. Thus π ( M ) < M , a contradiction to minimality of M . This proves m < m .Similarly we can show that m < m .To prove part (b), assume that n < n . Let π = π , . For i < n < n , Φ ⊆ u i and Φ ⊆ u i ,and hence π ( u i ) = u i . For i = n , we have u i = S ∪ S ∪ S , where S = Φ \{ x i } , S = Φ and S ⊆ Φ ∪ Φ ∪ Φ . Thus, π ( u i ) = π ( S ) ∪ π ( S ) ∪ π ( S ). But π ( S ) = Φ as S = Φ . Thus [ π ( u i )]ontains first three positions from Φ , whereas [ u i ] contains first two positions from Φ . Hence π ( u i ) < u i for i = n , and hence π ( M ) < M , a contradiction to minimality of M . This proves n < n .Parts (c) and (d) may be proved using the permuations π i = ( a i , b i , c i ). Informally, in the stringrepresentation of M , when an orbit element appears for the first time, we can always permute theorbit, so that it is the least element a i , for i th orbit. Similarly, when an orbit element leaves for thefirst time, we can always permute the elements so that it is the greatest element c i for i th orbit.For i ∈ { , , . . . , } , let N i be the simplicial complexes as defined in Example 3.1. Lemma 5.16.
Let M ∈ C be minimal with Λ( M ) ∼ = G , . Then M ∼ = N .Proof. Let X = ( x , x , . . . , x ) and Y = ( y , y , . . . , y ) be the tuple associated with M . We claimthe following:(a) ( y , y , y ) = ( a , a , a ).(b) ( x , x ) = ( c , c ).(c) x ∈ { b , b } .Recall that a set S of facets of M is critical in M if each component of Λ( M ) − S has less than 10vertices. Thus the set of facets u , v , w is critical. By Proposition 4.2, V = u ∪ v ∪ w . Since v = Φ( u ) and w = Φ ( u ), we conclude that u intersects each Φ-orbit. Since z = Φ ∪ Φ , and u \ z = { y , y , y } , we conclude that y , y and y each come from distinct orbits among Φ , Φ and Φ . By Parts (a) and (c) of Lemma 5.15, we must have ( y , y , y ) = ( a , a , a ). This proves(a).By Lemma 5.9, { x , x , x } ⊆ z ⊆ Φ ∪ Φ . Thus, by Parts (b) and (d) of Lemma 5.15, we have x ∈ Φ and further that x = c . We claim that x = l ( −−→ u u ) ∈ Φ . If possible, let l ( −−→ u u ) ∈ Φ .Then x = l ( −−→ u u ) ∈ Φ , otherwise u will not intersect Φ . Let us count the vertices in T x . Clearly T x contains z , u , u . Further notice that none of the edges oriented away from z on the paths z · · · v and z · · · w have the label x . Thus, { v , v , v , w , w , w } ⊆ V ( T x ). We have alreadyaccounted for 6 + 3 = 9 vertices of T x . By Part (b) of Lemma 5.7 v , w cannot be leaves of T x ,therefore we must have at least two more vertices in T x . Thus, T x contains at least 11 vertices, acontradiction. Therefore, x ∈ Φ , and hence by Part (d) of Lemma 5.15, x = c . This proves (b).We first show that if x ∈ Φ , then x = b and if x ∈ Φ , then x = b . Assume that x ∈ Φ and x = b . Then x = a . Let π := ( a , c )( a , c )( a , b )( a , b )( a , b ). We will show that π ( M ) < M . Let M ′ = π ( M ). For clarity, we will denote the facet of M corresponding to vertex u of the dual graph G , as M ( u ) and similarly facet of M ′ corresponding to vertex u of G , as M ′ ( u ). We notice that M ′ ( z ) = M ( z ) = a b c a b c . Since M ′ ( u ) is the lexicographically leastneighbor of M ′ ( z ), we see that M ′ ( u ) = π ( M ( v )) (Lexicographically least neighbor is the one,along which c leaves). It follows that M ′ ( u i ) = π ( M ( v i )) for 1 ≤ i ≤
3. From parts (a) and (b) wehave, str( M ) = a b c a b c + a b c a b a + a b a b a a + b a b a a a + · · · str( M ′ ) = [ M ′ ( z )] + [ M ′ ( u )] + [ M ′ ( u )] + [ M ′ ( u )] + · · · = [ π ( M ( z ))] + [ π ( M ( v ))] + [ π ( M ( v ))] + [ π ( M ( v ))] + · · · = a b c a b c + a b c a b a + a b a b a a + a a b a a a + · · · < str( M ) . This contradicts the minimality of M . Hence x ∈ Φ implies x = b . Similarly it can be shownthat x ∈ Φ implies x = b . This proves (c).Let us deduce more about the arrangement of the trees T x for x ∈ V . Because of the auto-morphism Φ, we only need to know the trees of a , a , a , a , a . We note that for x ∈ { a , a , a } , x z ∪ v ∪ w , and hence T x ⊆ Λ( M ) − { z , v , w } . Thus the trees T a , T a , T a are contained inthe union of the paths w w · · · u · · · u and u · · · u . Since y = a leaves along the edge −−→ u u , we u v w u v w u v w u v w u v w u v w u v w u v w c c x a a a a a a b b b a a a (a) Case 1: z u v w u v w u v w u v w u v w u v w u v w u v w c c x a a a a a a b b b a a a (b) Case 2: Figure 4:
Cases for Lemma 5.16conclude that T a is contained in the arc w · · · u · · · u . As T a has 10 vertices, we have followingcases: Case 1: T a = w · · · u · · · u . Immediately we have x = l ( −−→ u v ) = a and l ( −−−→ w w ) = a , andhence y = l ( −−→ u u ) = Φ( l ( −−−→ w w )) = b (see Figure 4(a)). Now T a contains the vertex u from thepath u · · · u , hence it must induce an arc of 9 vertices on outer cycle. By Lemma 5.7, it cannotshare its end points with T a . We see that the only possibility for T a is u u ∪ w · · · u · · · u .Similarly we have T a as u · · · u ∪ w · · · u · · · u . From these, we conclude as before that x = a , x = a , y = b and y = b . From Proposition 5.5, we must have u = { x , . . . , x } and v = { y , . . . , y } . Hence we have { x , x , x } = u \{ a , a , a } and { y , y , y } = v \{ b , b , b } .Putting x = x ∈ { b , b } , we have the following constraints,( x , x , x , x , x , x ) = ( c , c , x, a , a , a ) , { x , x , x } = u \{ a , a , a } , (7a)( y , y , y , y , y , y ) = ( a , a , a , b , b , b ) , { y , y , y } = v \{ b , b , b } . (7b)Above equation gives 2 choices for x , 3! = 6 each for ( x , x , x ) and ( y , y , y ). We examine the2 × × simpcomp [6]. We get the following solution. X = ( c , c , b , b , a , a , a , a , a ) , Y = ( a , a , a , b , b , b , c , b , b ) . The pair ( X , Y ) yeilds the complex N . Case 2: T a = w w · · · u · · · u . In this case, we have T a = u u ∪ w · · · u · · · u and T a = u · · · u ∪ w · · · u · · · u (see Figure 4(b)). Analyzing as in Case 1, we have the following constraints,( x , x , x , x , x , x ) = ( c , c , x, a , a , a ) , { x , x , x } = u \{ a , a , a } , (8a)( y , y , y , y , y , y ) = ( a , a , a , b , b , b ) , { y , y , y } = v \{ b , b , b } . (8b)where x = x ∈ { b , b } . Examining the possible 72 cases using simpcomp [6], we find no member of K ∗ (5). Thus N is the only minimal element of C with Λ( M ) ∼ = G , . This proves the lemma. Lemma 5.17.
Let M ∈ C be minimal with Λ( M ) ∼ = G , . Then M ∼ = N , N or N .Proof. Let X = ( x , x , . . . , x ) and Y = ( y , y , . . . , y ) be the tuple associated with M . We claimthe following:a) ( x , x ) = ( c , c ).(b) ( y , y , y ) = ( a , a , a ).(c) ( x , x ) ∈ { ( b , a ) , ( b , b ) , ( b , a ) , ( b , b ) } .(d) y ∈ { b , c , b , c } .We start by proving that u ∪ v ∪ w = V , which would imply that u intersects all Φ-orbits.By Proposition 4.3, z ∪ u ∪ v ∪ w = V . We show that z ⊆ u ∪ v ∪ w . Suppose not, andlet x ∈ z be such that x u ∪ v ∪ w . Thus T x ⊆ Λ( M ) − { u , v , w } . Since | V ( T x ) | = 10,we see that V ( T x ) = { z } ∪ ( S i =1 { u i , v i , w i } ). But then there is at most one such x ∈ V . Thus | u ∪ v ∪ w | ≥
14. Notcing that u ∪ v ∪ w is union of Φ-orbits, we conclude | u ∪ v ∪ w | = 15,and hence V = u ∪ v ∪ w . Thus u intersects all Φ-orbits.We now prove (a). The claim x = c is proved exaclty as in Lemma 5.16. We prove x ∈ Φ , and then by Lemma 5.15, Part (d) claim that x = c . Assume that x Φ . Then x ∈ Φ . By Lemma 5.9, { x , x , x , x } ∩ u = ∅ . Since u intersects Φ , we must have Φ x , x , x , x } . Since { x , x } ⊆ Φ , we conclude { x , x } ⊆ Φ . We will show that in this case | V ( T x ) | >
10. For the purpose of estimating | V ( T x ) | , assume ( x , x ) = ( a , b ). Then lookingoutwards from z , a leaves along −−→ u u and −−−→ w w , and does not leave along the path z · · · v . Thus { z , u , u , v , v , v , v , w , w , w } are vertices in T a . Since v cannot be a leaf, there must be atleast one more vertex in T a , and thus number of vertices exceeds 10, a contradiction. Note that wewould arrive at the same estimate for any other choice of ( x , x ). Thus x ∈ Φ and hence x = c .This proves (a).To prove (b), we claim that y , y , y belong to distinct Φ-orbits among Φ , Φ , Φ . Sincedistance between the pairs ( u , v ) and ( u , w ) in Λ( M ) is 5, there exist x, y ∈ V such that { u , u , u , z , v , v } ⊆ V ( T x ) and { u , u , u , z , w , w } ⊆ V ( T y ). We orient the two trees with z as root (edges oriented towards the root). Now if y i = y j for some 1 ≤ i < j ≤
3, we see that y j is repeated as a label leaving v i or w i towards z . Thus Lemma 5.8 is violated in T x or T y . Hence y , y , y are from distinct orbits, and by Lemma 5.15, we have ( y , y , y ) = ( a , a , a ). This proves(b).To prove (c), we prove that x = b if x ∈ Φ and x = b if x ∈ Φ . The proof is exacltythe same as of Claim (c) in Lemma 5.16. Together with part (a), and the observation that Φ i x , x , x , x } for i = 1 ,
2, we have ( x , x ) ∈ { ( b , a ) , ( b , b ) , ( b , a ) , ( b , b ) } . This proves (c).To prove (d), it is sufficient to show that y Φ . Suppose y ∈ Φ , say y = b . Then observethat l ( −−−→ w w ) = a . Since w and u are at a distance 5 in Λ( M ), there exists z ∈ V such that { w , w , w , w , z , u } ⊆ V ( T z ). Orient T z with z as the root (edges oriented towards the root).Then l ( −−→ u z ) = l ( −−−→ w w ) = a , which contradicts Lemma 5.8. A similar contradiction also follows ifwe assume y = c . Thus y Φ , which proves (d).Under conditions (a), (b), (c) and (d), we attempt to deduce the arrangement of trees T x . From(d), we have the following cases: Case 1: y ∈ { b , c } . Observe that in this case u ∩ Φ = { a } and u ∩ Φ = { a } . Consequently, a v ∪ w and a v ∪ w . Thus T a , T a ⊆ Λ( M ) − { z , v , w } . Therefore the trees T a and T a are confined to the union of the paths u · · · u and the arc w w · · · u · · · u . Since a leavesalong the edge −−→ u z , it contains vertices u , u , u . Hence, T a induces an arc with 7 vertices on theouter cycle. Similarly T a induces an arc with 9 vertices on the outer cycle. It can be seen that theonly possibilities for T a and T a are (see Figure 5), T a = u u u u ∪ w w w u u u u ,T a = u u ∪ w w w w u u u u u . (9)From (9), we conclude x = a , x = a , y = b , y = b . By Lemma, 5.9, we have { x , . . . , x } = u \ v and { y , . . . , y } = v \ u . Thus { x , x , x } = ( u \ v ) \{ a , a } , { y , y , y } =( v \ u ) \{ b , b } . Putting ( x , x ) = ( x, y ) and y = z , we have the following constraints,( x , x , x , x , x , x ) = ( c , c , x, y, a , a ) , { x , x , x } = ( u \ v ) \{ a , a } , (10a)( y , y , y , y , y , y ) = ( a , a , a , z, b , b ) , { y , y , y } = ( v \ u ) \{ b , b } . (10b) u v w u v w u v w u v w u v w u v w u v w u v w c c xy a a a z a a b b a a Figure 5:
Illustration for Lemma 5.17, Case 1Examining the cases using simpcomp [6], we find the following solutions: X = ( c , c , b , b , a , a , a , a , a ) , Y = ( a , a , a , b , b , b , c , b , b ) ,X = ( c , c , b , b , a , a , a , a , a ) , Y = ( a , a , a , b , b , b , c , b , b ) . The tuples ( X , Y ) and ( X , Y ) yeild the complexes N and N respectively. Case 2: y ∈ { b , c } . In this case we see that the trees T a and T a are contained in the union ofarc A = w w · · · u · · · u and the path P = u u · · · u . As T a contains 3 vertices on P , it mustinduce a path containing 8 vertices (including u ) on A . Similarly T a , which contains 4 vertices on P must induce a path containing 7 vertices on A . It can be seen that we have the following solutionsfor T a and T a . T a = u u u u ∪ w w w w u u u , T a = u u u ∪ w w w u u u u u ,T a = u u u u ∪ w w u u u u u , T a = u u u ∪ w w w w u u u u . (11)For the first solution in (11) as illustrated in Figure 6(a), we have x = a , x = a , y = b and y = b . Letting ( x , x ) = ( x, y ) and y = z we have the following constraints,( x , x , x , x , x , x ) = ( c , c , x, y, a , a ) , { x , x , x } = ( u \ v ) \{ a , a } , (12a)( y , y , y , y , y , y ) = ( a , a , a , z, b , b ) , { y , y , y } = ( v \ u ) \{ b , b } . (12b)Examining the above cases using simpcomp [6], we get the following solution, X = ( c , c , b , a , b , c , a , a , a ) , Y = ( a , a , a , c , b , b , b , c , b ) . The tuple ( X , Y ) yeilds the complex N .For the second solution in (11) as illustrated in Figure 6(b), we have x = a , x = a , y = b and y = b . Putting ( x , x ) = ( x, y ) and y = z , we have the following constraints,( x , x , x , x , x , x ) = ( c , c , x, y, a , a ) , { x , x , x } = ( u \ v ) \{ a , a } , (13a)( y , y , y , y , y , y ) = ( a , a , a , z, b , b ) , { y , y , y } = ( v \ u ) \{ b , b } . (13b)We do not get any solutions for minimal member of C with above constraints. Thus N , N , N arethe only minimal elements of C with G , as the dual graph. Lemma 5.18.
Let M ∈ C be minimal with Λ( M ) ∼ = G , . Then M ∼ = N , N , . . . , N or N . u v w u v w u v w u v w u v w u v w u v w u v w c c xy a a a z a a b b a a (a) z u v w u v w u v w u v w u v w u v w u v w u v w c c xy a a a z a a b b a a (b) Figure 6:
Illustration for Lemma 5.17, Case 2
Proof.
We begin by proving the following claims:(a) ( x , x , x ) = ( c , b , c ), ( x , x ) ∈ { ( a , b ) , ( b , a ) } .(b) ( y , y , y ) = ( a , a , a ), y ∈ { b , c } .(c) y ∈ { b , c , b , c } .(d) y = b ⇒ ( x , x ) = ( b , a ), y = c ⇒ ( x , x ) = ( a , b ).By Lemma 5.9, we know that { x , x , . . . , x } = z \ u . Since z = Φ ∪ Φ , we have Φ ⊆{ x , x , . . . , x } or Φ ⊆ { x , x , . . . , x } . By Lemma 5.15, we have x = c . Let 1 ≤ p < q < r ≤ { x p , x q , x r } is one of the orbits Φ or Φ . Then it can be seen that for x ∈ { x p , x q , x r } , T x contains p + q + r − p + q + r − p, q, r ) = (3 , , x = c , { x p , x q , x r } must be the orbit Φ = { a , b , c } . Hence x ∈ Φ .As in Claim (c) in Lemma 5.16, we conclude x = b . From Lemma 5.15, we further have x = c .Therefore { x , x } = { a , b } . This proves (a).For part (b), ( y , y , y ) = ( a , a , a ) follows exactly as in the proof of Claim (b) in Lemma 5.17.We now show that y ∈ Φ . Suppose y Φ . We show that in this case the trees T c and T a donot intersect. Since Φ ⊆ { x , x , . . . , x } , we see that T c is contained in the union of the threepaths z u · · · u , z v · · · v and z w · · · w . Thus T a and T c must intersect along one of the abovethree paths. Clearly T c and T a do not intersect along the path z u · · · u . However, if y Φ ,we see that T a does not contain any vertex on the paths z v · · · v and z w · · · w and hence T a cannot intersect T c , a contradiction to neighborliness of M . Thus y ∈ Φ .For part (c), it is enough to show that y Φ . If y ∈ Φ , then we see that Φ ⊆ u . HenceΦ ⊆ v and Φ ⊆ w . But then | u \ v | ≤
3. But this contradicts Lemma 5.9 along the path u u · · · v , as the four oriented edges −−→ u u , −−→ u u , −−→ u u , −−→ u v cannot all have distinct labels. Thisproves (c).For part (d), we again use the neighborliness of M . Suppose y = b . As before, the trees T a and T b must intersect along one of the paths z u · · · u , z v · · · v and z w · · · w . We see thatif y = b , T b does not contain any vertex from the path z w · · · w (As both z and w do notcontain b ). On the path z v · · · v , T b contains v , v , whereas T a contains z , v , v . Thus T a and T b must intersect along z u · · · u . Now y = b , implies only vertex on z u · · · u on T b is u . Hence T a must also contain u , and thus x = a . Thus ( x , x ) = ( b , a ). Similarly if y = c ,we can show that ( x , x ) = ( a , b ). ase 1: y ∈ { b , c } , y = b . From Claim (d), we must have ( x , x ) = ( b , a ) (see Figure7). Consider the tree T a . Notice that u ∩ Φ = { a } , and hence a v , w . Thus T a ⊆ Λ( M ) − { z , v , w } . In other words, T a is contained in the union of the path P = z u · · · u andthe arc A = w w · · · u · · · u . Since T a contains 4 vertices on P , it must induce a path containing7 vertices (including u ) on A . The only possible solution is T a = u u u u ∪ w w w u u u u .This gives us x = a , y = b . Putting y = x ∈ { b , c } , we get the following constraints( x , x , x , x , x , x ) = ( c , b , c , b , a , a ) , { x , x , x } = ( u \ v ) \{ a } , (14a)( y , y , y , y , y , y ) = ( a , a , a , b , x, b ) , { y , y , y } = ( v \ u ) \{ b } . (14b) z u v w u v w u v w u v w u v w u v w u v w u v w c b c b a a a a b x a b a Figure 7:
Illustration for Lemma 5.18, Case 1This case yeilds the following solution when examined using simpcomp [6] X = ( c , b , c , b , a , a , a , a , a ) , Y = ( a , a , a , b , b , b , b , c , c ) . The pair ( X , Y ) yeilds the complex N . Case 2: y ∈ { b , c } , y = c . This is similar to Case 1, except that y = c and hence from Claim(d), ( x , x ) = ( a , b ). Letting y = x ∈ { b , c } , we get the constraints,( x , x , x , x , x , x ) = ( c , b , c , a , b , a ) , { x , x , x } = ( u \ v ) \{ a } , (15a)( y , y , y , y , y , y ) = ( a , a , a , c , b , b ) , { y , y , y } = ( v \ u ) \{ b } . (15b)Using simpcomp [6] we get the following solution for this case X = ( c , b , c , a , b , a , a , c , a ) , Y = ( a , a , a , c , b , b , b , c , b ) . The pair ( X , Y ) gives the complex N . Case 3: y ∈ { b , c } , y = b . From Claim (d), we have ( x , x ) = ( b , a ). We notice that u ∩ Φ = { a } , and hence a z ∪ v ∪ w . Thus T a is contained in the union of the arc A = w w · · · u · · · u and the path P = u u · · · u . Since T a contains 5 vertices on the path P , itmust induce a 6 vertex path on A . We have the following possiblities for T a , T a = u u u u u ∪ w w u u u u ,T a = u u u u u ∪ w w w u u u . (16) u v w u v w u v w u v w u v w u v w u v w u v w c b c b a a a a b x a b a (a) Case 3.1 z u v w u v w u v w u v w u v w u v w u v w u v w c b c b a a a a b x a b a (b) Case 3.2 Figure 8:
Illustration for Lemma 5.18, Case 3For the first solution in (16), we get x = a and y = b (see Figure 8(a)). Setting y = x ∈{ b , c } , we have the following constraints,( x , x , x , x , x , x ) = ( c , b , c , b , a , a ) , { x , x , x } = ( u \ v ) \{ a } , (17a)( y , y , y , y , y , y ) = ( a , a , a , b , x, b ) , { y , y , y } = ( v \ u ) \{ b } . (17b)However, using simpcomp [6] we observe that the above constraints do not yeild any member of C .For the second solution in (16), we get x = a and y = b (see Figure 8(b)). Setting y = x ∈{ b , c } , we have the following constraints,( x , x , x , x , x , x ) = ( c , b , c , b , a , a ) , { x , x , x } = ( u \ v ) \{ a } , (18a)( y , y , y , y , y , y ) = ( a , a , a , b , x, b ) , { y , y , y } = ( v \ u ) \{ b } . (18b)Using simpcomp [6] we get the following solutions X = ( c , b , c , b , a , a , a , a , a ) , Y = ( a , a , a , b , b , b , c , c , b ) ,X = ( c , b , c , b , a , a , a , a , a ) , Y = ( a , a , a , b , b , b , c , c , b ) ,X = ( c , b , c , b , a , a , c , a , a ) , Y = ( a , a , a , b , c , b , c , b , b ) . The pairs ( X , Y ) , ( X , Y ) and ( X , Y ) yeild N , N and N respectively. Case 4: y ∈ { b , c } , y = c . From Claim (d), we must have ( x , x ) = ( a , b ). The rest ofanalysis is the same as Case 3, where the possiblities for T a are given by (16). For the first solutionin (16), we have x = a and y = b . Setting y = x ∈ { b , c } , we have the constraints,( x , x , x , x , x , x ) = ( c , b , c , a , b , a ) , { x , x , x } = ( u \ v ) \{ a } , (19a)( y , y , y , y , y , y ) = ( a , a , a , c , x, b ) , { y , y , y } = ( v \ u ) \{ b } . (19b)We obtain no solutions for members of C meeting above constraints.For the second solution for T a in (16), we have x = a and y = b . Setting y = x , we havethe constraints,( x , x , x , x , x , x ) = ( c , b , c , a , b , a ) , { x , x , x } = ( u \ v ) \{ a } , (20a)( y , y , y , y , y , y ) = ( a , a , a , c , x, b ) , { y , y , y } = ( v \ u ) \{ b } . (20b)sing simpcomp [6] we obtain the following solutions for members in C X = ( c , b , c , a , b , a , c , a , a ) , Y = ( a , a , a , c , b , b , c , b , b ) ,X = ( c , b , c , a , b , c , a , a , a ) , Y = ( a , a , a , c , b , b , b , c , b ) ,X = ( c , b , c , a , b , c , c , a , a ) , Y = ( a , a , a , c , c , b , b , b , b ) . The pairs ( X , Y ), ( X , Y ) and ( X , Y ) yeild the complexes N , N and N , respectively. Theabove cases complete the proof of the lemma. Lemma 5.19. If M ∈ C is minimal then, Λ( M ) = G , .Proof. Assume that M is a minimal member of C with Λ( M ) ∼ = G , . Let ( X, Y ) be the pairassociated with M . Now by Lemma 5.8, we have { x , x , x , x , x , x } = z = Φ ∪ Φ . Let p, q, r be such that { x p , x q , x r } = Φ . Then it can be seen that T a contains p + q + r − i, j, k be such that { x i , x j , x k } = Φ . Now T a contains i + j + k − T a , T a each contain 10 vertices, we must have p + q + r + i + j + k = 24. However, p, q, r, i, j, k is a permutationof 1 , , . . . ,
6, and hence we must have p + q + r + i + j + k = 21, a contradiction. This proves thelemma. Proof of Theorem 3.2.
The proof follows from Lemmata 5.6, 5.16, 5.17, 5.18, 5.19.
Proof of Theorem 3.3.
The proof follows from Theorem 3.2 and Proposition 2.4.
Acknowledgement:
The author would like to thank Basudeb Datta for useful comments andsuggestions. The author would also like to thank ‘IISc Mathematics Initiative’ and ‘UGC Centre forAdvanced Study’ for support.
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Graph Theory , New York, Springer, 2008.[4] B. Datta, N. Singh, An infinite family of tight triangulations of manifolds, arXiv:1210.1045v2,2013, 17 pages.[5] F. Effenberger, Stacked polytopes and tight triangulations of manifolds,
J. Combin. Theory (A) (2011), 1843-1862.[6] F. Effenberger, J. Spreer, simpcomp – a
GAP toolkit for simplicial complexes
Invent. math. (1987), 125–151.[8] W. K¨uhnel, Higher dimensional analogues of Cs´asz´ar’s torus, Results in Mathematics (1986),95–106.[9] F. H. Lutz, T. Sulanke, E. Swartz, f -vector of 3-manifolds, Electron. J. Comb. (2009), Adv. in Math. (2009), 2059–2084.[11] N. Singh, Non-existence of tight neighborly triangulated manifolds with β = 2, Adv. in Geom. (to appear).[12] D. W. Walkup, The lower bound conjecture for 3- and 4-manifolds,
Acta Math. (1970),75–107. ppendix A
We give a proof for the graph theoretic Lemma 5.3. For standard terminology on graphs see [3]. Fora vertex v in a graph G , d G ( v ) will denote the degree of the vertex v in G . For vertices u, v in G , d G ( u, v ) will denote the length of a shortest path between u and v in G . For a vertex a ∈ V ( G ) anda subset B of V ( G ), a path v v · · · v k such that v = a and { v , v , . . . , v k } ∩ B = { v k } is called an a - B path. The following is an easy consequence of the fan lemma (cf. [3, Chapter 9]). Lemma 5.20.
Let G be a two connected graph and let B ⊆ V ( G ) . If a B and | B | ≥ then, thereexist two a - B paths in G , which intersect only in a . We prove the following generalization of Lemma 5.3.
Theorem 5.21.
Let graphs G r,s and T r,s be as defined in Examples . and . , respectively. Let G be a two connected graph on n vertices with n + 2 edges. If Aut( G ) ⊇ Z then G ∼ = G r,s for some r, s > with r + s ) = n + 2 or G ∼ = T r,s for some r, s > with r + s = n .Proof. Let ϕ be an order three automorphism of G . Let Fix( ϕ ) = { v ∈ V ( G ) : ϕ ( v ) = v } denotethe set of vertices fixed by the automorphism ϕ . Let T be the set of vertices with degree three ormore in G . Since G is two connected, we have d G ( v ) ≥ v ∈ V ( G ). Then from the identity X v ∈ V ( G ) d G ( v ) = 2( n + 2) = 2 n + 4 , (A1)it follows that | T | ≤
4. We have the following cases:
Case 1: T Fix( ϕ ). Let u ∈ T be such that ϕ ( u ) = u . Let v = ϕ ( u ) and w = ϕ ( u ). As ϕ is anautomorphism of G , we have d G ( u ) = d G ( v ) = d G ( w ). Let k be the degree of u, v and w in G . Clearly k ≥
3. Now from (A1), it follows that k = 3, and there exists z
6∈ { u, v, w } with d G ( z ) = 3. Since ϕ orbits are either singleton or three element subsets and | T | ≤
4, we have ϕ ( z ) = z , or z ∈ Fix( ϕ ).Thus we have, d G ( z, u ) = d G ( z, v ) = d G ( z, w ) = r for some r ≥
1. Let p zu := zu · · · u r (= u ) be ashortest z - u path in G . Then p zv := zv · · · v r (= v ) and p zw := zw · · · w r (= w ) are z - v and z - w pathsrespectively, where v i = ϕ ( u i ) and w i = ϕ ( u i ) for 1 ≤ i ≤ r . As G is two connected, G ′ := G − z is a connected graph. Note that ϕ is an automorphism of G ′ . Let p uv := u r u r +1 · · · u r + s − v r be ashortest u - v path in G ′ , where s = d G ′ ( u, v ) = d G ′ ( v, w ) = d G ′ ( w, u ). Let p vw := v r v r +1 · · · v r + s − w r and p wu := w r w r +1 · · · w r + s − u r where v i = ϕ ( u i ) and w i = ϕ ( u i ) for r ≤ i ≤ r + s −
1. We claimthe following:(a) p zu ∩ p zv = p zv ∩ p zw = p zw ∩ p zu = { z } .(b) p uv ∩ p vw = { v } , p vw ∩ p wu = { w } and p wu ∩ p uv = { u } .(c) p zu ∩ p uv = p zu ∩ p wu = { u } , p zv ∩ p uv = p zv ∩ p vw = { v } and p zw ∩ p vw = p zw ∩ p wu = { w } .(d) p zu ∩ p vw = p zv ∩ p wu = p zw ∩ p uv = ∅ .We first prove (a). Let i > u i ∈ p zu ∩ p zv . Then u i = v j for some1 ≤ j ≤ r . Since i = d G ( z, u i ) = d G ( z, v j ) = j , we have i = j . Because u r = u = v = v r , wehave i < r . Further, as d G ( z, w ) = r > i , we have u i = w . Thus u i
6∈ { z, u, v, w } , and hence d G ( u i ) = 2. However by maximality of i , we have { v i − , v i +1 , u i +1 } as three distinct neighbors of u i , a contradiction. Therefore p zu ∩ p zv = { z } , and similar argument works for other pairs. Thisproves (a). Claim (b) can be proved in a manner similar to Claim (a).To prove Claim (c), we first show that p zu ∩ p uv = { u } . Clearly z p uv as p uv ⊆ G − z . Let0 < i < r be maximum such that u i is a vertex on p uv . By Claim (a), u i
6∈ { z, u, v, w } , and hence d G ( u i ) = 2. Again maximality of i implies that u i +1 is distinct from the two neighbors of u i on p uv ,a contradiction. Therefore p zu ∩ p uv = { u } . Similarly, we can show for other pairs. This proves (c).Claim (d) can be proved in a manner similar to Claim (c).Define the subgraph H := p zu ∪ p zv ∪ p zw ∪ p uv ∪ p vw ∪ p wu . bserve that d H ( v ) = d G ( v ) for all v ∈ V ( H ). Since G is connected, this implies G = H . It can beseen that G ∼ = G r,s and 3( r + s ) = n + 2. Case 2: T ⊆ Fix( ϕ ). Let u u · · · u r be a maximal path in G − Fix( ϕ ). Let x, y be neighbors of u and u r in G respectively, which are not on the path (such neighbors exist as d G ( v ) ≥ v ∈ V ( G )). By maximality of the path, we conclude that x, y ∈ Fix( ϕ ). Let p u := xu · · · u r y . Then,observe that p v := xv · · · v r y and p w := xw · · · w r y are also x - y paths in G , where v i = ϕ ( u i ) and w i = ϕ ( u i ) for 1 ≤ i ≤ r . We note that x = y , for otherwise we would have d G ( x ) = 6 and it canbe seen that G − x cannot be connected in this case ( G − x would be a graph on n − n + 2 − n − H := p u ∪ p v ∪ p w . We claim the following:(a) Paths p u , p v and p w are vertex-independent.(b) If V ( H ) = V ( G ) then G = H + xy . Further G ∼ = T r, .(c) If V ( H ) = V ( G ) then G = H ∪ p where p u , p v , p w and p are vertex-independent x - y paths.Further G ∼ = T r,s where s is the number of vertices in the path p .We first prove (a). Assume that p u and p v intersect in a vertex other than x or y . Let i be maximumsuch that u i ∈ p v . Then u i = v j for some 1 ≤ j ≤ r . Since u r = v r , we have min( i, j ) < r .Without loss assume i < r . Since T ⊆ Fix( ϕ ) and u i Fix( ϕ ), we have d G ( u i ) = 2. But we seethat v j − , v j +1 , u i +1 are three distinct neighbors of u i , a contradiction. Therefore p u and p v arevertex-independent. Similarly we can prove for other pairs.To prove (b), assume that V ( H ) = V ( G ). From Claim (a), we see that d H ( v ) = d G ( v ) = 2 for v ∈ V ( G ), v
6∈ { x, y } . It can be seen that to satisfy ((A1)), we must have G = H + xy . In this case,it is readily seen that G ∼ = T r, and 3 r + 2 = n . This proves (b).To prove (c), assume that V ( H ) = V ( G ). Let a ∈ V ( G ) \ V ( H ). By Lemma 5.20, there exist two a - V ( H ) paths p and p in G such that p ∩ p = { a } . Since d G ( v ) = 2 for all v ∈ V ( H ) \{ x, y } ,we conclude that the paths p and p meet H in vertices x and y . Without loss, let p be an a - x path and p be an a - y path. Then p = p ∪ p is an x - y path. Clearly p is vertex-independentto p u , p v and p w . Let H = H ∪ p . Then we observe that d H ( v ) = 2 for v ∈ V ( H ) \{ x, y } and d H ( x ) = d H ( y ) = 4. From (A1), it follows that d H ( v ) = d G ( v ) for all v ∈ V ( H ). Since G isconnected, we have G = H . It can be seen that G ∼ = T r,s in this case, where s is the number ofvertices in p . Further we have, 3 r + s = nn