Lower bound for Buchstaber invariants of real universal complexes
aa r X i v : . [ m a t h . A T ] J a n Lower bound for Buchstaber invariants of real universalcomplexes
Qifan Shen
Abstract.
In this article, we prove that Buchstaber invariant of 4-dimensional real universalcomplex is no less than 24 as a follow-up to the work of Ayzenberg [2] and Sun [14]. Moreover,a lower bound for Buchstaber invariants of n -dimensional real universal complexes is given asan improvement of Erokhovet’s result in [7]. Moment-angle complex and its real counterpart are fundamental objects in toric topologyas they construct links among algebraic geometry, sympletic geometry and combinatorics (seeDefinition 1). Moreover, they are equipped with certain group actions, yielding applications inboth non-equivariant and equivariant categories (see [4] for more details).For a given simplicial complex K on m vertices, the associated real moment-angle complex R Z K (resp. moment-angle complex Z K ) admits a natural Z m -action (resp. T m -action) bycoordinate-wise sign permutation (resp. rotation). However, these actions fail to be free unless K is the empty complex, leading to the definition of real Buchstaber invariant s R ( K ) (resp. Buchstaber invariant s ( K )) as the maximal rank of subgroup (resp. toric subgroup) that actsfreely on R Z K (resp. Z K ) (see Definition 2). These two types of invariants measure the degreeof symmetry of the corresponding complexes and was first introduced in [3] for simplicial sphereswith generalization in [8] for arbitrary simplicial complexes.Buchstaber asked for a combinatorial description of s ( K ) in [3], which turns out to be quitehard and remains open till today. As a matter of fact, calculation of s ( K ) is not completed evenfor the special case where K is dual cyclic polytopes and partial results can be found in [7].On the other hand, there exists a general bound for s R ( K ) and s ( K ): m − γ ( K ) ≤ s ( K ) ≤ s R ( K ) ≤ m − n where K is an ( n − m vertices and γ ( K ) stands for ordinarychromatic number of K . This formula can be derived from relations among generalized chromaticnumbers in a systematic manner, as shown in [1]. Indeed, with the help of real universal complex K n and universal complex K n introduced in [6], s R ( K ) and s ( K ) can be expressed as m − r R ( K )and m − r ( K ) respectively, where r R ( K ) and r ( K ) are minimal rank of certain colorings on K (see Section 2, also see [10]). Moreover, the upper bound of s ( K ) and s R ( K ) can be controlledby sum of rational Betti numbers since 2 m − n ≤ P i dim H i ( Z K ; Q ) = P i dim H i ( R Z K ; Q ) wasproved as a special case of Halperin-Carlsson conjecture (see [5] and [15]).The general inequality s ( K ) ≤ s R ( K ) follows from the fact that involutions on T m and Z K induced by complex conjugation have fixed point sets Z m and R Z K respectively. Thus, any free Mathematics Subject Classification (2020): 57S25, 52B05, 05E45Key words and phrases: Buchstaber invariant, Universal complex, Lower bound, Lifting problemPartially supported by the grant from NSFC (No. 11971112). r -action on Z K induces a free Z r -action on R Z K . Meanwhile, the special case s ( K ) = s R ( K )is closely related to Lifting problem (see Section 2) presented by L¨u at the conference on torictopology held in Osaka in November 2011 . Denote the difference ∆( K ) = s R ( K ) − s ( K ) = r ( K ) − r R ( K ), then the vanishing of ∆( K ) is necessary for the validity of Lifting problem on K .For real universal complex K n , we have the following monotonicity and universal property: Proposition 1 (Theorem 3.3 in [14]) ∆( K n ) ≤ ∆( K n +11 ). Proposition 2 (Proposition 4.1 in [14]) r R ( K ) ≤ r R ( K n ) = ⇒ ∆( K ) ≤ ∆( K n ).Therefore, the value of ∆( K n ) is significant as it gives out an upper bound for general cases. In[1] and [2], Ayzenberg showed that ∆( K n ) = 0 for n = 1 , , K ) > K ) = 1 was confirmed by Sun. Furthermore, an upper bound ∆( K n ) ≤ · n − − − n for n ≥ K n ), which can bededuced from construction of certain colorings. Since s ( K n ) + ∆( K n ) = 2 n − − n by definition(see Section 2), upper bound estimation of ∆( K n ) is equivalent to lower bound estimation of s ( K n ). Explicit construction of non-degenerate simplicial maps from K to K and from K n to K n − +12 for n ≥ Theorem 1 ∆( K ) ≤ , i.e., s ( K ) ≥ . Theorem 2
For n ≥ , ∆( K n ) ≤ n − + 1 − n , i.e., s ( K n ) ≥ · n − − . Remark 1
The construction process is equivalent to finding the solution to a system of non-linear Diophantine equations. These equations all belong to a certain type discussed in [12],where existence and classification problem of solutions to a single equation was solved. How-ever, existence problem of solutions to the system is much harder to deal with since the numberof equations grow rapidly as n increases. Remark 2
Vanishing of ∆( K n ) for n = 1 , , GL (3 , Z ) has integral determinant ±
1. For general n , upper bound estimation of ∆( K n ) isalso related to determinant calculation in both Z and Z . An upper bound ( n +1) ( n +1)2 n for abso-lute value of integral determinant of matrix in GL ( n, Z ) was given in [9]. This is a special caseof Hadamard maximum determinant problem which aims to calculate the maximal determinantof a square matrix with elements restricted in a given set S . It should be pointed out that thisproblem is far from being solved even for the simplest case S = { , } since whether the boundgiven above is sharp or not remains unknown except for n + 1 being power of 2. Computa-tional results in low dimensions ( n ≤
9) with the aid of computer were listed in [16] and recenttheoretical progress can be found in [13].This article is organized as follows. In Section 2, basic definitions and notations are listed.Section 3 is divided into four parts to prove Theorem 1 step by step. Section 4 includes theproof of Theorem 2 with an illustrative example. Preliminaries
In the first place, we shall give out the formal definition of (real) moment-angle complex and(real) Buchstaber invariant.
Definition 1
Given a simplicial complex K on [ m ] = { , . . . , m } , we can define the real moment-angle complex R Z K and the moment-angle complex Z K associated to K : R Z K = [ I ⊂ K ( D , S ) I ⊆ ( D ) m Z K = [ I ⊂ K ( D , S ) I ⊆ ( D ) m where ( X, A ) I = { ( x , . . . , x m ) ∈ X m , x i ∈ A if i / ∈ I } for A ⊆ X . Definition 2
For R Z K and Z K associated to a simplicial complex K on [ m ]:(1) The real Buchstaber invariant s R ( K ) is the maximal rank of a subgroup H ⊆ Z m s.t. therestricted action H y R Z K is free;(2) The Buchstaber invariant s ( K ) is the maximal rank of a toric subgroup G ⊆ T m s.t. therestricted action G y Z K is free. Example 1
Let K be the boundary of a square with vertices labeled as 1 , , , Z K = ( D × S ∪ S × D ) × ( D × S ∪ S × D ) = S × S while R Z K = ( D × S ∪ S × D ) × ( D × S ∪ S × D ) = S × S . Moreover, s ( K ) = s R ( K ) = 2follows from the fact that γ ( K ) = dim K + 1 = 2.Secondly, we introduce the (real) universal complex and the corresponding coloring to getan equivalent expression of (real) Buchstaber invariant. Definition 3
Let R nd = Z n when d = 1 and R nd = Z n when d = 2. The simplicial complex K nd is defined on the set of primitive vectors in R nd as follow:[ v , . . . , v k ] is a simplex of K nd ⇐⇒ { v , . . . , v k } is part of a basis of R nd . K n is called real universal complex while K n is called universal complex . Definition 4 A R rd -coloring on a simpicial complex K is defined as a non-degenerate simplicialmap λ : K → K rd . The non-degenerate condition means λ is an isomorphism on each simplex of K . Denote r R ( K ) as the minimum value of r s.t. there exists a R r -coloring on K . Anagolously, r ( K ) is the minimum value of r s.t. there exists a R r -coloring on K . Then it follows fromdefinition that r R ( K n ) = r ( K n ) = n . In addition, equivalent expressions s R ( K ) = m − r R ( K )and s ( K ) = m − r ( K ) were first proved in [10]. Since there are 2 n − K n ,we have s R ( K n ) = s ( K n ) + ∆( K n ) = 2 n − − n .With notations above, we can formally state the Lifting problem as follow: Lifting Problem (Remark 6 in [11])
For any given simplicial complex K and non-degeneratesimplicial map f : K → K s R ( K )1 , does there exist a lifting map e f : K → K s R ( K )2 s.t. the diagrambelow is commutative: K s R ( K )2 π (cid:15) (cid:15) K e f = = ④④④④④④④④④ f / / K s R ( K )1 where π : K s R ( K )2 → K s R ( K )1 is natural modulo 2 projection. Lower bound for s ( K ) The statement of Theorem 1 is equivalent to r ( K ) ≤
7, i.e., there exists a vertex mapΛ : vt ( K ) → vt ( K ) which induces a non-degenerate simplicial map from K to K . Indeed, itis even possible to construct Λ with additional restrictions: p j ◦ Λ = id j for j ∈ { , , , , } where p j is the projection onto the j th coordinate of Z and id j is the identity map of the j th coordinate.Let e represent the identity of Z and 1 represent the identity of Z to avoid confusion. Aslisted in the table below, we can take a partition vt ( K ) = V ⊔ V ⊔ V ⊔ V ⊔ V s.t. V i consistsof primitive vectors with (5 − i ) zeros and label 31 elements of vt ( K ) in lexicographic order. V V V V v = ( e, , , , v = ( e, e, , , v = ( e, e, e, , v = ( e, e, e, e, v = (0 , e, , , v = ( e, , e, , v = ( e, e, , e, v = ( e, e, e, , e ) v = (0 , , e, , v = ( e, , , e, v = ( e, e, , , e ) v = ( e, e, , e, e ) v = (0 , , , e, v = ( e, , , , e ) v = ( e, , e, e, v = ( e, , e, e, e ) v = (0 , , , , e ) v = (0 , e, e, , v = ( e, , e, , e ) v = (0 , e, e, e, e ) v = (0 , e, , e, v = ( e, , , e, e ) v = (0 , e, , , e ) v = (0 , e, e, e, v = (0 , , e, e, v = (0 , e, e, , e ) v = (0 , , e, , e ) v = (0 , e, , e, e ) v = (0 , , , e, e ) v = (0 , , e, e, e ) V : v = ( e, e, e, e, e )Denote φ = p ◦ Λ and ψ = p ◦ Λ for simplicity. Suppose for each i ∈ { , . . . , } , φ ( v i ) = s i ∈ Z and ψ ( v i ) = t i ∈ Z , then Λ( v i ) = ( v i , s i , t i ) ∈ Z . Furthermore, define Φ( v i ) = ( v i , s i ) ∈ Z and Ψ( v i ) = ( v i , t i ) ∈ Z that will be useful in the construction process. For convenience, weassume all vectors are understood as column vectors and det Z (-) represents determinant takenin Z while det(-) represents determinant taken in Z for the rest of this article.By the restriction of non-degenerate condition, it remains to verify that: ∀ { v i , v i , v i , v i , v i } ⊆ vt ( K ) satisfying det Z ( v i , v i , v i , v i , v i ) = e, ∃ α = ( a , . . . , a ) T ∈ Z and β = ( b , . . . , b ) T ∈ Z with the property : ( ⋆ )det(Λ( v i ) , Λ( v i ) , Λ( v i ) , Λ( v i ) , Λ( v i ) , α , β ) = ± . Denote α ′ = ( a , . . . , a , a ) T , β ′ = ( b , . . . , b , b ) T , A = ( v i , v i , v i , v i , v i ) and ( A ) j asthe matrix A with j th row replaced by ( s i , s i , s i , s i , s i ), ( A ) j as the matrix A with j th rowreplaced by ( t i , t i , t i , t i , t i ), then by basic linear algebra:det(Φ( v i ) , Φ( v i ) , Φ( v i ) , Φ( v i ) , Φ( v i ) , α ′ )= a det A − a det( A ) − a det( A ) − a det( A ) − a det( A ) − a det( A ) ;det(Ψ( v i ) , Ψ( v i ) , Ψ( v i ) , Ψ( v i ) , Ψ( v i ) , β ′ )= b det A − b det( A ) − b det( A ) − b det( A ) − b det( A ) − b det( A ) .
4y Chinese Remainder Theorem, the existence of α ′ for the first determinant being ± g.c.d. (det A, det( A ) , det( A ) , det( A ) , det( A ) , det( A ) ) = 1 . ( ⋆ β ′ for the second determinant being ± g.c.d. (det A, det( A ) , det( A ) , det( A ) , det( A ) , det( A ) ) = 1 . ( ⋆ ⋆
1) or ( ⋆
2) holds, then taking β = e or α = e as standard basis of Z yields the validityof ( ⋆ ). Specifically, there is nothing to prove if det( A ) = ± | det A | ≤ A ∈ GL (4 , Z ) and | det A | ≤ A ∈ GL (5 , Z ).In order to discuss 5-dimensional case, two lemmas in 4-dimensional case are needed. Lemma 1 (Lemma 3.2 in [14])
For M ∈ GL (4 , Z ) regarded as an integral matrix, det M = ± imples that: M = or after necessary permutation of rows and columns. Lemma 2
For M ∈ GL (4 , Z ) regarded as an integral matrix, if det M = ± , then M equals toone of the following types after necessary permutation of rows and columns: (1) x x x (2) (3) y y where x , x , x and y , y belong to { , } . Proof.
Similar to 5-dimensional case, we can take a partition vt ( K ) = W ⊔ W ⊔ W ⊔ W s.t. W i consists of primitive vectors with (4 − i ) zeros and label 15 elements of vt ( K ) in lexicographicorder. Define two matrices as equivalent if they differ from each other by permutation of rowsand columns. We analyze the columns of M = ( λ , λ , λ , λ ) in the sequel: Case 1.
There exists λ i ∈ W , then we can assume λ = (1 , , , T by equivalence. In thisway, M belongs to type (1) since (cid:16) (cid:17) is the only (3 ×
3) binary matrix with absolute value5f determinant equal to 2 up to equivalence.
Case 2.
There exists λ i ∈ W , then we can assume λ = (1 , , , T by equivalence. Note thatin this case the other column vectors can not belong to W since that is contradictory to Case 1 .Moreover, if λ ∈ W , then substract λ from λ will lead to contradiction by the same reason.Therefore, all three other vectors belong to W and type (2) is obtained after taking equivalence. Case 3.
According to Lemma 1, one column vector belongs to W and others belong to W ⊔ W in the remaining case. We call λ i is contained in λ j if λ ik ≤ λ jk is valid for every k . If thishappens, then we can assume λ = (1 , , , T and λ = (0 , , , T up to equivalence anddirect computation gives out the second matrix in type (3). Otherwise, every column vectorsmust belong to W , which leads to the first matrix of type (3). ✷ GL (5 , Z ) with integral determinant ± Claim 1 ∀ M = ( λ , λ , λ , λ , λ ) ∈ GL (5 , Z ) with integral determinant ± , there is a (4 × minor M ij corresponding to element m ij = 1 s.t. det M ij = ± . Proof.
If there exists one row or one column with more than two zeros, then expansion byminors on that row or column leads to a minor corrsponding to element 1 with determinant ±
3. Otherwise, after certain permutation of columns, it can be assumed that λ ∈ V ⊔ V isthe column with most zeros. If λ ∈ V , then M consists of all five elements in V or any fourelements in V plus v , neither of which has determinant ±
5. Now suppose λ ∈ V and allminors of M corresponding to element 1 do not equal to ±
3, then expansion on λ shows thatthe minors corresponding to element 1 in λ are { , , } up to equivalence. Combine Lemma 2and restrictions above, the only possible choice for M is: y y l l l n n n with y , y , l , l , l , n , n , n ∈ { , } .Note that if n = 1, then subtract λ from λ leads to contradiction. In addition, if l = 0,then the sum of row 2,3 and 4 is (2,2,2,2,2), leading to contradiction again. Therefore, M musthave the following type: y y
11 1 1 0 01 1 0 1 00 0 1 1 10 1 n n with y , y , n , n ∈ { , } .Subtracting λ from λ shows either minor M or minor M equals to ±
3, resulting in con-tradiction anyway. ✷ With the help of Claim 1 above, 5-dimensional binary matrices with integral determinant ± ±
3. As a matter of fact, there are only three equivalent classes in total.6 laim 2 If M = ( λ , λ , λ , λ , λ ) ∈ GL (5 , Z ) has integral determinant ± , then M isequivalent to one of the following three matrices: M = M = M = Proof.
Since elements in GL (4 , Z ) can not have integral determinant ±
5, none of the columnbelongs to V . It turns out for the same reason, none of the column belongs to V either. If not,fix λ = v , then the other columns do not belong to V and any two column vectors λ i , λ j belonging to V share one common zero coordinate since otherwise basic column operationswill lead to a 5-dimensional binary matrix with one column belonging to V . The remainingcases are equivalent to either λ ∈ V or { λ , λ , λ , λ } ⊆ V pairwise sharing a commonzero coordinate. In the former case, expansion by minors on λ yields contradiction while inthe latter case, there exists one row with four zeros, leading to contradition as well. Since thematrix with all five columns in V has determinant ±
4, we can assume that λ ∈ V ⊔ V is thecolumn with most zeros.If λ ∈ V , then combining Lemma 1 and Claim 1, M is equivalent to: n n n n or n ′ n ′ n ′ n ′ or n ′′ n ′′ n ′′ n ′′ with { n i , n ′ i , n ′′ i } i =1 belong to { , } . Direct computation of determinant yields M for the firstmatrix and M for the third matrix while there is no solution for the second matrix above.Similarly, if λ ∈ V , then M is equivalent to: n n n n or n ′ or n ′′ with n , n , n , n , n ′ , n ′′ ∈ { , } . Direct computation of determinant yields M for the firstmatrix while there is no solution for the second and third matrix above. ✷ GL (5 , Z ) with integral determinant ± In this subsection, suppose det M = ± λ having most zeros and m = 1 by equiva-lence. According to the proof of Claim 2, it is necessary for λ ∈ V ⊔ V ⊔ V here. The followingdiscussion is based on expansion by minors on λ with argument similar to Lemma 2. Class 1 λ ∈ V . 7y assumption and Lemma 1, λ = v and M is equivalent to one of the following types: N a = n n n n N b = n ′ n ′ n ′ n ′ with { n i , n ′ i } i =1 belong to { , } . Class 2 λ ∈ V .The corresponding minors of element 1 in λ are either ± (0 ,
3) or ± (1 ,
2) and it can be assumedthat M = ± ± M is equivalent to one ofthe following types: N a = n n n n N b = n ′ n ′ n ′ n ′ N c = n ′′ n ′′ n ′′ n ′′ with { n i , n ′ i , n ′′ i } i =1 belong to { , } satisfying n + 2 n = n + n , n ′ + n ′ + n ′ = n ′ and n ′′ + n ′′ = 2 n ′′ + n ′′ respectively.In the latter case, type (1) matrix in Lemma 2 gives out four types of equivalent classes for M : N a = N b = x x
00 1 1 0 00 1 0 1 00 0 1 1 11 1 n n N c = x ′ x ′
00 1 1 0 00 1 0 1 00 0 1 1 11 0 n ′ n ′ N d = x ′′ x ′′
00 1 1 0 00 1 0 1 00 0 1 1 11 1 n ′′ n ′′ with { x i , x ′ i , x ′′ i } i =2 and { n j , n ′ j , n ′′ j } j =3 belong to { , } satisfying x + x = n + n , x ′ + x ′ = n ′ + n ′ + 1 and x ′′ + x ′′ = n ′′ + n ′′ + 1 respectively.Similarly, the first matrix of type (2) in Lemma 2 gives out two types of equivalent classes for M : N a = N b = n n n n with { n i } i =1 belong to { , } satisfying n + n + 1 = n + n .While the second matrix of type (2) in Lemma 2 gives out three types of equivalent classes for M :8 a = n N b = N c = with n ∈ { , } .Similarly, the first matrix of type (3) in Lemma 2 gives out two types of equivalent classes for M : N a = n N b = n ′ n ′ n ′ with { n ′ , n ′ , n ′ , n } belong to { , } satisfying n ′ + 1 = n ′ + n ′ .And the second matrix of type (3) in Lemma 2 gives out two types of equivalent classes for M : N a = y y
01 1 1 0 01 1 0 1 00 0 1 1 1 n n n n N b = y ′ y ′
01 1 1 0 01 1 0 1 10 0 1 1 0 n ′ n ′ n ′ n ′ with { n i , n ′ i } i =1 and { y j , y ′ j } j =1 belong to { , } satisfying ( n − n − n ) + ( y + y )( n − n ) = 1and ( n ′ − n ′ − n ′ ) + ( y ′ − y ′ )( n ′ − n ′ ) = 1 respectively. Class 3 λ ∈ V .This class can be further divided into three cases with regard to the largest absolute value ofminors corresponding to element 1 in λ .If the minor equals to ±
3, then there are two types of equivalent classes induced by matrices inLemma 1: N a = n n n n N b = with { n i } i =1 belong to { , } satisfying n + n = 2 n + 2 n .For the other two cases, argument can be simplified by the following two claims: Claim 3 If M ∈ GL (5 , Z ) has integral determinant ± and λ ∈ V as the column with mostzeros s.t. there exists a minor corresponding to element 1 in λ equal to ± , then submatrixcorresponding to this minor must be the second matrix of type (3) in Lemma 2. Proof.
Without loss of generality, we can suppose det M = ±
2, then a straightforward checkon matrices listed in Lemma 2 verifies this claim.For type (1), λ has more zeros than λ , which is not allowed. For type (2), the sum offirst row to fourth row is (4,2,2,2,2), leading to a contradiction against det( M ) = ±
3. The firstmatrix in type (3) can not appear for the same reason. ✷ laim 4 If M ∈ GL (5 , Z ) has integral determinant ± and λ ∈ V as the column with mostzeros s.t. minors corresponding to element 1 in λ are ± (1 , , , then either M consists of fourcolumns in V plus v or M has a column in V contained in another column in V . Proof.
Since coexistence of an element in V and v is not allowed by determinant restriction,it suffices to discuss cases where M is consisted of elements all belong to V or V ⊔ V .In the former case, every element 1 in M must have minor ± λ = (0 , , , , T , λ = (0 , , , , T by equivalence,then the first row must be (1 , , , , v . Suppose λ = v = (1 , , , , T by equivalence again, then M must be equivalent to However, the (1 , V is no more than two, then there mustbe one column in V contained in another column belonging to V . If there are three columnsbelonging to V and none of them is contained in another column belonging to V , then theymust share two common nonzero coordinates i.e. M is equivalent to However, the (3 , −
3, resulting in contradiction. If four columnsof M is in V and none of them is contained in the rest column belonging to V , then we cansuppose λ = (0 , , , , T by equivalence. Expansion by minors on the first row leads to evendeterminant, which is not allowed. ✷ Combining Claim 3 and Claim 4 with Lemma 1, the remaining possible equivalent classesfor M can be listed below: N a = N b = n n n with { n i } i =1 belong to { , } . Remark 3
It should be pointed out that equivalent classes in this subsection may overlap. Forinstance, taking n ′ = n ′ = n ′ = 1 in N b leads to the same equivalent class as N a . In fact, thereare 51 equivalent classes according to the counting table in [16].10 .4 Construction of φ and ψ Now we are in the position to construct map φ and ψ for the validity of ( ⋆
1) or ( ⋆ φ and ψ to be constant on each V i , then it suffices to do verification on equivalent class level. Claim 5 φ ( v i ) = v i ∈ V v i ∈ V ⊔ V ⊔ V v i ∈ V and ψ ( v i ) = v i ∈ V ⊔ V v i ∈ V v i ∈ V ⊔ V guarantee the validityof ( ⋆ or ( ⋆ in all equivalent classes above. Proof.
The proof follows from straightforward calculation.For { M i } i =1 in Claim 2, det( M ) = −
2, det( M ) =2 and det( M ) = −
1, all coprime to ± ⋆ N a ) = − N b ) =2; det( N a ) = −
2, det( N b ) =1 and det( N c ) =2; det( N a ) = − N b ) = − −
1; det( N a ) = − N b ) = −
1; det( N a ) = − N b ) = − N a ) = − N b ) =1; det( N a ) = − N b ) =1. All these determinants arecoprime to ±
3, making ( ⋆
1) valid in these classes.The remaining parts are of N ∗ -type and N ∗ -type, in which ψ is needed. In fact, problem liesin N b , N c and N b since det( N a ) = − N d ) = − −
2; det( N a ) =1 and det( N c ) =1.For N b , if n + n = 0, then det( N b ) = − n + n = 1, then det( N b ) = −
2. However,det( N b ) j ≡ j ∈ { , , , , } if n + n = 2. Similarly for N c , if n ′ + n ′ = 0,then det( N c ) = − N c ) j ≡ n ′ + n ′ = 1. For N b , each det( N b ) j is alsodivisible by 3. To sum up, matrices that can not satisfy ( ⋆
1) with map φ defined above are: X = X = X = Note that X can be converted into X by row permutation σ = ( ) composed withcolumn permutation τ = ( ), i.e., they belong to the same equivalent class. Therefore,calculation results det( X ) = − X ) =1 complete the proof. ✷ Remark 4
The value of ψ on set V ⊔ V is irrelevant since matrices belonging to N ∗ -type and N ∗ -type do not include column vectors in V ⊔ V . Remark 5
The value of φ can be taken modulo 15 since only coprimeness to 3 and 5 is con-cerned. Similarly, the value of ψ can be taken modulo 3. s ( K n ) By Proposition 2, ∆( K n ) can be viewed as an upper bound for general cases. However, itremains open whether or not ∆( K n ) is bounded when n goes to + ∞ . On the other hand, bymapping 2 n − vt ( K n ) to standard basis of Z n − , one can easily verify that∆( K n ) ≤ n − − n . With some symmetric modifications, this upper bound can be improvedto 2 n − + 1 − n for n ≥
2, as stated in Theorem 2.11 roof.
Since n ≥
2, one can choose two arbitrary primitive vectors x , y ∈ vt ( K n ), then a parti-tion of vt ( K n ) is given by A = { x , y , x + y } and A i = { a i , a i + x , a i + y , a i + x + y } for i =1,. . . ,2 n − − Z . Define a vertex map Λ : vt ( K n ) → vt ( K n − +12 ) with the followingassignment: x e y e a i e i +2 x + y e + e a i + x e + e i +2 a i + y e + e i +2 a i + x + y e + e + e i +2 where { e j } n − +1 j =1 is the standard basis. It suffices to verify that Λ induces a non-degeneratesimplicial map e Λ from K n to K n − +12 . Apparently, for each simplex σ ∈ K n , A vt ( σ ). Denote ♯ ( X ) as the number of elements in set X for convenience, then there are three different cases: Case 1 ♯ ( A ∩ vt ( σ ))=2.By linear dependency, ♯ ( A i ∩ vt ( σ )) ≤ i ≥
1. Thus, images of vt ( σ ) are parts ofcolumns in the matrix equivalent to P = (cid:18) P ∗ I n − − (cid:19) where P =( ) or ( ) and I n − − stands for identity matrix of dimension 2 n − − Case 2 ♯ ( A ∩ vt ( σ ))=1.By linear dependency, there exists at most one index i ≥ ♯ ( A i ∩ vt ( σ ))=2 while ♯ ( A i ∩ vt ( σ )) ≤ i ≥
1. Take subtraction between columns if such i doexist, then images of vt ( σ ) are parts of columns in the matrix equivalent to Q = (cid:18) Q ∗ I n − − (cid:19) where Q =( ) or ( ) or (cid:0) −
10 1 (cid:1) . Case 3 ♯ ( A ∩ vt ( σ ))=0.Similar to Case 2 , either there exists at most one index j ≥ ♯ ( A j ∩ vt ( σ ))=3 while ♯ ( A j ∩ vt ( σ )) ≤ j ≥
1, or there are at most two indices j , j ≥ ♯ ( A j ∩ vt ( σ ))= ♯ ( A j ∩ vt ( σ ))=2 while ♯ ( A j ∩ vt ( σ )) ≤ j ≥
1. Take subtractionbetween columns if such j or j , j do exist, then images of vt ( σ ) can also be viewed as partsof columns in the matrix equivalent to Q .Since det( P ) = det( Q ) = 1, the induced map e Λ is non-degenerate as desired. ✷ Remark 6
In [7], Erokhovets gave an upper bound of r ( K ) for general simplicial complex K on[ m ] in terms of minimal non-simplices: r ( K ) ≤ P li =0 dim ω i if there exists a collection of minimalnon-simplices { ω i } li =0 such that ∪ li =0 ω i = [ m ]. Take K = K n and ω i = A i , then an upper bound3 · n − − − n is obtained for ∆( K n ). Theorem 2 can be regarded as an improvement of thisresult and it gives out sharp upper bound when n ≤ emark 7 Choosing one primitive vector in K n , an upper bound 2 n − − n can be obtained forany n ≥ K n are chosen at the beginning, then for anysimplex σ ∈ K n , the images of vt ( σ ) can be viewed as parts of columns in the matrix equivalentto R = (cid:18) R ∗ I n − − (cid:19) where R may equal to (cid:16) − − (cid:17) due to necessary column subtractions, leading to det( R )= − ±
1. Starting from choosing more linearly independent primitive vectors causes moreproblems like this.
Example 2
For n =4, take primitive vectors x , y as (1 , , , T and (0 , , , T respectively,then Λ : vt ( K ) → vt ( K ) is defined as follow: Since p j ◦ Λ = id j for j ≤
4, this map is different from the construction given in [14].
Acknowledgement
The author would like to thank professor Zhi L¨u for introducing this topic to him and makingvaluable discussions.
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