Yi SUN

1 Introduction

The Buchstaber invariant is an important combinatorial invariant of simplicial complexes.This number was first introduced by Buchstaber[3]to describe the maximal rank of a torus subgroup which acts freely on the moment-angle complex over a simple convex polytope.Later,Fukukawa and Masuda generalized this number to the case of finite simplicial complexes and 2-torus actions in[7].Therefore there exist two types of Buchstaber invariants:The complex(or ordinary)Buchstaber invariant s(K)and the real Buchstaber invariant sR(K).In a certain sense they measure the degree of symmetry of moment-angle complexes and real moment-angle complexes respectively.

These two kinds of numbers are closely related to the colorings or characteristic maps of a simplicial complex.In fact,a subgroup which acts freely defines a monomorphism,which corresponds to a coloring in such a way that they satisfy a short exact sequence(cf.Section 2).So Buchstaber invariants can be considered as an invariant of simplicial complexes,not only simple polytopes.

There is a general bound

where K is an(n−1)-dimensional simplicial complex with m vertices.

It is known from[1,Lemma 5]that s(K)=sR(K)for any simplicial complex K with dimK≤2.Moreover,if K is an(n−1)-dimensional simplicial sphere with n=2,3,then s(K)=sR(K)=m−n.For 2-dimensional spheres this follows from the four color theorem.

However,the calculation of Buchstaber invariants in general is quite difficult.Even for the skeleta of a simplex Buchstaber invariants are not completely computed(cf.[7]).The readers are referred to[3,5]for some basic properties.

It is known that the complex conjugation both induces involutions on ZKand torus Tmwith RZKand 2-torus(Z2)mas the fixed point sets respectively,for every Kn−1with m vertices.This means that a torus subgroup of maximal rank which acts freely on ZKactually determines a 2-torus subgroup which also acts freely on RZKvia the above involutions,although it may not be of the maximal rank.This implies that s(K)≤sR(K).Then how about the converse?The following is the lifting problem.

Lifting ProblemConsider the above correspondence from torus subgroups of maximal rank that act freely onZKto2-torus subgroups that act freely onRZKas a mapΘ.Does its image contain all the2-torus subgroups of maximal rank?

For example,if there exist quasitoric manifolds over a simple convex polytope Pn,then there exist small covers,too.Denote all of the small covers and quasitoric manifolds over P by S(P)and Q(P)respectively.Passing to the real part,as described above,defines a map from Q(P)to S(P),since the fixed point set of the complex conjugation on a quasitoric manifold is a small cover.We may ask if this map is surjective(up to equivariant homeomorphism or equivariant cobordism).This is still unknown.

Remark 1.1 A quasitoric manifold or a small cover over a simple convex polytope Pnwith m codimension-1 faces can be seen as the quotient space of ZKor RZKunder free actions of some torus subgroup or 2-torus subgroup of rank m−n respectively.

Obviously,Δ(K)=sR(K)−s(K)is an obstruction of the lifting problem:If Δ(K)0,the image of Θ is not 2-torus subgroups of maximal rank.Davis and Januszkiewicz[4]introduced two classes of simplicial complexesandwhich have universal properties for the category of toric spaces which they studied in their paper.Moreover,they are also closely related to the calculation of the colorings and Buchstaber invariants(cf.[9]).Δ(K)can be controlled by Δ()for some n.That is,if K has a(Z2)n-coloring(cf.Section 2),then

Ayzenberg[2]showed that Δ() ≥ 1.Next is the main theorem of this paper about such universal complexesand

Theorem 1.1Letbe a real universal complex.

Using this fact,we can deduce the following corollary.

Corollary 1.1 Δ()≥ 1,n ≥ 4.

The article is organized as follows.In Section 2,the definitions of Buchstaber invariants and universal complexes are introduced.We investigate their basic properties and relations,which give another description of Buchstaber invariants.We calculate Δ()when n ≤ 4 and prove that Δ()is non-decreasing for n in Section 3.In Section 4 we discuss several further problems on Buchstaber invariants.

2 Buchstaber Invariants and Universal Complexes

In this section,we give the original definition of Buchstaber invariants and an equivalent description,which is related to universal complexes.

Let K be an abstract simplicial complex on the vertex set[m]={1,···,m}.We can define a(complex)moment-angle complex associated with K:

whereif i∈σ and Xi=S1if not.

Since(D2,S1)is invariant under the action of S1and Tm=(S1)macts in coordinate-wise manner on(D2)m,ZKadmits a torus action Tmas the restriction to the subspace of(D2)m.

definition 2.1The(complex)Buchstaber invariants(K)is the maximal dimension of a subtorus ofTmwhich acts freely onZK.

Similarly,we can define a(real)moment-angle complex:

which admits a(Z2)m-action.And therefore we can get the following definition.

definition 2.2The(real)Buchstaber invariantsR(K)is the maximal rank of a subgroup of(Z2)mwhich acts freely onRZK.

For any simple convex polytope Pn,there exist corresponding definitions of s(P)and sR(P)by s(P)=s(∂(P∗))and sR(P)=sR(∂(P∗)).Here ∂(P∗)is the boundary of the simplicial polytope dual of P.

Example 2.1 For n-simplex Δn(considered as a simplicial complex),there is no subgroup of Tn+1or(Z2)n+1which acts freely on ZΔn=(D2)n+1or RZΔn=(D1)n+1.Therefore s(Δn)=sR(Δn)=0.

Consider its boundary.By construction we have Z∂Δn= ∂(D2)n+1=S2n+1and the diagonal subgroup of Tn+1acts freely on Z∂Δn.There are no larger subgroups of Tn+1acting freely on Z∂Δn,thus s(∂Δn)=1.Similarly,sR(∂Δn)=1.

Now let us give the definition of colorings.Let

definition 2.3LetKbe an(n−1)-dimensional simplicial complex on the vertex set[m].A-coloring ofKis a mapλfrom the vertex set ofKtoRrd,such that forσ =[i1,···,ik]∈ K,the subspace spanned byλ(i1),···,λ(ik)is a direct summand in.Λ =(λ(1),···,λ(m))is called the coloring matrix.

Remark 2.1 In the case that K is a simplicial sphere dual to a simple convex n-polytope P,such an-coloring is usually called a “characteristic map” of P.

The existence of a characteristic map of P is directly related to Buchstaber invariants.

Theorem 2.1(cf.[3,Proposition 7.34])Pnadmits a characteristic map if and only ifs(P)=m−n,wheremis the number of the facets ofP.

Actually,let H⊂Tmbe a subtorus of dimension l.We can write it in the form

where ti∈ R,i=1,···,l.The integer m × l-matrix S=(sij)defines a monomorphism Zl→ Zm,whose image is a direct summand in Zm.H acts freely if and only if S~i1,···,~inwhich is obtained by deleting the rows i1,···,inof S defines a monomorphism to a direct summand for every vertex v=Fi1∩···∩Fin(cf.[3,Lemma 7.32]).And if l=s(P)=m−n,there is a short exact sequence

where Λ =(λ1,···,λm),λiis the facet vector for each facet.

Then for the same reason there holds the following proposition.

Proposition 2.1LetH⊂Tmbe a subtorus of dimensionm−r,written in the form of(2.1).The corresponding matrix is denoted byS.IfHacts freely onZK,there is aZr-coloring ofKwhose coloring matrixΛr×mfi ts in the short exact sequence

Conversely,ifKadmits aZr-coloring(r≤ m)andΛis the coloring matrix,then there is a subtorusHof dimensionm−rwhich acts freely,such that the above exact sequence holds.

This result also holds for the real case.So we can get an equivalent statement of the Buchstaber invariants.

Proposition 2.2(cf.[9])Denote the minimal integer for which there is anRrd-coloring byrR(K)ifd=1andr(K)ifd=2.Then

Let us give some examples.

Example 2.2 Suppose that P is a simple convex polytope.The subtorus

surely acts freely.So s(P)≥1.

Example 2.3 For m＞n,there is an n-dimensional simplicial convex polytope with m vertices denoted by Cn(m)and called a cyclic polytope,defined as the convex hull of m points on the curve v(t)=(t,t2,···,tn).Its boundary is a simplicial complex,still denoted by Cn(m).Let v1,···,vmbe its vertices,where vi=v(ti)for t1＜ ···＜ tm.Shephard[11]showed that[vi1,vi2,vi3,vi4]∈Cn(m),i1＜i2＜i3＜i4if and only if i1+1=i2,i3+1=i4or i1=1,i4=m,i2+1=i3.We can give a Z4-coloring matrix Λ(C4(m))of C4(m)for m=6,7.

So s(C4(6))=2,s(C4(7))=3.

Moreover,by the works of Erokhovets[6],we have some more general results about cyclic polytopes:

Next is the main part of this paper.Associating to Rnd,(n−1)-dimensional simplicial complexis defined as follows:

(1)The vertex set ofis P,the set of lines in

(2)A k-simplex inis a collection of k+1 lines{l0,l1,···,lk},li∈ P,which span a(k+1)-dimensional unimodular subspace of

Remark 2.2 A subspace A ofis unimodular if there exists another subspace B such that=A⊕B.

As shown in[4,p.429],determines a universal-space(Gd=Z2if d=1 and Gd=S1if d=2),soalso called the real universal complex if d=1,or the complex universal complex if d=2.

According to the definitions,an-coloring of K is equivalent to a nondegenerate simplicial map from K to,so we have rR()=n and r()=n.

Remark 2.3 A nondegenerate simplicial map is a simplicial map which restricts to an isomorphism on each simplex.There is a natural nondegenerate simplicial map from Kn2to Kn1,denoted by Φ,which is induced by the mod 2 map from Znto(Z2)nmapping each integer vector(a1,···,an)to(a1,···,an)mod 2.

Now we can describe the lifting problem as the following statement(also cf.[10,Remark 6]).

Let K be a simplicial complex of dimension n−1.Then there is a nongenerate simplicial map f:Does there exist a nongenerate simplicial mapsuch that the following diagram is commutative?

So we can see that Δ(K)=sR(K)−s(K)=r(K)− rR(K)is an obstruction of the lifting problem in the sense that Δ(K)0 implies that there is no lifting in the above diagram.On the other hand,we may have Δ(K)=0 but the lifting still does not exist for some particular maps from K to

3 Computations of Δ()

We see that if there is a cross-section π :→of Φ for any n(i.e.,π is a nondegenerate simplicial map such that Φ ◦ π is an identity of),then any nondegenerate simplicial mapalways admits a lifting=π◦f.This is true for n≤3.

Theorem 3.1 Δ()=0forn=1,2,3.

Proof define the map π :→on the vertices by sending each binary vector to the corresponding integral vector with 0/1-coefficients.It is well known that whenever A∈GL(n,Z2)and n=1,2,3,then the corresponding integral 0/1-matrix lies in GL(n,Z).Thus π is a well-defined nondegenerate map of simplicial complexes.Obviously,it is a lift of Φ.

This statement was originally observed by Ayzenberg in[1,Lemma 5].My result is a special case.For n=4 there does not exist such a cross-section π :→by the following lemma.Lemma 3.1(cf.[2,Thereom 1])There is no nondegenerate simplicial map fromK41toK42.Now let us begin to compute Δ().

We know that according to Hadamard’s maximum determinant problem(cf.[8]),for A ∈GL(n,Z2),if we regard A as an integral matrix,its determinant does not exceedFor the case of n=4,detA=±1 or±3.

Let Vi⊂(Z2)4be the set of all vectors which have exactly i non-zero coordinates,i=1,2,3,4.

Lemma 3.2ForA∈GL(4,Z2),regard it as an integral matrix.IfdetA=±3,then there are only two possible cases:

(1)The row vectors ofAare just the four vectors ofV3;

(2)One row vector ofAbelongs toV3and the others belong toV2.

Proof Firstly,A has no row vector in V1.Otherwise,detA is equal to the determinant of some matrix in GL(3,Z2),which is certainly±1 if regarded as an integral matrix.Next,A has no row vector in V4.Otherwise,suppose that the 1st row vector of A is(1,1,1,1).For the same reason,other three row vectors do not belong to V3.If not so,the 1st row of A can be turned into a vector in V1by a row transformation with the determinant unchanged.So the three row vectors of A except(1,1,1,1)must belong to V2.However,this is also impossible,since detA will be even.

Therefore according to the above arguments,the row vectors of A can be only contained in V2and V3.We only need to eliminate the two cases:

(1)Two row vectors belong to V2and other two belong to V3.

(2)Only one row vector belongs to V2and other three belong to V3.

Consider the case(1).Without loss of generality,suppose that the 1st and 2nd rows of A are(1,1,1,0)and(1,1,0,1)respectively.There is only one vector in V2whose 3rd and 4th coordinates are 1.So one of the remaining row vectors can necessarily remove two non-zero coordinates of(1,1,1,0)or(1,1,0,1)by a row transformation,which is in contradiction with the 1st statement.Similarly,the case(2)is also impossible.

Remark 3.1 Actually,we can obtain the specific matrix of the case:One row vector of A belongs to V3and the others belong to V2as the following:

Without loss of generality,suppose that the 1st row of A is(1,1,1,0).Furthermore,suppose that three other row vectors’4th coordinates are 1,otherwise by subtracting the one whose 4th coordinate is 0 from the 1st row,the resulting matrix does not change the determinant but its 1st row belongs to V1.This is in contradiction with Lemma 3.2.So

B is a 0/1-matrix in GL(3,Z)whose rows are all vectors with only one nonzero coordinate.Multiply the sum of the 2nd,3rd and 4th row by−1 and add it to the 1st row,we get a matrix

Its determinant is 3·detB= ±3.

Now we can calculate Δ(K41).

Theorem 3.2 Δ(K41)=1.

Proof From Lemma 3.1,it is sufficient to find a nondegenerate simplicial map from K41to,which is equivalent to a map from(Z2)4to Z5such that every basis of(Z2)4is mapped to a part of a basis of Z5.A natural idea is to add a 5th coordinate to vectors in(Z2)4,and then regard them as vectors in Z5.We try to find a function f:(Z2)4→ Z,such that for any basisis a part of a basis of Z5,which means that there exists β =(b1,···,b5)T∈ Z5,such that

Expand this determinant along the 5th column,and let A denote the matrix(α1,···,α4).Replace the i-th row of A by(f(α1),···,f(α4)),and denote the derived matrix by Ai.Then

Denote it by Γ.It is easy to see that Γ =1 if and only if detA,detA1, ···,detA4are relatively prime,i.e.,

Claim If we set f(α)=1 for all α ∈ (Z2)4,then(3.1)holds for any A ∈ GL(4,Z2).

We have mentioned that the determinants of matrices in GL(4,Z2),regarded as integral matrices,must be±1 or±3.

If detA=±1,(3.1)naturally holds.

If detA=±3,by Lemma 3.2 and Remark 3.1,there are exactly two possible cases and they can be uniquely determined up to permutation of rows and columns.That is,

There must exist one of detA1,···,detA4that is relatively prime to detA in the above two cases,for

So(3.1)holds.

For n ≥ 5,we can not determine the exact value of Δ(Kn1).But a non-decreasing relation holds for general situations.

Theorem 3.3 Δ

Proof It is sufficient to prove that rbe N.Then there exists some nondegenerate simplicial map f fromWithout loss of generality,we can assume that f maps e1=(1,0,···,0)T∈ (Z2)n+1to(1,0,···,0)T∈ ZN.So f can be written as the following form:

Similarly,vectors of ZNare supposed to be written as

Let W denote the set of the vectors in(Z2)n+1whose 1st coordinates are zero.The full subcomplex KWofspanned by the vertex set W is isomorphic to

For every(k−2)-simplex v2v3···vk∈ KW,e1v2v3···vkis a(k−1)-simplex of Kn+11.f(e1),f(v2),f(v3), ···,f(vk)form a(k − 1)-simplex of KN2,which implies that ∃uk+1,uk+2, ···,uN∈ZN,such that

Expanding the above determinant along the 1st column,we have

Namely,h(v2),···,h(vk)must be a part of a basis of ZN−1.Hence h is a nondegenerate simplicial map from KWto

Remark 3.2 This theorem can also be deduced from[1,Proposition 8]from another point of view.

Using Lemma 3.1,we can directly get the following corollary.

Corollary 3.1There is no nondegenerate simplicial map from

4 Conclusion and Further Problems

For a given family F of simplicial complexes,we may ask whether Δ(K)=0 for every K∈F.The answer is negative for the family of all simplicial complexes since we have a counterexample Kn1for n≥4.As proved in[4],Kn1is Cohen-Macaulay,thus the answer for the family of all Cohen-Macaulay complexes is negative as well.This question for the family of simplicial spheres is open.

Problem 4.1WhetherΔ(K)=0or not for any simplicial spheresK?

We know that any simple convex polytope determines a dual simplicial sphere.If a simple convex polytope Pnwith m codimension-1 faces admits a small cover(equivalently,sR(P)=m−n),then the above problem is just equivalent to the existence of quasitoric manifolds over it(equivalently,s(P)=m−n).A further discussion is the lifting problem mentioned in Section 1.A special case follows.

Problem 4.2Can we obtain a quasitoric manifold over a simple convex polytope with a given small cover as a real part?

Another problem is the estimation of the upper bound of Δ()when n goes to+∞.

It follows easily from the definition.

Proposition 4.1If a simplicial complexKadmits a(Z2)n-coloring for somen,then

Proof r=rR(K)≤n.There exists a nondegenerate simplicial map from K to Kr1and a nondegenerate simplicial map from.Their composition is then a nondegenerate simplicial map from K to

Hence,the estimation of Δ()helps to estimate the general cases.By Theorem 3.3,we know that Δ()is a non-decreasing function of n.Thus,its upper bound is significant.

Finally we give a conjecture on this upper bound.For more discussion the reader is referred to[2,6–7].

Conjecture 4.1 Δ()is unbounded whenngoes to+∞.

AcknowledgementsThe authors would like to thank the editors and the anonymous referees for their valuable comments and helpful suggestions that helped to improve the quality of the paper.Moreover,the author would like to thank his supervisor Prof.Kefeng Liu for his constant encouragement and help.

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