Hideya KUWATAMikiya MASUDAHaozhi ZENG

1 Introduction

A toric variety is a normal complex algebraic variety of complex dimension n with an algebraic action of(C∗)nhaving a dense orbit.A toric variety is not necessarily compact and may have singularity.The famous theorem of Danilov-Jurkiewicz gives an explicit description of the integral cohomology ring of a compact smooth toric variety in terms of the associated fan.It in particular says that the integral cohomology groups are torsion-free and concentrated in even degrees.

The analogous result holds for a compact simplicial toric variety X(simplicial means that X is an orbifold)but with rational coefficients.Fischli and Jordan studied the integral cohomology groups H∗(X)in their dissertations[7,11]using spectral sequences.Their results gave an explicit computation of Hk(X)and H2n−k(X)for k ≤ 3 under some conditions.Based on their results,Franz developed Maple package torhom(see[8])to compute those cohomology groups.One can see that H∗(X)has torsion in general while it has no torsion when X is a weighted projective space(see[12]).Therefore we are naturally led to ask when H∗(X)has torsion or no torsion.

The orbit space Q of a compact simplicial toric variety X by the restricted action of the n-dimensional compact torus T is a nice manifold with corners(sometimes called a manifold with faces).All faces of Q(even Q itself)are contractible and Q is often homeomorphic to a simple polytope as manifolds with corners.MacPherson showed that X is homeomorphic to the quotient space(Q×T)/∼under some equivalence relation∼defined using the primitive vectors in the one-dimensional cones in the fan of X(see[9]).The one-dimensional cones correspond to the facets of Q so that one can think of the primitive vectors as a map

The map v satisfies some linear independence condition and a map satisfying the condition is called a characteristic function on Q(see definition 2.1 in Section 2).Note that there are many characteristic functions which do not arise from compact simplicial toric varieties.

Bahri,Sarkar and Song[1]considered the quotient space X(Q,v)=(Q×T)/∼.Although they restricted their concern to Q being a simple polytope,the characteristic function v used to define the equivalence relation∼is arbitrary;so the quotient space does not necessarily arise from a compact simplicial toric variety.They gave a sufficient condition for H∗(X(Q,v))to be torsion-free in terms of Q and v.They also gave a Danilov-Jurkiewicz type description for the ring structure of H∗(X(Q,v))when it is torsion-free.

In this paper,we also consider the quotient space X=X(Q,v)=(Q×T)/∼where v is arbitrary as above but our Q is a compact connected nice manifold with corners and not necessarily a simple polytope.When Q has a vertex(equivalently X has a T- fixed point),our X is a torus orbifold in the sense of[10].We give an explicit description of Hk(X)and H2n−k(X)for k ≤ 2 under some condition on Q.Motivated by the explicit description of H2n−1(X),we introduce a positive integer µ(QI)depending on the characteristic function v for eachwhere I is a subset of{1,···,m}and we understand QI=Q when I= Ø and µ(QI)=1 when QI= Ø.The µ(QI)’s are all one when X has no singularity.Here is a summary of our results,which follows from Propositions 6.1,8.1–8.3.

Theorem 1.1LetQbe a connected nice manifold with corners of dimensionn≥1.Letpbe a prime number and suppose that every face ofQ(evenQitself)is acyclic withZ/pcoefficients.IfH∗(X(Q,v))has nop-torsion,thenµ(QI)is coprime topfor everyQI.The converse holds when the face poset ofQis isomorphic to the face poset of one of the following:

(1)the suspension◇nof the(n−1)-simplexΔn−1,i.e.,◇nis obtained fromΔn−1×[−1,1]by collapsingΔn−1× {1}andΔn−1× {−1}to a point respectively,

(2)Δn,

(3)Δn−1×[−1,1].

Remark 1.1(1)When n≥3,there are many nice manifolds with corners Q which have the same face posets as◇n,Δnor Δn−1×[−1,1]but not homeomorphic to them.For instance,one can produce such Q by taking connected sum of them and integral homology n-spheres with non-trivial fundamental groups.

(2)The n-simplex Δnand the prism Δn−1× [−1,1]can be obtained from the suspension◇nby performing a vertex cut once and twice respectively.So,the reader might think that the converse mentioned in the theorem above would hold for Q obtained from◇nby performing a vertex cut repeatedly.However,we shall see in Section 9 that this is not true for Q obtained from◇3by performing a vertex cut four times.

The paper is organized as follows.In Section 2,we set up notations.In Section 3,we compute H2n−k(X)(k ≤ 2)for the quotient space X=(Q×T)/∼ using the idea in Yeroshkin’s paper[17].Namely,we delete a small neighborhood of the singular set in X to obtain a smooth manifold and investigate the relation of the cohomology groups between X and the smooth manifold.In Section 4,we show that the quotient map X→Q induces an isomorphism on their fundamental groups when Q has a vertex.In Section 5,we apply the results in Sections 3–4 to the case when n=2 and 3.In Section 6,we introduceµ(QI)and find a necessary condition for H∗(X)to have no p-torsion.In Section 7,we recall the theorem on elementary divisors and deduce two facts used in Section 8.In Section 8,we prove that the necessary condition obtained in Section 6 is sufficient for Q mentioned in the theorem above.Section 9 gives an example mentioned in the remark above.In the appendix we shall observe that a result of Fischli or Jordan on H2n−1(X)and the torsion part of H2n−2(X)agrees with our Proposition 3.1 when X is a compact simplicial toric variety.

2 Setting and Notation

In this section,we set up some notations and give some remarks.Let Q be a connected manifold with corners of dimension n(see[6,p.180]for the precise definition of a manifold with corners).Then faces are defined and a codimension-one face is called a facet.We assume that Q is nice,which means that every codimension-k face is a connected component of intersections of k facets.The teardrop,which is homeomorphic to the 2-disk,is a manifold with corners but not nice(see[6,p.181]).A simple polytope is a nice manifold with corners and any intersection of faces is connected unless it is empty.However,intersections of faces of a nice manifold with corners are not necessarily connected.For instance,a 2-gon,that is the suspension◇2in the theorem in the introduction,is a nice manifold with corners but the intersection of the two facets consists of two vertices.

Let S1be the unit circle group of the complex numbers C and T be an n-dimensional connected compact abelian Lie group.As is well-known,T is isomorphic to(S1)n.We set

Let Q have m facets and we denote them by Q1,···,Qm.

definition 2.1A functionv:{Q1,···,Qm} → Nis called a characteristic function onQif it satisfies the following two conditions:

(1)v(Qi)is primitive for eachi∈ [m]:={1,···,m},and

(2)wheneveris nonempty forI ⊂ [m],v(Qi)’s(i∈ I)are linearly independentoverQ.

We denote bybNthe sublattice ofNgenerated byv1,···,vm.

We call v(Qi)’s the characteristic vectors and abbreviate v(Qi)as vi.Condition(2)above implies that when Q has a vertex,rank=n.It also implies that when QIØ,the toral subgroup of T generated by vi(S1)’s(i∈ I),denoted by TI,is of dimension|I|where|I|is the cardinality of I.

To the pair(Q,v)we associate a quotient space

with the equivalence relation∼on the product Q×T defined by

(q,t)∼ (q′,t′)if and only if q=q′and t−1t′∈ TI,

where I is the subset of[m]such that QIis the smallest face of Q containing q=q′.The space X(Q,v)has a T-action induced from the natural T-action on Q×T.The orbit space of X(Q,v)by the T-action is Q and the quotient map

is induced from the projection map Q×T→Q.Then it is not difficult to see the following facts(see[15]for example).A T- fixed point in X(Q,v)corresponds to a vertex of Q,so X(Q,v)has a T- fixed point if and only if Q has a vertex.If vi’s(i∈ I)are a part of a basis of N for every I with QIØ,then X(Q,v)is a manifold but otherwise X(Q,v)is an orbifold.The singularity of X(Q,v)lies in the union of π−1(QI)over all I with|I|≥ 2.

As mentioned in the introduction,if X is a compact simplicial toric variety of complex dimension n so that X has an algebraic action of(C∗)nhaving a dense orbit,then the orbit space Q of X by the compact n-dimensional subtorus T of(C∗)nis a nice manifold with corners and X is homeomorphic to X(Q,v)where vi’s are primitive edge vectors of the fan associated to X.Moreover,faces of Q(even Q itself)are all contractible,which follows from the existence of the residual action of(C∗)n/T on Q=X/T.

3 H2n−k(X(Q,v))for k ≤ 2

In this section,we abbreviate X(Q,v)as X and all(co)homology groups will be taken with Z-coefficients unless otherwise stated.When n=1,Q is a closed interval if Q has a vertex and a circle otherwise,and X is homeomorphic to S2or a torus accordingly.We shall assume n≥2 in this section.Remember that π:X → Q is the quotient map.

Let Q(n−2)be the union of QIover all I with|I|≥ 2 and we assume Q(n−2)Ø.The singular set of X lies in π−1(Q(n−2))as remarked in Section 2.Let Q′be a “small closed tubular neighborhood” of Q(n−2)of Q and set X′:= π−1(Q′).

Lemma 3.1 H2n−k(X)Hk(XIntX′)fork ≤ 2.

ProofNote that Hr(X′)=0 for r≥ 2n−3 because X′is homotopy equivalent to π−1(Q(n−2))and dimπ−1(Q(n−2))=2n−4.Therefore,the exact sequence in cohomology for the pair(X,X′)yields an isomorphism

On the other hand,

Note that XIntX′is a manifold with boundary ∂X′.The lemma follows from(3.1)and(3.2).

Proposition 3.1for everyi,then

Remark 3.1 When Q has a vertex,rank=n as remarked in Section 2.Moreover,when Q has a vertex and n=2,the last term ∧2N/∧N above is zero.Indeed,since we may assume N=Z2and= 〈e1,ae2〉with some integer a,∧ N= 〈e1∧ e2〉= ∧2N,where{e1,e2}denotes the standard base of Z2.

Proof The statement for H2n(X)follows immediately from Lemma 3.1.

We shall prove the statement for H2n−1(X).Let Q0:=(IntQ)∩ (QQ′)and Q1be the intersection of(QQ′)and a small open neighborhood of∂Q in Q.

Since

the Mayer-Vietoris exact sequence in homology for the triple(XX′,π−1(Q0),π−1(Q1))yields the following exact sequence:

As is easily seen,f0is injective;so

We write f1as(ψ1,ϕ1)according to the decomposition of the target space.Since

which is f1composed with the projection on the second factor,is surjective,one has

Since H1(Y ×T)=H1(Y)⊕H1(T)for any topological space Y,elements in kerϕ1are of the form(c1v1,···,cmvm)with integers ci,where H1(T)is identified with N=Hom(S1,T)in a natural way.It follows that

The statement for H2n−1(X)in the proposition follows from(3.4)–(3.6)and Lemma 3.1.

The computation of H2n−2(X)is similar to that of H2n−1(X).We write f2as(ψ2,ϕ2)similarly to f1.Since H1(Qi)=0 for any i by assumption,kerf1is a free abelian group of rank m−rankas is easily seen;so it follows from(3.3)that

Similar to ϕ1,the map

is surjective;so

Here,

for any topological space Y by the K¨unneth formula.Therefore,since H1(Qi)=0 by assumption,it follows from(3.8)and(3.10)that kerϕ2is contained in.We note that H2(T)and H2(T/vi(S1))can be identified with ∧2N and ∧2(N/〈vi〉)respectively and the kernel of the projection ∧2N → ∧2(N/〈vi〉)is 〈vi〉∧N.Therefore

This together with(3.7)and(3.9)proves the statement for H2n−2(X)in the proposition.

4 Fundamental Groups

For a subset I of[m],we define

and consider a space

where∼eis the equivalence relation on the product Q×Tmdefined by

and I is the subset of[m]such that QIis the smallest face of Q containing q=q′.

We note that ZQlocally admits a smooth structure.Indeed,since Q is a manifold with corners,any point of Q has a neighborhood U homeomorphic to(R≥0)r× Rn−rfor some 0≤r≤n and it follows from the construction of ZQthat the inverse image of U by the projection map κ:ZQ→ Q is homeomorphic to Cr× Rn−r× Tm−r.Therefore ZQlocally admits a smooth structure and hence is a topological manifold.

Remark 4.1 When Q is a simple polytope,ZQis called a moment-angle manifold and it is known that ZQadmits a smooth structure and is 2-connected(see[3–4]).Moreover,the moment-angle manifold ZQis homotopy equivalent to Cm−Z defined in[5](see[4,Theorem 4.7.5]),where Z is the union of coordinate subspaces in Cmdetermined by Q.

Lemma 4.1The projection mapκ:ZQ→ Qinduces an isomorphismκ∗:π1(ZQ)π1(Q)on the fundamental groups.

Proof In a way similar to the above argument,one can see that κ−1(Qi),where Qiis a facet of Q,is a locally smooth closed manifold.Moreover,it is a locally smooth codimension two submanifold of ZQ.Indeed,a closed tubular neighborhood of Qiin Q can be identified with Qi× [0,1],and ρi:κ−1(Qi× {1})→ κ−1(Qi),where ρiis induced from((q,1),t)→ (q,t)for q∈Qi=Qi×{0}⊂Qi×[0,1]⊂Q and t∈Tm,is a principal S1-bundle,and the total space Eiof the associated complex line bundle can be identified with a closed tubular neighborhood of Zi:= κ−1(Qi)in ZQ.

Since Ziis a locally smooth closed codimension two submanifold of ZQ,the transversality argument can be applied.Therefore,if a continuous map f:S1→ZQmeets Zi,then one can slightly push f in the fiber direction of Eiso that the deformed f does not meet Zi.Applying this deformation to f for every i,we see that f is homotopic to a continuous map whose image lies in κ−1(IntQ)=IntQ × Tm.This means that the inclusion map ι:IntQ × Tm→ ZQinduces an epimorphism

Since IntQ is homotopy equivalent to Q,we may replace IntQ by Q above and we have a sequence

where the composition κ∗◦ ι∗agrees with the projection on the first factor,so that the kernel of ι∗is contained in the second factor π1(Tm).

Let Sibe the i-th S1-factor of Tmand choose a point qi∈(Qi×{1})∩IntQ.Then ι({qi} × Si)is a fiber of the principal S1-bundle ρi:κ−1(Qi× {1}) → Zi= κ−1(Qi),so it shrinks to a point in Zi.Therefore π1(Tm)is in the kernel of the epimorphism ι∗and this implies the lemma.

We recall a result from Bredon’s book[2].

Lemma 4.2 (see[2,Corollary 6.3,p.91])IfXis an arcwise connectedG-space,Gcompact Lie,and if there is an orbit which is connected(e.g.,Gconnected orXGØ),then the quotient mapX→X/Ginduces an epimorphism on their fundamental groups.

The characteristic map v:{Q1,···,Qm} → Hom(S1,T)defines a homomorphism Tm→ T,denoted v again.Note that v(Tm)is a subtorus of T of dimension rankbN,in particular,v is surjective if and only if rank=rankN(this is the case when Q has a vertex).The product map id×v:Q×Tm→Q×T induces a continuous map

and it further induces an injective continuous map so thatis a homeomorphism if v is surjective since the spaces are compact and Hausdorff.

Proposition 4.1IfQhas a vertex,thenπ∗:π1(X)π1(Q).

Proof We have a sequence

Since κ∗is an isomorphism by Lemma 4.1,it suffices to prove that V∗is surjective.

Since Q has a vertex,rank=rankN and the homomorphism v:Tm→T is surjective;so the map:ZQ/kerv→X above is a homeomorphism.Sinceis a sublattice of N of finite index,there is a finite covering homomorphism ρ:→ T corresponding to,whereis also a compact connected abelian Lie group of dimension n(precisely speaking,ρ∗(π1())=when N is regarded as π1(T))and the characteristic function v uniquely determines a characteristic function:{Q1,···,Qm} → Hom(S1,)such that ρ∗(v(Qi))=v(Qi)for any i.Then we have

andinduces a homomorphism Tm→denotedgain similarly to v,and=ZQ/kerMoreover,we have X=/kerρ.Namely,the quotient map V:ZQ→ X factors as the composition of two quotient maps

The theorem on elementary divisors(see Section 7)implies that since(Qi)’s spanthe homomorphism:Tm→composed with a suitable automorphism of Tmcan be viewed as a projection map if we take a suitable identification ofwith Tn;so keris connected and hence α∗:π1(ZQ)→ π1()is surjective by Lemma 4.2.The action ofonhas a fixed point since Q has a vertex and kerρ is contained in,so the action of kerρ onhas a fixed point.Therefore β∗:π1()→ π1(X)is also surjective again by Lemma 4.2.

Remark 4.2 As mentioned in the introduction,even if Q is a simple polytope,X=ZQ/kerv is not necessarily a compact toric orbifold because the characteristic map v is not necessarily coming from primitive vectors of a complete simplicial fan.

Corollary 4.1IfQhas a vertex andH1(Q)=H2(Q)=0,thenH1(X)=0andH2(X)Zm−n.

Proof By Proposition 4.1,π1(X)π1(Q)and hence H1(X)H1(Q).Therefore H1(X)=0 since H1(Q)=0 by assumption and hence H1(X)=0 and H2(X)has no torsion by the universal coefficient theorem.On the other hand,since X is an orbifold,Poincar´e duality holds with Q-coefficients.Therefore the rank of H2(X)is equal to that of H2n−2(X),that is m−n by Proposition 3.1 and its subsequent remark.

5 Low Dimensional Cases

A nice manifold with corners Q is called face-acyclic(see[13])if every face of Q(even Q itself)is acyclic.We note that if Q is face-acyclic,then Q must have a vertex.Indeed,let F be a face of Q of minimum dimension.Then F has no boundary because the boundary of F must consist of faces of smaller dimensions,so F is a closed manifold.But since F is acyclic,this means that F is a point.Therefore Q has a vertex.

We shall apply the previous results when Q is face-acyclic and n=dimQ is 2 or 3.The following corollary follows from Proposition 3.1 and Corollary 4.1.

Corollary 5.1Suppose thatQis face-acyclic anddimQ=2,that is,Qis anm-gon(m≥2).Then we have

Example 5.1 Let a be a positive integer.Take Q to be a 2-simplex,N=Z2and

Then= 〈ae1,e2〉and N/Z/a.The space X is not a weighted projective space when a≥2 since it has torsion in cohomology,where{e1,e2}denotes the standard base of Z2as before.

Corollary 5.2Suppose thatQis face-acyclic anddimQ=3.Then

Proof Since Q is face-acyclic,Q has a vertex as remarked at the beginning of this section,all the statements except for j=3 follows from Proposition 3.1 and Corollary 4.1.In order to prove the statement for j=3,it suffices to show H3(X;Q)=0 and this is equivalent to show that the Euler characteristic of X is 2m−4(note that we know the rank of Hj(X)except for j=3).

Since Q is face-acyclic and of dimension 3,the boundary of Q is a 2-sphere,every 2-face of Q is a 2-disk and the number of 2-faces is m by definition.Let V be the number of vertices of Q.Then the number of edges of Q isand hence we obtain an identity V−+m=2 by Euler’s formula,which implies V=2m − 4.On the other hand,it is known that the Euler characteristic of X is equal to that of the T- fixed point set XT(see[2,Theorem 10.9,p.163]).In our case XTis isolated and corresponds to the vertices of Q.Therefore,the Euler characteristic of X is equal to V,that is 2m−4.

Example 5.2 It happens that∧N=∧2N even ifN.For instance,take Q to be a 3-simplex,N=Z3and

Then

where{e1,e2,e3}denotes the standard base of Z3.

Corollary 5.2 says that if=N,then Hj(X)has no torsion except j=3.However,H3(X)can be nontrivial(so,a nontrivial torsion group)whenbN=N.We shall give such an example below.One can also find many such examples using Maple package Torhom.

Example 5.3 Let a be a positive integer and take the following five primitive vectors in Z3:

Then=N.We consider the complete simplicial fan Δ having the following six 3-dimensional cones

where ∠v∈vivj(∈∈ {+,−},i,j ∈ {1,2,3})denotes the cone spanned by v∈,viand vj.Let X be the compact simplicial toric variety associated to the fan Δ.Let ρ be the projection of R3on the line R corresponding to the last coordinates of R3.Then the vectors v1,v2,v3are in the kernel of ρ and ρ(v±)are primitive vectors and determine the complete 1-dimensional fan.This means that we have a fibration F→X→CP1,where the fiber F is the compact simplicial toric variety associated to the fan obtained by projecting the fan Δ on the plane R2corresponding to the first two coordinates of R3.The E2-terms of the Serre spectral sequence of the fibration are

and=0 unless p=0,2 and q=0,2,3,4 by Corollary 5.1.Therefore all the differentials except

are trivial.Here,=H0(CP1;H3(F))=H3(F)is trivial or a torsion group by Corollary 5.1 while=H2(CP1;H2(F))=H2(F)is a free abelian group again by Corollary 5.1,somust be trivial.Therefore=.Sincewith p+q3 vanishes unless(p,q)=(0,3),we obtain an isomorphism H3(X)H3(F).Here H3(F)=Z/a again by Corollary 5.1(see Example 5.1)and hence we have H3(X)Z/a.On the other hand,sinceNb=N as remarked above,Hj(X)has no torsion for j3 by Corollary 5.2.

6 A Necessary Condition for no p-Torsion

Let I be a subset of[m]with QIØ.Although QIis not necessarily connected,we understand that QIstands for a connected component of QIin this section for notational convenience.Then the characteristic function v associates a characteristic function vIon QIas follows.Since vi’s(i ∈ I)are linearly independent over Q,they span a|I|-dimensional linear subspace of N⊗R and its intersection with N is a rank|I|sublattice of N,denoted NI.Then N(I):=N/NIis a free abelian group of rank n−|I|and we denote the projection map from N to N(I)by πI.If QI∩ Qjis nonempty for j ∈ [m]I,then its connected components are facets of QI,and any facet of QIis of this form.The element πI(vj) ∈ N(I)is not necessarily primitive and we define vI(QI∩Qj)to be the primitive vector in N(I)which has the same direction as πI(vj),where QI∩ Qjalso stands for a connected component of QI∩ Qj.Then one can see that vIis a characteristic function on QI.In a way similar tobN,one can define a sublattice(I)of N(I)using vI.We allow I= Ø and understand QØ=Q,N(Ø)=N and(Ø)=We define

Here|N(I)/(I)|is not necessarily finite.For instance,take Q=S1Detailed explanation about this assertion.Since the set of isotropy groups of X is finite,there is a positive integer r such that XGk=XGrfor every k≥r.Since Gris a subgroup of TI,we have XGr⊃XTI.We shall prove the opposite inclusion.Let x∈XGr.The isotropy subgroup Txat x contains Gkfor every k≥r because XGk=XGr,but since Txis a closed subgroup of T,Txmust contain the closure that is TI.Therefore x∈XTIand hence XGr=XTI.× [−1,1]and assign characteristic vectors(1,0)and(−1,0)to the facets S1×{1}and S1×{−1}respectively.Then N/is an in finite cyclic group and hence|N(I)/(I)|is in finite for I=Ø.One can easily construct a similar example such that|N(I)/(I)|is in finite for some IØ.

Remark 6.1 When|I|=n,N(I)={0};soµ(QI)=1.When|I|=n−1,N(I)is of rank one and(I)is generated by a primitive vector;so(I)=N(I)and henceµ(QI)=1 in this case too.Another case which ensuresµ(QI)=1 is the following.Let q be a vertex of Q.Then there is a subset J of[m]with|J|=n such that q ∈ QJ.If{vj}j∈Jis a base of N,thenµ(QI)=1 for every subset I of J,which easily follows from the definition ofµ(QI).

We note that for a prime number p,H∗(X(Q,v);Z)has no p-torsion if and only if

which follows from the universal coefficient theorem(see[14,Corollary 56.4]).

Lemma 6.1 (see[2,Theorem 2.2,pp.376–377])Let a group G of prime order p act on a finite dimensional space X with A⊂X closed and invariant.Suppose that G acts trivially on H∗(X,A;Z).Then

Proposition 6.1 If Hodd(X(Q,v);Z/p)=0,then H1(QI;Z/p)=0 andµ(QI)is finite and coprime to p for every I.

Proof We abbreviate X(Q,v)as X as before.Since Hodd(X;Z/p)=0,we have Hodd(XG;Z/p)=0 for every p-subgroup G of TIby repeated use of Lemma 6.1.In fact,let G be an order p subgroup of S1.The induced action of G on H∗(X)is trivial because G is contained in the connected group S1.Then rk Hodd(XG;Z/p)≤rk Hodd(X;Z/p)by Lemma 6.1 applied with A=Ø.Therefore,Hodd(XG;Z/p)=0 by assumption.Repeating the same argument for XGwith the induced action of S1/G,which is again a circle group,we conclude that Hodd(XG;Z/p)=0 for any p-subgroup G of S1Detailed explanation about this assertion.Since the set of isotropy groups of X is finite,there is a positive integer r such that XGk=XGrfor every k≥r.Since Gris a subgroup of TI,we have XGr⊃XTI.We shall prove the opposite inclusion.Let x∈XGr.The isotropy subgroup Txat x contains Gkfor every k≥r because XGk=XGr,but since Txis a closed subgroup of T,Txmust contain the closure that is TI.Therefore x∈XTIand hence XGr=XTI..

For a positive integer k,let Gkbe the p-subgroup of TIconsisting of all elements of order at most pk.Then Gk⊂ Gk′for k ≤ k′and the unionis dense in TI.Therefore XGk=XTIif k is sufficiently large.1Since XI= π−1(QI)is a connected component of XTI,this shows that Hodd(XI;Z/p)=0.But H2(n−|I|)−1(XI)is isomorphic to H1(QI)⊕N(I)/bN(I)by Proposition 3.1 and hence the universal coefficient theorem implies the proposition.

When Hodd(X(Q,v);Z/p)=0,Proposition 6.1 gives a constraint on the topology of QI,that is H1(QI;Z/p)=0.It is proved in[13]that if X(Q,v)is a manifold and Hodd(X(Q,v);Z)=0,then Q is face-acyclic.This implies that there will be more constraints on the topology of QIwhen Hodd(X(Q,v);Z/p)=0,to be more precise,we expect that Q is face p-acyclic which means that(every component of)QIis acyclic with Z/p-coefficients for every I.Therefore,in order to consider the converse of Proposition 6.1,it would be appropriate to assume that Q is face p-acyclic.We shall prove in Section 8 that the converse holds in some cases while we shall see in Section 9 that the converse does not hold in general.

7 Theorem on Elementary Divisors

We recall the theorem on elementary divisors and deduce two facts from it,which will play a role in the next section.

Theorem 7.1 (Theorem on Elementary Divisors,see[16])LetN′be a submodule of rankn′inN=Zn.Then there are bases{,···,}ofN′and{u1,···,un}ofNsuch that=∈iuiwith some integer∈ifori=1,2,···,n′and∈1|∈2|···|∈n′.Moreover,ifA=(a1,···,ak)is ann × kinteger matrix whose column vectorsa1,···,akgenerateN′and

thenδi= δi−1∈ifori=1,2,···,n′.In particular,ifn′=n,thenδn=|N/N′|.

We deduce two facts from Theorem 7.1.

Lemma 7.1LetAbe ann×ninteger matrix of ranknandeA:Rn/Zn→Rn/Znbe the epimorphism induced fromA.ThenkereAcokerA.

Proof By Theorem 7.1 we may think of A as the diagonal matrix with diagonal entries∈1,···,∈n.Then one easily sees that kerand cokerA are both isomorphic toproving the lemma.

Let a1,···,an+1be elements of Znwhich generate a sublattice 〈a1,···,an+1〉of rank n and set di:=|det((aj)ji)|for i∈[n+1].It follows from Theorem 7.1 that

Suppose that an+1is primitive. Let(kn+1)be the projection image of akon Zn/〈an+1〉and letbe the primitive vector in the quotient lattice Zn/〈an+1〉which has the same direction aswhenis nonzero,andbe the zero vector when so is.Set:=det(,···,,···,).With this understood we have the following lemma.

Lemma 7.2 gcd(d1,···,dn)|dn+1,i.e.,gcd(d1,···,dn)=gcd(d1,···,dn+1).Moreover,gcd(···,)|gcd(d1,···,dn+1).

Proof Theorem 7.1 applied with N′generated by an+1says that there is a basis{u1,···,un}of N=Znsuch that an+1= ∈1u1with some integer ∈1.But since an+1is primitive,we have∈1= ±1.Therefore,we may assume that an+1=(0,···,0,1)Tthrough a linear transformation of Zn.We have

whereis the(n,j)entry of the matrix(a1,···,an)andis its cofactor.Since an+1=(0,···,0,1)T,agrees with dj=|det(a1,···,,···,an+1)|up to sign.Thereforeis divisible by gcd(d1,···,dn)for every j and this together with(7.2)implies the former statement in the lemma.

Since an+1=(0,···,0,1)T,Zn/〈an+1〉can naturally be identified with Zn−1and we have

where(k=1,2,···,n)is the projection image of akon Zn/〈an+1〉=Zn−1.Sinceis a positive scalar multiple of,=|det(···,···,)|divides the latter term in(7.3)above and hence dj.This together with the former statement in the lemma implies the latter statement in the lemma.

8 Converse of Proposition 6.1 in Three Cases

In this section we show that if Q is face p-acyclic and has the same face poset as one of the following:

Case 1 the suspension ◇nof an(n−1)-simplex Δn−1(see the introduction),

Case 2 the n-simplex Δn,

Case 3 the prism Δn−1× [−1,1],

then the converse of Proposition 6.1 holds,i.e.,ifµ(QI)is finite and coprime to p for every I,then Hodd(X(Q,v);Z/p)=0.

First we establish Case 1.Then we reduce Case 2 to Case 1 by collapsing a face of Q to a point.In Case 3,according to the characteristic function v,we collapse one or two faces of Q to a point reducing Case 3 to Case 2 or Case 1.The argument then becomes much more complicated than that reducing Case 2 to Case 1.It would be interesting to see whether this inductive argument works for an arbitrary product of simplices.

Let q be a vertex of Q.Then q lies in QIfor some I⊂[m]with|I|=n.We set

where vi=v(Qi)as before.

Case 1 In this case Q has two vertices,say q and q′,and dQ(q)=dQ(q′)= µ(Q).

Proposition 8.1Suppose thatQis facep-acyclic,has the same face poset as◇nandµ(Q)is coprime top.ThenX(Q,v)has the same cohomology asS2nwithZ/p-coefficients,in particularHodd(X(Q,v);Z/p)=0.

Proof When n=1,Q is a closed interval and X(Q,v)is homeomorphic to S2;so the proposition holds when n=1.In the following we assume n≥2,so that Q has n facets.

Let Tn=(S1)n.Then Hom(S1,Tn)is naturally isomorphic to Znand we identify them.be the standard basis of Znand e:{Q1,···,Qn} → Zn=Hom(S1,Tn)be the characteristic function assigning eito Qi.Then we have a Tn-space X(Q,e)which is actually a manifold becauseis a basis of Zn.

The characteristic vectors vi∈N=Hom(S1,T)define an epimorphism:Tn→T sending(h1,···,hn)One can see that the surjective map from Q×Tnto Q×T sending(q,t)to(q, ev(t))descends to a-equivariant map from X(Q,e)to X(Q,v)and further descends to a homeomorphism

Here|ker|=|N/|by Lemma 7.1 and it is coprime to p by assumption.Moreover,since keris a subgroup of the connected group Tnacting on X(Q,e),the induced action of ker ev on H∗(X(Q,e);Z/p)is trivial.Therefore we have

(see[2,Theorem 2.4 in p.120])and hence it suffices to prove that X(Q,e)has the same cohomology as S2nwith Z/p-coefficients.

Since Q has the same face poset as◇nand every face of◇nis contractible,there is a face preserving map f:Q→◇nwhich induces an isomorphism on the face posets.Since Q is face p-acyclic,f induces an isomorphism on cohomology with Z/p-coefficients at each face.In a way similar to the definition of e,one has a characteristic function on◇n,also denoted by e.Then the map from Q×Tnto◇n×Tnsending(q,t)to(f(q),t)descends to a map

which induces an isomorphism on cohomology with Z/p-coefficients.Since X(◇n,e)is homeomorphic to S2n,this proves the desired result.

Case 2 Since Q has the same face poset as the n-simplex Δn,Q has n+1 facets Q1,···,Qn+1and n+1 vertices q1,···,qn+1.We number them in such a way that qiis the unique vertex not contained in Qi.It follows from(7.1)and Lemma 7.2 that

In fact,the former identity in(8.1)follows from(7.1).The latter identity with i=n+1 follows from Lemma 7.2 but the same proof of Lemma 7.2 works for any i and proves the desired identity.Similarly,the last assertion in(8.1)also follows from(the proof of)Lemma 7.2.

Proposition 8.2Suppose thatQis facep-acyclic,has the same face poset asΔnandµ(Q)is coprime top.ThenHodd(X(Q,v);Z/p)=0.

Proof We abbreviate X(Q,v)as X.We prove the proposition by induction on n.When n=1,Q is a closed interval and X is homeomorphic to S2;so the proposition holds in this case.We assume that the proposition holds for any face p-acyclic(n−1)-dimensional manifold with corners satisfying the assumption in the proposition.For every i,Qihas the same face poset as Δn−1and µ(Qi)|µ(Q)by(8.1),so Hodd(Xi;Z/p)=0 by the induction assumption,where Xi= π−1(Qi)and π :X → Q is the quotient map.On the other hand,sinceµ(Q)=gcd(dQ(q1),···,dQ(qn+1))is coprime to p by assumption,dQ(qi)is coprime to p for some i.For such i,Q/Qiis face p-acyclic,has the same face poset as◇nandµ(Q/Qi)=dQ(qi)is coprime to p,so Hodd(X/Xi;Z/p)=0 by Proposition 8.1.These together with the exact sequence

show Hodd(X;Z/p)=0.

Case 3 We denote the facets of Q corresponding to Δn−1×{±1}by Q±and the others by Q1,···,Qn.Accordingly,we abbreviate the characteristic vectors v(Q±)as v±and v(Qi)as vi.We denote the vertices in Q∈by,···,for∈=±in such a way thatis not contained in Qi.

Lemma 8.1Suppose thatQis facep-acyclic and has the same face poset asΔn−1×[−1,1].Ifµ(Q)is coprime topand eitherµ(Q+)orµ(Q−)is coprime top,then there is a vertexqofQsuch thatdQ(q)is coprime top.

We shall prove this lemma later.It suffices to prove the following for our purpose in Case 3.

Proposition 8.3Suppose thatQis facep-acyclic,has the same face poset asΔn−1×[−1,1]andµ(Q),µ(Q±)are coprime top.ThenHodd(X(Q,v);Z/p)=0.

Proof We abbreviate X(Q,v)as X and denote by X∈(∈=+or−)the inverse image of Q∈by the quotient map π:X → Q.Since Q∈is face p-acyclic,has the same face poset as Δn−1and µ(Q∈)is coprime to p by assumption,we have

by Proposition 8.2.

By Lemma 8.1 there is a vertex q of Q such that dQ(q)is coprime to p.Without loss of generality,we may assume q=,i.e.,dQ()is coprime to p.Since we have(8.2)and the exact sequence

it suffices to prove

We consider two cases.

Case a The case where det(v1,···,vn)0.In this case,the characteristic function v on Q induces a characteristic function on Q/Q+,denoted v+,and X/X+=X(Q/Q+,v+).We note that Q/Q+is face p-acyclic and has the same face poset as Δnsince Q is face pacyclic and has the same poset as Δn−1×[−1,1].Moreover,sinceis a vertex of Q/Q+and dQ/Q+()=dQ()is coprime to p,µ(Q/Q+)is coprime to p.Therefore,(8.3)follows from Proposition 8.2.

Case b The case where det(v1,···,vn)=0.

Claim There is a vertex q of Qnsuch that dQn(q)is coprime to p,soµ(Qn)is coprime to p.

Proof WriteSince vnis primitive,we may assume vn=(0,···,0,1)Tby Theorem 7.1.Denote byandthe projection images of viand v−on Zn/〈vn〉and byandthe primitive vectors which have the same directions asandrespectively.Then

by definition and hence

On the other hand,since vn=(0,···,0,1)T,we have

and the left-hand side above is zero by assumption.It follows that

where the second identity above is the expansion of det(v1,···,vn−1,v−)with respect to the n-th row.By(8.4)gcd(dQn(),···,dQn())divides the last term above.Since dQ()is coprime to p,this means that dQn()is coprime to p for some i,proving the claim.

Now we shall prove(8.3)by induction on the dimension n of Q.When n=1,Q is a closed interval,X is S2and X+is a point;so(8.3)holds in this case.We assume n≥2 in the following.Let Xnbe the inverse image of Qnby the quotient map π:X → Q.The face poset of Qnis the same as that of Δn−2×[−1,1]and Qnis face p-acyclic.The facets corresponding to Δn−2×{±1}are Qn∩Q±andµ(Qn∩Q±)are coprime to p by(8.1)becauseµ(Q±)are coprime to p by assumption.Moreover,µ(Qn)is also coprime to p by the claim above.Therefore

by the induction assumption.

The quotient Q/(Qn∪Q+)=:is face p-acyclic andhas the same face poset as◇n.The characteristic function v on Q induces a characteristic function ondenotedbecauseis a vertex ofand d~Q()=dQ(q)is coprime to p,in particular nonzero.The quotient space Xn/(Xn∩X+)is a subspace of X/X+and

Sinceis coprime to p,Hoddby Proposition 8.1.This together with(8.6),(8.5)and the exact sequence

implies(8.3).

Now it remains to prove Lemma 8.1.

Proof of Lemma 8.1 We may assume thatµ(Q+)is coprime to p.We may also assume that v+=(0,···,0,1)Tby Theorem 7.1 through some identification of N with Zn.Suppose that

and we shall deduce a contradiction in the following.

By Lemma 7.2,det(v1,···,vn)is divisible by gcd(dQ(),···,dQ(q)),so it follows from(8.7)that

We writefor i=1,2,···,n.

Claim 1 There is an i∈[n]such that p|for all jn.

Proof Since v+=(0,···,0,1)T,we naturally identify the quotient lattice Zn/〈v+〉with Zn−1and then the projection imageof vion the quotient latticeSetTheni/si=:is primitive.Since dQ(q)is assumed to be divisible by p for all vertices q of Q,we have

Here,since v+=(0,···,0,1)T,we have

Now suppose that siis not divisible by p for any i.Then it follows from(8.9)–(8.10)thatfor every subset{i1,···,in−1}of[n].Sinceµ(Q+)agrees with the greatest common divisor of all deby(7.1),this shows that p|µ(Q+)which contradicts the assumption thatµ(Q+)is coprime to p.Therefore p|sifor some i,proving the claim.

Claim 2 p|det(vi1,···,vin−2,v−,v+)for every subset{i1,···,in−2}of[n].

Proof Since v+=(0,···,0,1)T,we have

whereis the projection image of v−on the quotient Zn/〈v+〉=Zn−1.We shall observe that the right-hand side in(8.11)is divisible by p.Without loss of generality,we may assume that the i in Claim 1 is n,so that p|for all jn.We consider two cases.

Case a The case where n ∈ {i1,···,in−2}.Sinceand p|for all jn,the right-hand side in(8.11)is divisible by p.

Case b The case where n/∈{i1,···,in−2}.In this case,we consider the expansion ofwith respect to the last column.Sinceand p|for all jn,we have

Here the left-hand side above is dQ(q)forso it is divisible by p by(8.7).Moreover,vnnis not divisible by p because otherwise every entry of vnis divisible by p and this contradicts the fact that vnis primitive.It follows from(8.12)that the right-hand side in(8.11)is divisible by p in this case,too.

This completes the proof of the claim.

Now(8.7)–(8.8)and Claim 2 show that all n×n minors of(v1,···,vn,v−,v+)are divisible by p and hence p|µ(Q)(=|N/bN|)by Theorem 7.1.This contradicts the assumption thatµ(Q)is coprime to p,and the lemma is proved.

9 Example

In this section we shall give an example of a compact simplicial toric variety showing that the converse of Proposition 6.1 does not hold in general.

Let Q be the 3-dimensional simple polytope with the 7 facets Q+,Q−,Q1,···,Q5,where Q4and Q5are triangles obtained by cutting two vertices of a prism,shown in Figure 1 below.The polytope Q can be obtained from◇3by performing a vertex cut four times.

Figure 1

Let d be a positive integer.To the 7 facets Q1,···,Q5,Q+,Q−,we respectively assign the following vectors

giving a characteristic function v on Q.There are ten vertices in Q.At each vertex,there are exactly three facets meeting and the determinant of the three vectors assigned to the facets is nonzero,indeed their absolute values are as follows:(Precisely speaking,the vectors are regarded as column vectors here by taking transpose.)Therefore,at each vertex,the cone spanned by the three vectors is 3-dimensional and has the origin as the apex.One can also check that

Since d is a positive integer,this shows that−v+is in the cone∠v1v2v3,v4is in the cone∠v1v2v+while v−is in the cone∠v1v2v3,and v5is in the cone∠v1v2v−(see Figure 2),where∠uvw denotes the cone spanned by vectors u,v,w.This implies that the ten 3-dimensional cones have no overlap and cover the entire R3,giving a complete simplicial fan so that the quotient space X=X(Q,v)is homeomorphic to a compact simplicial toric variety.

Figure 2 Each vector viis denoted by a point in R2∪{∞}and a segment connecting vi,vjcorresponds to the 2-dimensional cone spanned by them and a triangle formed by vi,vj,vkcorresponds to the 3-dimensional cone spanned by them.

We shall check thatµ(QI)=1 for each face QIof Q,whereµ(QI)is defined in Section 6.As remarked in Section 6,µ(QI)=1 when|I|=2 or 3.ClearlybN=N(=Z3).Therefore it suffices to checkµ(QI)=1 when|I|=1.At vertices Q1∩Q4∩Q+,Q2∩Q4∩Q+and Q1∩Q5∩Q−,we have

and henceµ(QI)=1 for every I with|I|=1 except I={3}again by the remark in Section 6.In order to seeµ(Q3)=1,we note that{v3,v4,v+}is a base of N and

Therefore,the images of v1and v2by the quotient map π{3}:N → N({3})=N/〈v3〉are(−d,0)and(2d,−d)with respect to the base{π{3}(v4),π{3}(v+)}.Thus the corresponding primitive vectors are(−1,0)and(2,−1)which form a base of N({3}).Hence µ(Q3)=1.

We shall compute H3(X).Take a plane in R3which meets the facets Q1,Q2,Q3transversally and does not meet the other facets of Q.Cutting Q along the plane,we divide Q into two polytopes,denoted P+and P−containing Q+and Q−respectively.Let π:X → Q be the quotient map and set

The quotient space P∈/P can be regarded as a prism.The characteristic function v on Q induces a characteristic function on P∈/P,denoted w∈,and X/Y+=Y−/Y(resp.X/Y−=Y+/Y)is homeomorphic to X(P−/P,w−)(resp.X(P+/P,w+)).The same argument as above shows thatµ takes 1 on all faces of the prism P∈/P,so

by Proposition 8.3.

Letbe a nice manifold with corners obtained from Q by collapsing Q4∪Q+and Q5∪Q−to a point respectively.TheeQ has three facets coming from Q1,Q2,Q3and the characteristic function v on Q induces a characteristic functionon.Since

one can see that H4(X())Z/d by Corollary 5.2,and since X()is homeomorphic to the suspension of Y,we obtain

Now,consider the exact sequence in cohomology for the pair(Y+,Y):

Since H3(Y+,Y)=0 and H4(Y+,Y)is torsion free by(9.1)and H3(Y)is a torsion group by(9.2),it follows from the exact sequence(9.3)that

Next,consider the exact sequence in cohomology for the pair(X,Y+):

Similarly to the above argument,H3(X,Y+)=0 and H4(X,Y+)is torsion free by(9.1)and H3(Y+)is a torsion group by(9.4),so it follows from the exact sequence(9.5)that

Thus X=X(Q,v)is the desired example when d≥2.

1 0 Appendix

In this appendix,we observe that when X is a compact simplicial toric variety of complex dimension n,a result of Fischli[7]or Jordan[11]implies that H2n−1(X)N/and TorH2n−2(X)∧2N/(∧ N),where TorH2n−2(X)denotes the torsion part of H2n−2(X).This result agrees with Proposition 3.1 since Q is contractible in this case.

Let Δ be a simplicial complete fan of dimension n and let X be the associated compact simplicial toric variety.Let M be the free abelian group dual to N.Since N=Hom(S1,T),M can be thought of as Hom(T,S1).According to[7,Theorem 2.3]or[11,Theorem 2.5.5],

where

is the sum of inclusion maps with signs,Δ(1)denotes the set of one-dimensional cones in Δ and τ⊥denotes the subspace of M ⊗ R which vanish on τ.

We shall interpret the above in terms of N.Let σ be a cone of dimension n − k in Δ.Then we have

where Nσis the intersection of N with the subspace of N ⊗R spanned by σ.The last isomorphism above is given as follows.Choose a base ρ1,···,ρn−kof Nσ.Since Nσis of rank n − k,∧n−kNσis a free abelian group of rank one and ρ1∧···∧ρn−kis its generator.For w ∈ N,we denote by[w]the element of N/Nσdetermined by w.Then the following correspondence

is well defined and gives the desired isomorphism from ∧k−ℓ(N/Nσ)to(∧n−kNσ)∧ (∧k−ℓN).This isomorphism is independent of the choice of the base ρ1,···,ρn−kup to sign.Namely,the isomorphism(10.2)depends only on the choice of orientations on M(or N)and σ.

Applying(10.2)to σ = τ∈ Δ(1)and σ =0,we obtain

Since δris the sum of inclusion maps with signs,the image of δ1(resp. δ2)in(10.1)can be identified with(resp.∧N)and hence

AcknowledgementsWe thank Tony Bahri,Soumen Sarkar and Jongbaek Song for their interest and useful comments on the paper.We also thank Matthias Franz for his comments and for his development of the Maple package torhom which was very useful in our research.Finally we thank the anonymous referee for helpful comments to improve the presentation of the paper.

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