The Complement of Polyhedral Product Spaces and the Dual Simplicial Complexes
Qibing Zheng
School of Mathematical Science and LPMC, Nankai University
Tianjin 300071, China
[email protected] Supported by Natural Science
Foundation of China, grant No. 11071125 and No. 11671154
Key words and phrases: complement space; dual complex;
polyhedral product complex; universal algebra.
Mathematics subject classification: 55N10
Abstract
In this paper, we define and prove basic properties
of complement polyhedral product spaces, dual complexes and polyhedral product complexes.
Then we compute the universal algebra of polyhedral product complexes under certain split conditions
and the Alexander duality isomorphism on certain polyhedral product spaces.
TableofContents
Section 1 Introduction
Section 2 Complement Spaces, Dual Complexes and Polyhedral Product Complexes
Section 3 Homology and Cohomology Group
Section 4 Universal Algebra
Section 5 Duality Isomorphism
1 Introduction
The polyhedral product theory, especially the homotopy type of polyhedral product
spaces, is developing rapidly nowadays. The first known polyhedral product space was the moment-angle
complex introduced by Buchstaber and Panov [8] and
was widely studied by mathematicians in the area of toric topology and geometry
(see [1],[9],[12],[13],[14],[15]).
Later on, the homotopy types of polyhedral product spaces were studied by
Grbić and Theriault [13],[14],[15], Beben and Grbić [7],
Bahri, Bendersky, Cohen and Gitler [3],[4],[5] and many others
([6],[10],[11]).
The cohomology ring of homology split polyhedral product spaces and the
cohomology algebra over a field of polyhedral product spaces were computed in [17].
In this paper, a polyhedral product space Z(K;X,A) with
(X,A)={(Xk,Ak)}k=1m is general than usual in that each (Xk,Ak) is
a topological pair but not a CW-complex pair (see Definition 2.1).
Then for M=Z(K;X,A),
is the complement space Mc=(X1×⋯×Xm)∖M a polyhedral product space?
In Theorem 2.4, we show that Mc=Z(K∘;X,Ac), where K∘
is the dual complex of K relative to [m] and Akc=Xk∖Ak is the complement space of Ak.
Let Z(K;Y,B),
(Y,B)={(Yk,Bk)}k=1m be the polyhedral product space
defined as follows. For each k, (Yk,Bk) is a pair of polyhedral product spaces given by (sk=n1+⋯+nk)
[TABLE]
where (Xk,Ak) is a simplicial pair on [nk].
In theorem 2.9, we prove that
Z(K;Y,B) is also a polyhedral product space
Z(S(K;X,A);U,C),
where S(K;X,A) is the polyhedral product complex
defined in Definition 2.7.
When all (Xk,Ak)=(Δnk,Kk), the simplicial complex S(K;X,A) is
denoted by S(K;K1,⋯,Km) which is just
the composition complex K(K1,⋯,Km) defined by Anton in Definition 4.5 of [2].
In Section 3,
we compute the reduced (co)homology group and the (right) total (co)homology group of polyhedral product complexes
from the point of view of split inclusion (Theorem 3.9 and Theorem 3.11) .
In Example 3.10, we show that if K and all Lk are homology spheres, then
S(K;L1,⋯,Lm) is a homology sphere. This result is in accordance
with that of [2], where the homotopy type of S(K;L1,⋯,Lm)
is studied by Anton.
In section 4, we compute cohomology algebra of a wide class of complexes in Theorem 4.5
including the cohomology algebra of Z(K;Y,B) mentioned above in Example 4.6.
In Theorem 5.6, we compute the Alexander duality isomorphism on the pair (X1×⋯×Xm,Z(K;X,A)),
where all Xk’s are orientable manifolds and all Ak’s are polyhedra.
2 Complement Spaces, Dual Complexes
and Polyhedral Product complexes
Conventions and Notations For a finite set S, ΔS is the simplicial complex with only one maximal simplex S,
i.e., it is the set of all subsets of S including the empty set ∅.
Define ∂ΔS=ΔS∖{S}.
For [m]={1,⋯,m}, Δ[m] is simply denoted by Δm.
Specifically, define
Δ0=Δ∅={∅} and ∂Δ0={}.
The void complex {} with no simplex is inevitable in this paper.
For a simplicial complex K on [m] (ghost vertex {i}∈/K is allowed) and σ⊂[m] (σ∈/K is allowed),
the link of σ with respect to K is the simplicial complex
linkKσ={τ∣σ∪τ∈K,σ∩τ=∅}.
This implies linkKσ={∅} if σ is a maximal simplex of K and linkKσ={} if σ∈/K.
Specifically, if K={}, then linkKσ={} for all σ.
Definition 2.1
For a simplicial complex K on [m]
and a sequence of topological (not CW-complex!) pairs (X,A)={(Xk,Ak)}k=1m,
the polyhedral product space Z(K;X,A)
is the subspace of X1×⋯×Xm defined as follows.
For a subset τ of [m], define
[TABLE]
Then Z(K;X,A)=∪τ∈KD(τ).
Empty space ∅ is allowed in a topological pair and ∅×X=∅ for all X.
Define Z({};X,A)=∅.
Notice that D(σ)=Z(Δσ;X,A),
D(∅)=A1×⋯×Am=Z({∅};X,A) and
D([m])=X1×⋯×Xm=Z(Δm;X,A).
But ∅=Z({};X,A) has no corresponding D(−).
Example 2.2 For Z(K;X,A),
let S={k∣Ak=∅}. Then
[TABLE]
where (X′,A′)={(Xk,Ak)}k∈/S and link is as defined in conventions.
Definition 2.3 Let K be a simplicial complex with vertex set a subset of S=∅.
The dual of K relative to S is the simplicial
complex
[TABLE]
It is obvious that
(K∘)∘=K, (K1∪K2)∘=(K1)∘∩(K2)∘ and
(K1∩K2)∘=(K1)∘∪(K2)∘. Specifically, (ΔS)∘={}
and (∂ΔS)∘={∅}.
Theorem 2.4 For Z(K;X,A), the complement space
[TABLE]
*where (X,Ac)={(Xk,Akc)}k=1m with Akc=Xk∖Ak
and K∘ is the dual of K relative to [m]. The polyhedral product space Z(K∘;X,Ac)
is called the complement of Z(K;X,A).
*
Proof For σ⊂[m] but σ=[m] (σ=∅ is allowed),
[TABLE]
So for K=Δm or {},
[TABLE]
For K=Δm or {}, the above equality holds naturally. □
Example 2.5 Let F be a field and V be a linear space over F with base e1,⋯,em.
For a subset σ={i1,⋯,is}⊂[m], denote by F(σ) the subspace of V with base ei1,⋯,eis.
Then for F=R or C and a simplicial complex K on [m], we have
[TABLE]
[TABLE]
This example is applied by Grujić and Welkerin in Lemma 2.4 in [16].
Theorem 2.6 Let K and K∘ be the dual of each other relative to [m].
The index set Xm={(σ,ω)∣σ,ω⊂[m],σ∩ω=∅}.
For (σ,ω)∈Xm, define simplicial complex
Kσ,ω=linkKσ∣ω={τ⊂ω∣σ∪τ∈K}
(so Kσ,ω={} if σ∈/K or K={}).
Then for any (σ,ω)∈Xm such that ω=∅,
[TABLE]
*where (Kσ,ω)∘ is the dual of
Kσ,ω relative to ω.
*
Proof Suppose σ∈K. Then
[TABLE]
If σ∈/K, then (K∘)σ~,ω=Δω=(Kσ,ω)∘.
□
A sequence of simplicial pairs (X,A)={(Xk,Ak)}k=1m in this paper means that
the vertex set of Xk is a subset of [nk] (nk>0) which is the subset
[TABLE]
of [n] with n=n1+⋯+nm.
For simplicial complexes Y1,⋯,Ym such that the vertex set of Yk is a subset of [nk],
the union simplicial complex is
[TABLE]
Definition 2.7 Let K be a simplicial complex on [m] and
(X,A) be as above.
The polyhedral product complex S(K;X,A)
is the simplicial complex on [n] defined as follows. For a subset τ⊂[m],
define
[TABLE]
Then S(K;X,A)=∪τ∈KS(τ).
Void complex {} is allowed in a simplicial pair and {}∗X={} for all X.
Define Z({};X,A)={}.
Example 2.8 For S(K;X,A),
let S={k∣Ak={}}. Then
[TABLE]
where (X′,A′)={(Xk,Ak)}k∈/S and link is as defined in conventions.
Theorem 2.9 Let Z(K;Y,B),
(Y,B)={(Yk,Bk)}k=1m be the polyhedral product space
defined as follows. For each k, (Yk,Bk) is a pair of polyhedral product spaces given by
[TABLE]
where (Xk,Ak) is a simplicial pair on [nk].
Then
[TABLE]
where (U,C)={(Uk,Ck)}k=1n,
n=n1+⋯+nm.
Proof If K={} or Uk=∅ for some k (this implies Yl=∅ for some l), then
Z(K;Y,B)=Z(S(K;X,A);U,C)=∅.
So we suppose K={} and Uk=∅ for all k in the remaining proof.
We first prove the case Ck=∅ for all k.
If Xk={} for some k, then Yk=∅ and S(K;X,A)={}.
So Z(K;Y,B)=Z(S(K;X,A);U,C)=∅.
Suppose Xk={} for all k.
Let S={k∣Ak={}}={k∣Bk=∅}. If S∈K, then linkKS={}.
From Example 2.2 and Example 2.8 we have
Z(K;Y,B)=Z(S(K;X,A);U,C)=∅.
Suppose S∈K.
Let Zkτ=Yk if k∈τ and Zkτ=Bk if k∈/τ.
For τk⊂[nk], Wtτk=Uk if t∈τk and Wtτk=Ck if k∈[nk]∖τk. Then
[TABLE]
where τ,τ1,⋯,τm are taken over all subsets such that τ∈K and S⊂τ, τk∈Xk if k∈τ
and τk∈Ak if k∈/τ.
Now we prove the case σ={k∣Ck=∅}=∅.
Let σk=σ∩[nk]. Then from Example 2.2 we have
[TABLE]
where (Uk′,Ck′)={(Uk,Ck)}k∈[nk]∖σk
and (Y′,B′)={(Yk′,Bk′)}k∈/σ.
Denote link(X,A)σ={(linkXkσk,linkAkσk)}k=1m.
Then S(K;link(X,A)σ)=linkS(K;X,A)σ
(this equality is a special case of Theorem 2.10 for (σ,ω)=(σ,[n]∖σ) and so
(−)σ,ω=link(−)σ). So
[TABLE]
where (U′,C′)={(Uk,Ck)}k∈/σ.
□
With this theorem we see that to compute the cohomology algebra of Z(K;Y,B),
we have to compute the universal algebra of S(K;X,A), which is the central work
of this paper.
Theorem 2.10 * For the S(K;X,A) in Definition 2.7 and
(σ,ω)∈Xn ( the simplicial complex (−)σ,ω is as defined in Theorem 2.6),*
[TABLE]
where (Xσ,ω,Aσ,ω)={((Xk)σk,ωk,(Ak)σk,ωk)}k=1m,
σk=σ∩[nk], ωk=ω∩[nk].
Proof By definition, (K∪L)σ,ω=Kσ,ω∪Lσ,ω
and (Y1∗⋯∗Ym)σ,ω=(Y1)σ1,ω1∗⋯∗(Ym)σm,ωm.
Let Ykτ=Xk if k∈τ and Ykτ=Ak if k∈/τ (the void complex {} is allowed).
Then
[TABLE]
□
Notice that the dual of S(K;X,A) relative to [n] is in general not a polyhedral product complex.
But the following special type of polyhedral product complexes is closed with respect to duality.
Definition 2.11 S(K;L1,⋯,Lm) is the polyhedral product complex S(K;X,A)
such that each pair (Xk,Ak)=(Δnk,Lk).
The complex S(K;K1,⋯,Km) is the composition complex K(K1,⋯,Km) in Definition 4.5 of [2].
Theorem 2.12 Let S(K;L1,⋯,Lm)∘ be the dual of S(K;L1,⋯,Lm) relative to [n].
Then
[TABLE]
where K∘ is the dual of K relative to [m] and Lk∘ is the dual of Lk relative to [nk].
So if K and all Lk are self dual (X=X∘ relative to its non-empty vertex set),
then S(K;L1,⋯,Lm) is self dual.
Proof For σ⊂[m] but σ=[m] (σ=∅, Lk=Δnk or {} are allowed),
[TABLE]
So for K=[m] or {},
[TABLE]
For K=Δm or {}, the equality holds naturally.
□
3 Homology and Cohomology Group
This is a paper following [17].
All the basic definitions such as indexed groups and (co)chain complexes, diagonal tensor product,
etc., are as in [17].
In this section, we compute the reduced simplicial (co)homology group and
the (right) total (co)homology group of polyhedral product complexes uniformly from the point of view of split inclusion.
Conventions In this paper, a group A∗Λ=⊕α∈ΛA∗α indexed by Λ
is simply denoted by A∗ when there is no confusion.
So is the (co)chain complex case.
The diagonal tensor product A∗Λ⊗ΛB∗Λ in [17] is simply denoted by
A∗Λ\mathaccent866⊗B∗Λ in this paper (so Λ can not be abbreviated in this case).
Definition 3.1 Let A∗=⊕α∈ΛA∗α, B∗=⊕α∈ΛB∗α
be two groups indexed by the same set Λ.
An indexed group homomorphism f:A∗→B∗ is the direct sum
f=⊕α∈Λfα such that each fα:A∗α→B∗α
is a graded group homomorphism. Define groups indexed by Λ as follows.
[TABLE]
For indexed group homomorphism f=⊕α∈Λfα and g=⊕β∈Γgβ,
their tensor product f⊗g is naturally an indexed group homomorphism with
f⊗g=⊕(α,β)∈Λ×Γfα⊗gβ.
For indexed group homomorphism f=⊕α∈Λfα and g=⊕α∈Λgα indexed by
the same set,
their diagonal tensor product f\mathaccent866⊗g is the indexed group homomorphism
f\mathaccent866⊗g=⊕α∈Λfα⊗gα.
Similarly, we have the definition of indexed (co)chain homomorphism
by replacing the indexed groups in the above definition by indexed (co)chain complexes.
Definition 3.2 An indexed group homomorphism θ:U∗→V∗
is called a split homomorphism if kerθ, cokerθ and imθ are all free groups.
An indexed chain homomorphism
ϑ:(C∗,d)→(D∗,d) with induced homology group homomorphism θ:U∗→V∗
(U∗=H∗(C∗), V∗=H∗(D∗)) is called a split inclusion if C∗ is a chain subcomplex of the free complex D∗ and
θ is a split homomorphism.
For a topological pair (X,A), let ϑ:S∗(A)→S∗(X) be the singular chain complex inclusion.
Regard this inclusion as an indexed chain homomorphism such that the index set has only one element.
Then ϑ is a split inclusion if and only if (X,A) is homology split as defined in Definition 2.1 in [17].
The work of this and the next section is just to generalize all the work in [17] from
the singular chain complex case to indexed total chain complex case.
Definition 3.3 Let θ:U∗→V∗ be a split homomorphism
with dual homomorphism θ∘:V∗→U∗.
The index set X=X1 (Xm is as defined in Definition 2.6) and R={{∅,∅},{∅,{1}}}⊂X.
\SS=X or R.
The indexed groups H∗\SS(θ)=⊕s∈\SSH∗s(θ)
and its dual groups H\SS∗(θ∘)=⊕s∈\SSHs∗(θ∘)
are given by
[TABLE]
[TABLE]
The indexed chain complexes (C∗\SS(θ),d)=⊕s∈\SS(C∗s(θ),d)
and its dual cochain complexes (C\SS∗(θ∘),δ)=⊕s∈\SS(Cs∗(θ∘),δ)
are given by
[TABLE]
[TABLE]
where d is trivial on C∗∅,∅(θ) and C∗{1},∅(θ)
and is the desuspension isomorphism on C∗∅,{1}(θ).
Let θ=⊕α∈Λθα.
Then H∗\SS(θ) is also a group indexed by Λ
and so denoted by H∗\SS(θ)=H∗\SS;Λ(θ)=⊕s∈\SS,α∈ΛH∗s;α(θ) with
[TABLE]
Other cases are similar.
Theorem 3.4 For a split inclusion ϑ:(C∗,d)→(D∗,d)
with induced homology homomorphism θ:U∗→V∗,
there are quotient chain homotopy equivalences q and q′
satisfying the following commutative diagram
[TABLE]
where ϑ′ is the inclusion by identifying U∗=kerθ⊕coimθ with
kerθ⊕imθ⊂C∗X(θ) (d is trivial on U∗).
There are also isomorphisms ϕ and ϕ′ of chain complexes indexed by X
satisfying the following commutative diagram
[TABLE]
where S∗X and T∗X are as in Definition 4.2 and Theorem 4.3 in [17], 1* is the identity and i is the inclusion.*
If θ is an epimorphism, then
H∗X(θ)=H∗R(θ)=U∗ by identify imθ
with coimθ and so all X is replaced by R.
Proof Take a representative ai in C∗ for every generator of kerθ
and let ai∈D∗ be any element such that dai=ai.
Take a representative bj in C∗ for every generator of imθ.
Take a representative ck in D∗ for every generator of cokerθ.
So we may regard U∗ as the chain subcomplex of C∗ freely generated by all ai’s and bj’s
and regard (C∗X(θ),d) as the chain subcomplex of D∗ freely generated by all
ai’s, ai’s, bj’s and ck’s.
Then we have the following commutative diagram of short exact sequences of chain complexes
[TABLE]
Since all the complexes are free, i’s have group homomorphism inverse. H∗(C∗/U∗)=0 and H∗(D∗/C∗X(θ))=0
imply that the inverse of i’s are complex homomorphisms. So we may take q,q′ to be the inverse of i’s.
ϕ is defined as shown in the following table.
[TABLE]
□
Definition 3.5 For k=1,⋯,m,
let ϑk:((Ck)∗,d)→((Dk)∗,d) be a split inclusion with induced
homology group homomorphism θk:(Uk)∗→(Vk)∗.
Denote ϑ={ϑk}k=1m, θ={θk}k=1m
and their dual ϑ∘={ϑk∘}k=1m, θ∘={θk∘}k=1m.
The index set \SS=X or R.
The indexed group H∗\SSm(θ) and its dual group H\SSm∗(θ∘)
(the index set \SSm as in Theorem 2.6) are given by
[TABLE]
Denote H∗\SSm(θ)=⊕(σ,ω)∈\SSmH∗σ,ω(θ),
H\SSm∗(θ∘)=⊕(σ,ω)∈\SSmHσ,ω∗(θ∘). Then
[TABLE]
[TABLE]
The indexed chain complex (C∗\SSm(θ),d) and its dual cochain complex
(C\SSm∗(θ∘),δ) are given by
[TABLE]
Definition 3.6 Let K be a simplicial complex on [m]
and everything else be as in Definition 3.5.
The indexed chain complex (C∗\SSm(K;θ),d) is the subcomplex of
(C∗\SSm(θ),d) defined as follows.
For a subset τ of [m], define
[TABLE]
Then (C∗\SSm(K;θ),d)=(+τ∈KH∗(τ),d). Define
(C∗\SSm({};θ),d)=0.
So the dual cochain complex (C\SSm∗(K;θ∘),δ) of
(C∗\SSm(K;θ),d) is a quotient complex of (C\SSm∗(θ∘),δ).
The chain complex (C∗(K;ϑ),d) is the subcomplex of ((D1)∗⊗⋯⊗(Dm)∗,d)
defined as follows.
For a subset τ of [m], define
[TABLE]
Then (C∗(K;ϑ),d)=(+τ∈KE∗(τ),d).
Define (C∗({};ϑ),d)=0.
So the dual cochain complex (C∗(K;ϑ∘),δ) of (C∗(K;ϑ),d)
is a quotient complex of ((D1)∗⊗⋯⊗(Dm)∗,δ).
Theorem 3.7 For the K, ϑ and θ in Definition 3.5 and Definition 3.6,
there is a quotient chain homotopy equivalence ( Xm neglected)
[TABLE]
and an isomorphism of chain complexes indexed by Xm
[TABLE]
So we have (co)homology group isomorphisms
[TABLE]
[TABLE]
where T∗Xm(K), TXm∗(K), H∗Xm(K) and HXm∗(K) are as in Definition 4.5 and Theorem 4.7 in [17].
If each θk is an epimorphism, then
H∗X(θk)=H∗R(θk)=(Uk)∗, HX∗(θk∘)=HR∗(θk∘)=(Uk)∗
and so all X is replaced by R.
Proof Denote by qk and qk′ the chain homotopy equivalence q and q′ in Theorem 3.4 for
ϑ=ϑk.
For σ⊂[m], let φσ=p1⊗⋯⊗pm,
where pk=qk if k∈σ and pk=qk′ if k∈/σ.
So φσ is a chain homotopy equivalence.
Then φ(K;ϑ)=+σ∈Kφσ is also a chain homotopy equivalence.
Denote by ϕk and ϕk′ the isomorphism ϕ and ϕ′ in Theorem 3.4 for
θ=θk.
For σ⊂[m], let ϕσ=r1⊗⋯⊗rm,
where rk=ϕk if k∈σ and rk=ϕk′ if k∈/σ.
So ϕσ is an isomorphism.
Then ϕ(K;θ)=+σ∈Kϕσ is also an isomorphism.
□
Definition 3.8 A polyhedral product complex S(K;X,A) is homology split
if the reduced simplicial homology homomorphism
[TABLE]
induced by inclusion is split
for k=1,⋯,m.
A polyhedral product complex S(K;X,A) is total homology split
if the reduced simplicial homology homomorphism
[TABLE]
induced by inclusion is split
for k=1,⋯,m and all (σk,ωk)∈Xnk.
Theorem 3.9* For homology split S(K;X,A),*
[TABLE]
[TABLE]
where H∗Xm(−)=⊕(σ,ω)∈XmH∗σ,ω(−),
HXm∗(−)=⊕(σ,ω)∈XmHσ,ω∗(−) with
[TABLE]
[TABLE]
[TABLE]
[TABLE]
where ιk is as in Definition 3.8 with dual ιk∘ and
Σ means suspension.
If each ιk is an epimorphism, then all X is replaced by R and we have H∗R(Xk,Ak)=H∗(Ak).
If the reduced simplicial (co)homology is taken over a field, then the conclusion holds for all
polyhedral product complexes.
Proof A corollary of Theorem 3.7 by
taking ϑ={ϑk}k=1m with split inclusion
ϑk:(Σ\mathaccent869C∗(Ak),d)→(Σ\mathaccent869C∗(Xk),d)
the suspension reduced simplicial complex inclusion. Regard this graded group inclusion as
an indexed chain homomorphism such that the index set has only one element.
Then (C∗(K;ϑ),d)=(Σ\mathaccent869C∗(S(K;X,A)),d)
and H∗X(θk)=H∗X(Xk,Ak).
□
Example 3.10 For S(K;X,A)=S(K;L1,⋯,Lm) such that all
H∗(Lk) is free,
each ιk:\mathaccent869H∗(Lk)→\mathaccent869H∗(Δnk)(=0) is an epimorphism.
By definition, H∗∅,∅(Δnk,Lk)=0,
H∗∅,{1}(Δnk,Lk)=\mathaccent869H∗−1(Lk),
[TABLE]
[TABLE]
So by Theorem 3.9,
[TABLE]
[TABLE]
If K and all Lk are homology spheres (\mathaccent869H∗(−)≅Z, so {∅}
is a homology sphere but {} is not),
then S(K;L1,⋯,Lm) is a homology sphere. The homotopy type
of S(K;L1,⋯,Lm) is discussed in [2].
We have ring isomorphism H∗(S(K;X,A))≅H∗(∣S(K;X,A)∣), where ∣⋅∣ means geometrical
realization.
So \mathaccent869H∗(S(K;X,A)) is
a ring by adding a unit to it.
This ring is not considered in this paper.
Theorem 3.11* For a total homology split S(K;X,A), we have*
[TABLE]
[TABLE]
where \SSn=\SSn1×⋯×\SSnm,
\SS=X or R (\SSn1=Xn1, \SSn2=Rn2 is possible).
For (σ^,ω^)∈Xm, (σ,ω)∈\SSn,
σk=σ∩[nk], ωk=ω∩[nk],
[TABLE]
[TABLE]
[TABLE]
[TABLE]
where ισk,ωk is as in Definition 3.8
with dual ισk,ωk∘ and Σ means suspension.
If each ισk,ωk is an epimorphism, then all X is replaced by R and we have
H∗R;\SSnk(Xk,Ak)=H∗\SSnk(Ak),
HR;\SSnk∗(Xk,Ak)=H\SSnk∗(Ak).
If the (right) total (co)homology group is taken over a field, then the theorem holds for all
polyhedral product complexes.
Proof A corollary of Theorem 3.7
by taking ϑ={ϑk}k=1m with split inclusion
ϑk:T∗\SSnk(Ak)→T∗\SSnk(Xk)
the (right) total chain complex (as in Definition 4.6 in [17]) inclusion.
Then (C∗(K;ϑ),d)=(T∗\SSn(S(K,X,A)),d)
and H∗X(θk)=H∗X;\SSk(θk)=H∗X;\SSk(Xk,Ak).
□
Example 3.12 Apply Theorem 3.11 for S(K;X,A)=S(K;L1,⋯,Lm),
Lk={} for k=1,⋯,m.
So either all H∗Rnk(Lk) are free or the (co)homology is taken over a field.
For \SS=R, θk:H∗Rnk(Lk)→H∗Rnk(Δnk)≅Z
is an epimorphism. So
[TABLE]
[TABLE]
For \SS=X, θk:H∗Xnk(Lk)→H∗Xnk(Δnk)≅Z
is not an epimorphism.
To simplify notation, H∗−;−(Δnk,Lk) is abbreviated to H∗−;−.
For ωk=∅, H∗σk,ωk(Δnk)=0.
For σk∈Lk, H∗σk,∅(Lk)≅H∗σk,∅(Δnk)≅Z.
For σk∈/Lk, H∗σk,∅(Lk)=0, H∗σk,ωk(Δnk)≅Z.
So H∗∅,{1};σk,ωk≅H∗σk,ωk(Lk) for ωk=∅,
H∗∅,∅;σk,∅≅Z for σk∈Lk and
H∗{1},∅;σk,∅≅Z for σk∈/Lk. So
we have that for (σ,ω)∈Xn,
[TABLE]
[TABLE]
where σ^={k∣σk∈/Lk,ωk=∅}, ω^={k∣ωk=∅},
σk=σ∩[nk], ωk=ω∩[nk].
4 Universal Algebra
In this section, we compute the (right) universal (normal, etc.) algebra of total homology split polyhedral product complexes.
The (co)associativity is not required for a (co)algebra as in [17].
Theorem 4.1 Let ϑ:(C∗,d)→(D∗,d) be a split inclusion
with induced homology homomorphism θ:U∗→V∗
such that ϑ is also a coalgebra homomorphism ϑ:(C∗,ψC)→(D∗,ψD)
with induced homology coalgebra homomorphism θ:(U∗,ψU)→(V∗,ψV).
Then the group C∗X(θ) in Definition 3.3 has a unique character coproduct
\mathaccent866ψ(ϑ) satisfying the following three conditions.
i) \mathaccent866ψ(ϑ) makes the following diagram (q, q′ and ϑ′ as in Theorem 3.4).
{\scriptstyle C_{*}}$${\scriptstyle U_{*}}$${\scriptstyle C_{*}{\otimes}C_{*}}$${\scriptstyle U_{*}{\otimes}U_{*}}$${\scriptstyle D_{*}}$${\scriptstyle C_{*}^{\mathscr{X}\!}}$${\scriptstyle D_{*}{\otimes}D_{*}}$${\scriptstyle C_{*}^{\mathscr{X}\!}{\otimes}C_{*}^{\mathscr{X}\!}}$${}^{q^{\prime}}ϑ**q{}^{\vartheta^{\prime}}$${}^{q^{\prime}{\otimes}q^{\prime}}ϑ⊗ϑ**q⊗q{}^{\vartheta^{\prime}{\otimes}\vartheta^{\prime}}$${}^{\psi_{C}}$${}^{\psi_{U}}$${}^{\psi_{D}}$${}^{\mathaccent 866{\psi}(\vartheta)}$${\scriptstyle(C_{*}^{\mathscr{X}}=\,C_{*}^{\mathscr{X}}\!(\theta))}
commutative except the homotopy commutative (q⊗q)ψD≃\mathaccent866ψ(ϑ)q.
ii) \mathaccent866ψ(ϑ) is independent of the choice of ψC,ψD up to homotopy,
i.e., if ψC,ψD are replaced by ψC′,ψD′ such that ψC′≃ψC, ψD′≃ψD
and we get \mathaccent866ψ′(ϑ) for ψC′ and ψD′,
then \mathaccent866ψ′(ϑ)=\mathaccent866ψ(ϑ).
iii) Denote by α=cokerθ, β=Σkerθ,
γ=kerθ, η=imθ.
Then \mathaccent866ψ(ϑ) satisfies the following four conditions.
(1) \mathaccent866ψ(ϑ)(η)⊂η⊗η⊕γ⊗η⊕η⊗γ⊕γ⊗γ.
(2) \mathaccent866ψ(ϑ)(γ)⊂γ⊗γ⊕γ⊗η⊕η⊗γ.
(3) \mathaccent 866{\psi}(\vartheta)(\beta)\subset\big{(}\beta{\otimes}\gamma\oplus\beta{\otimes}\eta\oplus\eta{\otimes}\beta\big{)}\oplus\big{(}\alpha{\otimes}\alpha\oplus\alpha{\otimes}\eta\oplus\eta{\otimes}\alpha\oplus\eta{\otimes}\eta\big{)}.
(4) \mathaccent866ψ(ϑ)(α)⊂α⊗α⊕α⊗η⊕η⊗α⊕η⊗η.
Proof Denote N⊕(α⊕η)⊗(α⊕η)=C∗X⊗C∗X.
By Künneth theorem, H∗(N)=0 and H∗(C∗X⊗C∗X)=(α⊕η)⊗(α⊕η).
From the construction of q and q′ in Theorem 3.4 we have that there is a subcomplex F∗ of D∗ such that
H∗(F∗)=0 and
(D∗,d)=(F∗,d)⊕(C∗X,d). q is the projection such that
ϑ(kerq′)⊂F∗.
This implies that if we take f to be the composite
[TABLE]
then f makes the cubic diagram of the theorem commutative except the homotopy commutative
(q⊗q)ψD≃fq.
Construct coproduct chain homomorphism
ψ:(C∗X,d)→(C∗X⊗C∗X,d)
and homotopy s:(C∗X,d)→(ΣC∗X⊗C∗X,d)
such that ds+sd=f−ψ and ψ is independent of the choice of f
just as in the proof of Theorem 2.8 in [17] by regarding the graded group α,β,γ,η
in that theorem as the indexed groups with the same symbol in this theorem.
□
Definition 4.2 For the ϑ in Theorem 4.1, all the chain complexes in Theorem 3.4 are coalgebras defined as follows.
The character coalgebra complex of ϑ is (C∗X(θ),\mathaccent866ψ(ϑ)).
If θ is an epimorphism, then C∗X(θ)=C∗R(θ) and (C∗R(θ),\mathaccent866ψ(ϑ))
is called the right character coalgebra complex of ϑ.
The homology coalgebra of ϑ is (H∗X(θ),ψ(ϑ)) with coproduct
defined as follows.
(1) ψ(ϑ)(x)=\mathaccent866ψ(ϑ)(x) for all x∈α⊕η.
(2) For a generator x∈γ, there is a unique generator x∈β such that dx=x.
Suppose \mathaccent866ψ(ϑ)(x)=z+y with z∈β⊗γ⊕β⊗η⊕η⊗β
and y∈α⊗α⊕α⊗η⊕η⊗α⊕η⊗η.
Then define ψ(ϑ)(x)=\mathaccent866ψ(ϑ)(x)+y.
If θ is an epimorphism, then (H∗R(θ)(=H∗X(θ)),ψ(ϑ)) is the right homology coalgebra of ϑ.
The group isomorphisms H∗R(θ)≅V∗ is in general not an algebra isomorphism.
For \SS=X or \SS, the dual algebra (HR∗(θ∘),π(ϑ∘)))
of (H∗\SS(θ),ψ(ϑ)) is the (right) cohomology algebra of ϑ.
The index coalgebra complex
(T∗X,\mathaccent866ψϑ) of ϑ is defined as follows.
Let symbols x,x1′,x2′′,⋯ be one of α,β,γ,η
that are both the generators of T∗X and the summand groups of C∗X(θ) in
Theorem 4.1. If the summand group x satisfies \mathaccent866ψ(ϑ)(x)⊂⊕ixi′⊗xi′′
such that no summand xi′⊗xi′′ can be canceled,
then the generator x satisfies \mathaccent866ψϑ(x)=Σixi′⊗xi′′.
If θ is an epimorphism, then (T∗R,ψϑ) is
the right index coalgebra complex of ϑ.
Theorem 4.3 For the coalgebras in Definition 4.2,
all the chain homomorphisms in Theorem 3.4 induce cohomology algebra isomorphisms.
Proof q and q′ induce isomorphisms by Theorem 4.1. The proof that ϕ
is a coalgebra isomorphism is the same as that of Theorem 6.4 in [17].
□
Definition 4.4 Let K, ϑ, θ be
as in Definition 3.5 and Definition 3.6
such that each ϑk:((Ck)∗,ψCk)→((Dk)∗,ψDk)
and θk:((Uk)∗,ψUk)→((Vk)∗,ψVk)
satisfy the condition of Theorem 4.1. \SS=X or R.
Then all the chain complexes in Theorem 3.7 are coalgebras defined as follows.
The subgroup C∗(K;ϑ) of (D1)∗⊗⋯⊗(Dm)∗ is a subcoalgebra and
is called the (right) coalgebra complex of (K;ϑ). Its cohomology algebra is denoted by
(H∗(C∗(K;ϑ∘)),∪(K;ϑ∘)).
Define (C∗\SSm(ϑ),\mathaccent866ψ(ϑ))=(C∗\SS(ϑ1)⊗⋯⊗C∗\SS(ϑm),\mathaccent866ψ(ϑ1)⊗⋯⊗\mathaccent866ψ(ϑm)).
Then the subgroup C∗\SSm(K;ϑ) of C∗\SSm(ϑ) is a subcoalgebra and
is called the (right) character coalgebra complex of (K;ϑ).
The (right) homology coalgebra of ϑ is
(H∗\SSm(θ),ψ(ϑ))=(H∗\SS(θ1)⊗⋯⊗H∗\SS(θm),ψ(ϑ1)⊗⋯⊗ψ(ϑm)). The (right) cohomology algebra of ϑ is the dual algebra
(H\SSm∗(θ∘),π(ϑ∘))=(H\SS∗(θ1∘)⊗⋯⊗H\SS∗(θm∘),π(ϑ1∘)⊗⋯⊗π(ϑm∘)).
Define (T∗\SSm,\mathaccent866ψϑ)=(T∗\SS⊗⋯⊗T∗\SS,\mathaccent866ψϑ1⊗⋯⊗\mathaccent866ψϑm).
Then the subgroup T∗\SSm(K) of T∗\SSm (as defined in Definition 4.5 in [17]) is a subcoalgebra
and is called the (right) index coalgebra complex of K induced by ϑ.
Its cohomology algebra is called the (right) index cohomology algebra of K induced by ϑ and
is denoted by (H\SSm∗(K),∪(K;ϑ∘)).
Theorem 4.5 For the coalgebras in Definition 4.4,
all the chain homomorphisms in Theorem 3.7 induce cohomology algebra isomorphisms.
So we have cohomology algebra isomorphism
[TABLE]
Proof The φσ and ϕσ in the proof of Theorem 3.7 as a tensor product induce isomorphisms by
Theorem 4.3.
So φ(K,ϑ)=+σ∈Kφσ and ϕ(K,θ)=+σ∈Kϕσ
also induce isomorphisms.
□
Example 4.6 Let everything be as in Theorem 2.9.
We compute the cohomology algebra of Z(K;Y,B) over a field.
By Theorem 6.9 and 7.11 in [17], we have
[TABLE]
where \SSnk=Xnk or Rnk (if H∗(Ci)→H∗(Ui) induced by inclusion is an epimorphism for i∈[nk])
and ∪Xk and ∪Ak may be a (right) universal (or (right) normal, etc.) product appearing in the theorems.
Take ϑk:(T∗\SSnk(Ak),ψk)→(T∗\SSnk(Xk),ψk) to be the (right) total
chain complex inclusion and apply Theorem 4.5 for ϑ={ϑk}k=1m.
Then we have
[TABLE]
[TABLE]
where \SSn=\SSn1×⋯×\SSnm (\SSn1=Xn1, \SSn2=Rn2 is possible).
Specifically, take each (Xk,Ak)=(Δnk,Lk) and suppose H∗(Ci)→H∗(Ui)
induced by inclusion is an epimorphism for all i.
Then the coalgebra homomorphism θk:H∗Rm(Lk)→H∗Rm(Δnk)≅Z
is an epimorphism. By Theorem 4.1 for ϑ=ϑk,
η≅H∅,∅∗(Lk)≅H∅,∅∗(Δnk)≅Z,
γ≅⊕ωk=∅H∅,ωk∗(Lk), α=0.
Since η⊗η is non-zero only at degree [math] and β is zero at degree [math],
we have that η⊗η can not be a summand of \mathaccent866ψ(ϑk)(β). So
\mathaccent866ψ(ϑk)(η)⊂η⊗η.
\mathaccent866ψ(ϑk)(γ)⊂γ⊗γ⊕γ⊗η⊕η⊗γ.
\mathaccent866ψ(ϑk)(β)⊂β⊗γ⊕β⊗η⊕η⊗β.
So \mathaccent866ψϑk is the right strictly normal coproduct in Definition 7.4 in [17]
and the group isomorphism H∗R;Rnk(Δnk,Lk)≅H∗Rnk(Lk) is an algebra isomorphism.
So we have cohomology algebra isomorphism
[TABLE]
where HRm∗(K) is the right strictly normal algebra of K.
5 Duality Isomorphism
In this section, we compute the Alexander duality isomorphism
on some special type of polyhedral product spaces.
Theorem 5.1 Let (X,A)={(Xk,Ak)}k=1m be a sequence of topological
pairs satisfying the following conditions.
1) Each homology group homomorphism ik:H∗(Ak)→H∗(Xk)
induced by inclusion is a split homomorphism.
2) Each Xk is a closed orientable manifold of dimension rk.
3) Each Ak is a proper compact polyhedron subspace of Xk.
Let (X,Ac)={(Xk,Akc)}k=1m with Akc=Xk∖Ak.
Then for all (σ,ω)∈Xm, there are duality isomorphisms ( r=r1+⋯+rm )
[TABLE]
[TABLE]
where \mathaccent869σ=[m]∖(σ∪ω),
H∗σ,ω(−) and Hσ,ω∗(−) are as in Theorem 3.9.
If the (co)homology is taken over a field, then the conclusion holds for
(X,A) satisfying the following conditions.
1) Each Xk is a closed manifold of dimension rk orientable with respect to the homology theory over the field.
*2) Each Ak is a proper compact polyhedron subspace of Xk.
*
Proof We have the following commutative diagram of exact sequences
[TABLE]
where αk,βk are the Alexander duality isomorphisms
and γk is the Poncaré duality isomorphism.
So we have the following group isomorphisms
[TABLE]
Define θk:H∗X(Xk,Ak)→HX∗(Xk,Akc)
to be the direct sum of the above three isomorphisms.
Then θ1⊗⋯⊗θm=⊕(σ,ω)∈Xmγσ,ω.
□
Theorem 5.2 Let K and K∘ be the dual of each other relative to [m].
Then for all (σ,ω)∈Xm, ω=∅,
there are duality isomorphisms
[TABLE]
[TABLE]
*where \mathaccent869σ=[m]∖(σ∪ω), ∣ω∣ is the cardinality of ω.
*
Proof Let (C∗(Δω,Kσ,ω),d) be the relative simplicial chain complex.
Since \mathaccent869H∗(Δω)=0, we have a boundary isomorphism
∂:H∗(Δω,Kσ,ω)⟶≅\mathaccent869H∗−1(Kσ,ω)=H∗σ,ω(K).
C∗(Δω,Kσ,ω) has a set of generators consisting of all non-simplices of Kσ,ω, i.e., Kσ,ωc={η⊂ω∣η∈Kσ,ω} is a set of generators of C∗(Δω,Kσ,ω).
So we may denote (C∗(Δω,Kσ,ω),d) by (C∗(Kσ,ωc),d), where
η∈Kσ,ωc has degree ∣η∣−1 with ∣η∣ the cardinality of η.
The correspondence η→ω∖η for all η∈Kσ,ωc induces a dual complex isomorphism
ψ:(C∗(Kσ,ωc),d)→(\mathaccent869C∗((Kσ,ω)∘),δ).
Since (Kσ,ω)∘=(K∘)σ~,ω, we have induced homology group isomorphism
ψˉ:H∗(Δω,Kσ,ω)→Hσ~,ω∣ω∣−∗−1(K∘).
Define γK,σ,ω=ψˉ∂−1.
□
Notice that for σ∈K, [m]∖σ may not be a simplex of K∘.
In this case, there is no isomorphism from H∗σ,∅(K)=Z
to Hσ~,∅∗(K∘)=0.
Example 5.3 For the S(K;L1,⋯,Lm) and index sets σ,ω,σ^,ω^,σk,ωk in Example 3.12,
γS(K;L1,⋯,Lm),σ,ω=γK,σ^,ω^⊗(⊗ωk=∅γLk,σk,ωk).
Definition 5.4 For homology split M=Z(K;X,A), let
i:H∗(M)→H∗(\mathaccent869X) and i∗:H∗(\mathaccent869X)→H∗(M)
be the singular (co)homology homomorphism induced by the inclusion map from M to
\mathaccent869X=X1×⋯×Xm.
From the long exact exact sequences
[TABLE]
we define
[TABLE]
[TABLE]
Theorem 5.5 For a homology split space M=Z(K;X,A),
we have the following group decompositions
[TABLE]
and direct sum group decompositions
[TABLE]
where Xm={(σ,ω)∈Xm∣ω=∅}.
*The conclusion holds for all polyhedral product spaces if the (co)homology group is taken over a field.
*
Proof By definition, i=⊕(σ,ω)∈Xmiσ,ω with
iσ,ω:H∗σ,ω(K)⊗H∗σ,ω(X,A)−⟶i⊗1H∗σ,ω(Δm)⊗H∗σ,ω(X,A),
where i is induced by inclusion and 1 is the identity.
H∗σ,ω(Δm)=0 if ω=∅,
H∗σ,∅(K)=Z if σ∈L
and H∗σ,∅(K)=0 if σ∈/L.
So
[TABLE]
[TABLE]
The relative group case is similar.
□
Theorem 5.6 For the space M=Z(K;X,A) such that (X,A)
satisfies the condition of Theorem 5.1,
the Alexander duality isomorphisms
[TABLE]
satisfy α=α^⊕α, α∗=α^∗⊕α∗,
where
[TABLE]
[TABLE]
are as follows. Identify all the above groups with the direct sum groups in Theorem 5.5.
Then
[TABLE]
[TABLE]
*where γσ,ω, γσ,ω∗ are as in Theorem 5.1
and γK,σ,ω, γK,σ,ω∗ are as in Theorem 5.2.
*
Proof Denote by α=αM, α^=α^M,
α=αM.
Then for M=Z(K;X,A) and N=Z(L;X,A),
we have the following commutative diagrams of exact sequences
[TABLE]
[TABLE]
For (σ,ω)∈Xm,
A=Hlσ,ω(X,A),
B=Hσ,ωr−∣ω∣−l(X,Ac),
γ1=γK∩L,σ,ω, γ2=γK,σ,ω⊕γL,σ,ω,
γ3=γK∪L,σ,ω, we have the commutative diagram
[TABLE]
The direct sum of all the above diagrams is the following diagram.
[TABLE]
(1), (2) and (3) imply that
if the theorem holds for M and N and M∩N, then it holds for M∪N.
So by induction on the number of maximal simplices of K,
we only need prove the theorem for the special case that K has only one maximal simplex.
Now we prove the theorem for M=Z(ΔS;X,A) with S⊂[m].
Then
[TABLE]
So (\mathaccent869X,Mc)=(X1,Y1c)×⋯×(Xm,Ymc).
By identifying coimqk∗ and Σ−1im∂k∗ respectively with kerpk∗
and cokerpk∗ in the following commutative diagram
[TABLE]
we have H∗(Xk,Akc)=coimqk∗⊕Σ−1im∂k∗=kerpk∗⊕cokerpk∗⊂HX∗(Xk,Akc).
So the following diagrams are commutative
[TABLE]
where θk,αk,γk are as in the proof of Theorem 5.1.
This implies that the following diagram is commutative
[TABLE]
where the Σ∣ω∣ of γσ,ω comes from
the desuspension isomorphism Σ−1im∂k∗≅cokerpk∗
and the Σ of HX∗(X1,A1c)⊗⋯⊗HX∗(Xm,Amc) comes from
the isomorphism H∗(\mathaccent869X,Mc)≅ΣH∗(Mc).
For σ⊂S, H∗σ,ω(ΔS)=0 if ω∩S=∅ and
H∗σ,ω(ΔS)≅Z if ω∩S=∅.
So γΔS,σ,ω:0→0 if ω∩S=∅
and γΔS,σ,ω:Z→Z if ω∩S=∅.
For ω∩S=∅, identify H∗σ,ω(K)⊗H∗σ,ω(X,A) and
Hσ~,ω∗(K∘)⊗Hσ~,ω∗(X,Ac) respectively with
H∗σ,ω(X,A) and Σ∣ω∣−1Hσ~,ω∗(X,Ac), then we
have the following commutative diagram
[TABLE]
The direct sum of the above isomorphisms for all σ⊂S and ω∩S=∅ is just
the third row of (4).
α^M is the direct sum of
the above isomorphisms for all σ⊂S and ω=∅.
αM is the direct sum of the above isomorphisms for all σ⊂S, ω=∅ and
ω∩S=∅.
So αM=α^M⊕αM for the special case M=Z(ΔS;X,A).
□
Example 5.7 Regard Sr+1 as one-point compactification of Rr+1.
Then for q\mathchar13366r, the standard space pair (Sr+1,Sq) is given by
Sq={(x1,⋯,xr+1)∈Rr+1⊂Sr+1∣x12+⋯+xq+12=1,xi=0,ifi>q+1}.
Let M={\cal Z}_{K}\Big{(}\!\begin{array}[]{ccc}\scriptstyle{r_{1}{+}1}\!&\!\scriptstyle{\cdots}\!&\!\scriptstyle{r_{m}{+}1}\\
\scriptstyle{q_{1}}\!&\!\scriptstyle{\cdots}\!&\!\scriptstyle{q_{m}}\end{array}\!\Big{)}={\cal Z}(K;\underline{X},\underline{A}) be the polyhedral product space
such that (Xk,Ak)=(Srk+1,Sqk).
Since Sr−q is a deformation retract of Sr+1∖Sq,
the complement space Mc=Z(K∘;X,Ac)
is homotopic equivalent to {\cal Z}_{K^{\circ}}\Big{(}\begin{array}[]{ccc}\scriptstyle{r_{1}+\,1}&\scriptstyle{\cdots}&\scriptstyle{r_{m}+\,1}\\
\scriptstyle{r_{1}{-}q_{1}}&\scriptstyle{\cdots}&\scriptstyle{r_{m}{-}q_{m}}\end{array}\Big{)}.
Since all H∗σ,ω(X,A)≅Z,
we may identify H∗σ,ω(K)⊗H∗σ,ω(X,A)
with ΣtH∗σ,ω(K),
where t=Σk∈σ(rk+1)+Σk∈ωqk.
For σ⊂[m], let Zσ be the free group generated by
σ with degree [math]. Then
[TABLE]
[TABLE]
Dually, the cohomology of the complement space Mc is
[TABLE]
[TABLE]
In this case, the direct sum of γK,σ,ω:H∗σ,ω(K)→Hσ~,ω∣ω∣−∗−1(K∘)
over all (σ,ω)∈Xm (regardless of degree) is the isomorphism
H∗(M)≅Hr−∗−1(Mc).
Specifically, {\cal Z}(K;S^{2n+1},S^{n})={\cal Z}_{K}\Big{(}\!\begin{array}[]{ccc}\scriptstyle{2n{+}1}&\scriptstyle{\cdots}&\scriptstyle{2n{+}1}\\
\scriptstyle{n}&\scriptstyle{\cdots}&\scriptstyle{n}\end{array}\!\Big{)}.
Then we have
[TABLE]
