3.6 Derived and Barycentric Subdivisions 175 By Lemma 3.6.1, we have
KC =K∂C∪ {vC} ∪
vCσσ∈K∂C
⊳F(C) andK∂C is full inKC. Thus, we have a simplicial complex
Kn′ =Kn′−1∪
C∈K(n)\Kn−1
KC.
It is easy to see thatKn′ ⊳Kn,L′ is a full subcomplex inKn′, and Kn′(0) =K0′(0)∪ {vC|C∈K(n)\L}.
By induction, we have Kn′ ⊳ Kn, n ∈ N, with the above conditions.
Observe thatK′=
n∈NKn′ is the desired simplicial subdivision. ⊓⊔ WhenL = ∅ in Proposition 3.6.2 above, the obtained subdivision K′ is called aderived subdivisionofK. This is written as follows:
K′ =
vC1, . . . , vCnC1<· · ·< Cn∈K ,
and K′ is an ordered simplicial complex with the natural order on K′(0) defined as follows: vC vD if C D. If K is simplicial and vσ = ˆσ for each σ ∈ K, the derived subdivision K′ of K is called the barycentric subdivisionofK, which is denoted by SdK.
L K
SdLK
Fig. 3.5.Definition of SdLK
WhenLis simplicial andL′=Lin Proposition 3.6.2, we callK′aderived subdivisionofK relative toL, where
K′=L∪
vC1, . . . , vCnC1<· · ·< Cn ∈K\L
∪
v1, . . . , vm, vC1, . . . , vCnv1, . . . , vm ∈L, C1<· · ·< Cn ∈K\Lwithv1, . . . , vm< C1
.
IfKandLare simplicial andvσ = ˆσfor eachσ∈K\L, the derived subdivision ofKrelative toLis called thebarycentric subdivisionofKrelative toL,
which is denoted by SdLK (cf. Figure 3.5). Note thatLis a full subcomplex of SdLK by Proposition 3.6.2.
For simplicial complexesKandL, since the barycentric subdivisions SdK and SdLare ordered simplicial complexes, we can define the product simplicial complex SdK×sSdL, that consists of simplexes
(ˆσ1,τˆ1), . . . ,(ˆσn,τˆn), σ1· · ·σn∈K, τ1· · ·τn∈L.
On the other hand, by givingvσ×τ ∈rint(σ×τ) for eachσ×τ ∈K×cL, we can define a derived subdivision of the product cell complex K×cL, which consists of simplexes
vσ1×τ1, . . . , vσn×τn, σ1×τ1<· · ·< σn×τn∈K×cL.
Ifvσ×τ = (ˆσ,τˆ) for eachσ×τ ∈K×cL, the derived subdivision ofK×cL is simply SdK×sSdL.
Applying these derived subdivisions, we prove the following:
3.6.3 Theorem LetK,K0, andLbe simplicial complexes such thatK0 ⊂ K∩L, |L| ⊂ |K|, andL\K0 is finite. Then,K has a simplicial subdivision K′ such that K0 is a full subcomplex of K′ and L is subdivided by some subcomplex ofK′.
Proof. We show the theorem by induction onn= card(L\K0). In the case n= 0, a derived subdivisionK′ ofKrelative toL=K0is the desired one. In the case n >0, let σbe a maximal dimensional simplex ofL\K0. Sinceσis a principal simplex ofL,L1=L\ {σ}is a subcomplex ofL. By the inductive assumption, we have K′ ⊳ K such that K0 is a full subcomplex of K′ and L′1 ⊳ L1 for some L′1 ⊂ K′. Then, ∂σ is triangulated by the subcomplex L′∂σ={τ∈L′1|τ⊂∂σ}ofL′1 (⊂K′). Let
L′σ =
τ∩στ ∈K′ such thatτ∩σ=∅ .
Since τ∩σ′ τ for each σ′ < σ and τ ∈K′ with τ ∩σ′ =∅, L′σ is a cell complex by Proposition 3.2.12. Then, |L′σ|=σ and L′∂σ ⊂L′σ. It should be noted that ifτ∩σ=∅ but rintτ∩σ=∅, thenτ∩σ=τ′∩σfor someτ′< τ with rintτ′∩σ=∅.
Now, consider the subcomplex K1′ = {τ ∈ K′ | τ ∩rintσ = ∅} of K′. Then,K0∪L′1⊂K1′ andσ∩ |K1′|=|L′∂σ|=∂σ. For eachτ∈K′\K1′, choose vτ ∈rintτ so that
rintτ∩rintσ=∅ ⇒ vτ∈rintτ∩rintσ,
where it should be noted that rintτ∩rintσ=∅implies rintτ′∩rintσ=∅for some τ′ < τ because τ∩rintσ=∅. Using these points vτ, we define K′′ as a derived subdivision ofK′ relative toK1′. Then,K0∪L′1⊂K′′. SinceK0is full inK1′ andK1′ is full inK′′, it follows thatK0 is full inK′′.
3.6 Derived and Barycentric Subdivisions 177
K′\K1′ σ
τ vτ
Fig. 3.6.Definition ofK′′
On the other hand, for each cellC∈L′σ\L′∂σ, letvC=vτC, whereτC∈K′ is the unique cell such thatC=τC∩σand rintτC∩σ=∅. Then,vC∈rintC.
Indeed, since rintτC∩rintσ=∅, we have rintC= rintτC∩rintσby 3.1.9(2).
Using these points vC, we have the derived subdivision L′′σ ofL′σ relative to L′∂σ. Then,L′′σis a triangulation ofσ, which is simply the subcomplex ofK′′
consisting of simplexes with vertices in σ, that is,L′′σ ={τ ∈ K′′ | τ ⊂ σ}. Thus, we have a subcomplex L′′=L′1∪L′′σ ofK′′such thatL′′⊳L. ⊓⊔
To prove that proper PL maps are simplicial with respect to some simpli-cial subdivisions, we need the following:
3.6.4 Lemma Let C1, . . . , Cn be cells contained in a cell C and K0 be a triangulation of∂Csuch that ifCi∩∂C=∅thenCi∩∂Cis triangulated by a subcomplex ofK0. Then,C has a triangulationK such thatK0 ⊂K and C1, . . . , Cn are triangulated by subcomplexes ofK.
Proof. It suffices to prove the casen= 1. Indeed, assume thatChas triangula-tionsK1, . . . , Kn such thatK0⊂Ki andCi is triangulated by a subcomplex of Ki. By Corollary 3.2.13, we have a common simplicial subdivision K of K1, . . . , Kn withK0⊂K, which is the desired triangulation ofC.
The casen= 1 can be shown as follows: Consider the cell complexF(C1) and its subcomplex L1 ={D ∈ F(C1) | D ⊂ ∂C}, where |L1| =C1∩∂C.
Indeed, for eachx∈C1∩∂C, takeDC1withx∈rintD. Then,D=Dx⊂ Cx. Sincex∈rintC impliesCx< C, we haveD⊂∂C, that is,D∈L1. Note that K0 gives the simplicial subdivisionL′1 ofL1. We can apply Proposition 3.6.2 to obtain a simplicial subdivision K1 of F(C1) with L′1 ⊂ K1. Then, K1∪K0is a triangulation ofC1∪∂C. On the other hand,F(C) has a simplicial subdivisionKwithK0⊂Kby Proposition 3.6.2. Applying Theorem 3.6.3, we can obtain a simplicial subdivisionK1′ ofK such thatK0⊂K1′ andK1∪K0
is subdivided by a subcomplex ofK1′. ⊓⊔ Now, we shall show the following:
3.6.5 Proposition LetKandLbe cell complexes. A proper mapf :|K| →
|L|is PL if and only iff is simplicial with respect to some simplicial subdivi-sionsK′⊳Kand L′⊳L.
Proof. The “if” part is obvious, where the properness off is not necessary.
To see the “only if” part, replaceK with a subdivision so thatf|C can be assumed to be affine for each C ∈ K. For each cell D ∈ L, let KD be the smallest subcomplex of K such thatf−1(D)⊂ |KD|, which also can be defined as
KD={C∈K| ∃C′∈K such that CC′, rintC′∩f−1(D)=∅}. Sincef−1(D) is compact by the properness off, it follows thatKD is a finite subcomplex ofK. According to the definition,KD′ ⊂KD forD′< D.
By induction on dimD, we can apply Lemma 3.6.4 to obtain a triangula-tionLD of eachD ∈L such that f(C)∩D is triangulated by a subcomplex of LD for each C ∈ KD and LD′ ⊂LD for everyD′ < D. Indeed, assume that LD′ has been obtained for every D′ < D. Then, L∂D =
D′<DLD′ is a triangulation of∂D. Applying Lemma 3.6.4, we have a triangulationLDof D such thatL∂D⊂LD andf(C)∩Dis triangulated by a subcomplex ofLD
for everyC∈KD.
Now, we have a simplicial subdivision L′ =
D∈LLD of L, where f(C) is triangulated by a subcomplex of L′ for every C ∈ K. For each C ∈ K and τ ∈ L′ with τ ∩f(C) = ∅, C ∩f−1(τ) = (f|C)−1(τ) is a cell and (f|C)−1(τ)x=Cx∩f−1(τf(x)) forx∈C∩f−1(τ) by 3.1.9(3). By the analogy of Proposition 3.2.12, we can show that, for eachC, C′ ∈K andτ, τ′ ∈L,
rint(C∩f−1(τ))∩rint(C′∩f−1(τ′))=∅ ⇒ C∩f−1(τ) =C′∩f−1(τ′).
Thus, the following is a cell complex:
C∩f−1(τ)C∈K, τ ∈L′, C∩f−1(τ)=∅ ,
which is a subdivision ofK. By Theorem 3.2.10, we have a simplicial subdi-vision K′ of this complex with the same vertices. Then, f is simplicial with respect to K′ and L′. ⊓⊔
As we saw at the end of§3.4, the properness off is essential in the “only if” part of Proposition 3.6.5. By Propositions 3.4.6 and 3.6.5, we have the following:
3.6.6 Corollary LetK1,K2, andK3be simplicial complexes. For each sim-plicial map f :K1→K2 and each proper PL mapg:|K2| → |K3|, there are simplicial subdivisions K1′ ⊳ K1 and K3′ ⊳K3 such that gf : K1′ → K3′ is simplicial.
Proof. Using Proposition 3.6.5, we can find simplicial subdivisionsK2′ ⊳K2
andK3′ ⊳K3such thatg:K2′ →K3′ is simplicial. Then, by Proposition 3.4.6, K1 has simplicial subdivisionsK1′ ⊳K1 such thatf :K1′ →K2′ is simplicial, whence gf:K1′ →K3′ is simplicial. ⊓⊔
3.6 Derived and Barycentric Subdivisions 179 Remark 7.As observed, a PL map cannot be defined as a mapf :|K| → |L| that is simplicial with respect to some simplicial subdivisions K′ ⊳ K and L′ ⊳L. If we adopted such a definition, then we could not assert that the composition of PL maps is PL. In fact, even if f : K1 →K2 is a simplicial isomorphism and g : |K2| → |K3| is simplicial with respect to some simpli-cial subdivisions, the composition gf is not simplicial with respect to any simplicial subdivisions. For example, let K = {n, [n, n+ 1] | n ∈ ω} and I ={0, 1, I}be the natural triangulations of [0,∞) and I, respectively. We defineK′⊳K as follows:
K′ =
n, n+ 2−(n+1), [n, n+ 2−(n+1)], [n+ 2−(n+1), n+ 1]n∈ω . Letf :K→K′ andg:K→I be the simplicial maps defined by
f(2n) =n, f(2n+ 1) =n+ 2−(n+1), g(2n) = 0 and g(2n+ 1) = 1.
Then,f :K→K′ is a simplicial isomorphism andg:|K′|=|K| → |I|=Iis a PL map. Observe thatgf(4n+ 1) = 2−(2n+1)andgf(4n+ 3) = 1−2−(2n+2) for eachn∈ω, whencegf(ω) is infinite. Since every subdivision ofKcontains ω as vertices but every subdivision ofI contains only finitely many vertices, gf is not simplicial with respect to any simplicial subdivisions ofK andI.
As we saw in§3.5, a subdivision generally changes the metric topology but the barycentric subdivision does not.
3.6.7 Theorem For each simplicial complexK,|SdK|m=|K|mas spaces.
Proof. When K is finite, the result follows from Proposition 3.5.7 and the compactness of |SdK|m. We may assume that K is infinite. By Proposition 3.5.7, it suffices to show that id :|K|m→ |SdK|m is continuous. Letx∈ |K| andk= dimcK(x). For eachε >0, chooseδ >0 so thatδ <(2k+ 3)−1εand
βKv (x)=βKv′(x) ⇒ δ < 1
2|βvK(x)−βvK′(x)|.
The last condition implies that δ < 12βvK(x) for every v ∈ cK(x)(0) because βKv′(x) = 0 for some v′ ∈ K(0). For each y ∈ |K| with ρK(x, y) < δ, the following hold:
βvK(x)> βvK′(x) ⇒ βKv (y)> βvK′(y) ; βvK(x) = 0 ⇔ βvK(y)< δ.
Note that the first implication holds even ifβvK′(x) = 0, hencecK(x)cK(y).
SinceβKv (y)βKv′(y) impliesβKv (x)βKv′(x), we can write cK(x) =v0, . . . , vk, cK(y) =v0, . . . , vn, kn, βvK0(x)· · ·βvKk(x)>0 and βvK0(y)· · ·βvKn(y)>0.
For eachi = 0, . . . , n, let σi = v0, . . . , vi. Then, σ0 < σ1 < · · · < σn, σk=cK(x), andσn=cK(y). Observe that
x=βvKk(x)vk+βvKk−1(x)vk−1+· · ·+βvK1(x)v1+βvK0(x)v0
= (k+ 1)βvKk(x)ˆσk+k
βvKk−1(x)−βvKk(x) ˆ σk−1+
· · ·+ 2
βKv1(x)−βvK2(x) ˆ σ1+
βKv0(x)−βvK1(x) ˆ σ0. Hence,x∈ σˆ0, . . . ,σˆk ∈SdK,βσSdˆkK(x) = (k+ 1)βvKk(x) and
βSdˆσi K(x) = (i+ 1)
βvKi(x)−βvKi+1(x)
fori= 0, . . . , k−1.
Similarly, we havey∈ σˆ0, . . . ,ˆσn ∈SdK,βSdˆσnK(y) = (n+ 1)βvKn(y) and βσSdˆiK(y) = (i+ 1)
βvKi(y)−βKvi+1(y)
fori= 0, . . . , n−1.
Then, it follows that ρSdK(x, y) =
k
i=0
βσSdˆiK(x)−βσSdˆiK(y)+
n
i=k+1
βSdˆσiK(y)
k
i=0
(2i+ 1)βvKi(x)−βvKi(y)+ (k+ 1)βKvk+1(y) + (k+ 2)βKvk+1(y) +
n
i=k+2
βvKi(y)
(2k+ 3)
n
i=0
βvKi(x)−βKvi(y)
= (2k+ 3)ρK(x, y)<(2k+ 3)δ < ε.
Thus, id :|K|m→ |SdK|mis continuous. ⊓⊔ We have the following generalization of 3.2.16(1):
3.6.8 Proposition For each infinite cell complex K, dens|K| = cardK(0). IfK is simplicial,dens|K|= dens|K|m= cardK(0).
Sketch of Proof.As in 3.2.16(1), we can construct a dense setDin|K|with cardD = cardK = cardK(0), hence dens|K| cardK(0). On the other hand, letK′ be a derived subdivision ofK. Then,{OK′(v)|v∈K(0)}is a pair-wise disjoint collection of open sets in|K|, hence cardK(0) c(|K|) dens|K|.
If K is simplicial, the above D is also dense in |K|m. Moreover, {OSdK(v) | v ∈ K(0)} is a pair-wise disjoint collection of open sets in
|K|m.
It should be remarked thatw(|K|m) = dens|K|m = cardK(0) for every infinite simplicial complexK(see§0.1) but, as we saw in 3.2.16(4),w(|K|)= cardK(0) in general (w(|K|)dens|K|= cardK(0)).