MAS61015 Algebraic Topology

18. Subdivision

Video (Introduction to subdivision)

Consider the following pictures:

On the left we have the 1-simplex Δ1, divided into two pieces. In Lemma 10.27 we showed that if u and v are joinable paths in X, then u*v=u+v(modB1(X)). Equivalently, if we start with a path w and split it in the middle to get two paths u and v, then w=u+v(modB1(X)). Thus, there is a sense in which subdivision of paths acts as the identity in homology.

In the middle picture we have divided the simplex Δ2 into 6 pieces. In the right hand picture, and the interactive demonstration, we have divided Δ3 into 24 pieces. It will again turn out that this kind of subdivision acts as the identity in homology. To prove this, we need to study the combinatorics of the subdivision process.

We subdivide Δ1 by introducing a new vertex in the middle (which we call the barycentre), giving two copies of Δ1. To subdivide Δ2, we first subdivide each of the 3 edges in the same way as Δ1, giving 3×2=6 edges altogether. We then take each of these subdivided edges and connect it to the barycentre of Δ2; this divides Δ2 into 2×3 copies of Δ2. Next, we divide all 4 of the faces of Δ3 in the same way as Δ2, giving 4×3×2=24 triangles on the surface of Δ3. We connect all of these to the barycentre of Δ3; this divides Δ3 into 4×3×2 copies of Δ3. We can continue in the same way to divide Δn into (n+1)! copies of Δn. We now start to make this more formal.

Video (Definition 18.1 to Proposition 18.10)

Definition 18.1.
  • (a)

    The barycentre of Δn is the point bn=(1,,1)/(n+1)Δn (so b3=(14,14,14,14), for example). We will write b instead of bn if there is no danger of confusion.

  • (b)

    Given any linear k-simplex u=a0,,akCk(Δn), we define


    More generally, if u=n1u1++nrur with n1,,nr and each ui being a linear k-simplex, we define β(u)=n1β(u1)++nrβ(ur).

Remark 18.2.

It is possible to define β for nonlinear k-simplices, but a little work is required to check that the resulting maps Δk+1Δn are always continuous. We do not need the general case so we omit it.

Lemma 18.3.

Let u be a linear combination of linear k-simplices in Δn with k>0. Then β(u)+β(u)=u.


We can easily reduce to the case where u is a single linear k-simplex, say u=a0,,ak. Let ui be the same as u except that ai is omitted, so (u)=i(-1)iui, so β(u)=i(-1)iβ(ui). On the other hand, we have β(u)=b,a0,a1,,ak. For the initial term in β(u) we omit the b and we have a sign (-1)0; this just gives us u. For each of the remaining terms in β(u) we omit the ai appearing in position i+1 of β(u), and multiply by (-1)i+1; this gives -(-1)iβ(ui), which cancels with a term in β(u). Putting everything together gives β(u)+β(u)=u as claimed. ∎

We now want to define certain elements θnCn(Δn) for all n. The idea is that we subdivide Δn into smaller copies of Δn as sketched previously, and take θn to be the sum of these smaller copies with suitable ±-signs to make the orientations match up correctly. We can mark some points in Δ1 and Δ2 as follows:

It will work out that

θ0 =e0
θ1 =e01,e1-e01,e0
θ2 =e012,e12,e2-e012,e12,e1-e012,e02,e2+e012,e02,e0+e012,e01,e1-e012,e01,e0

The general picture is as follows.

Definition 18.4.

We start with θ0=e0C0(Δ0). Now suppose that we have n>0 and we have already defined an element θn-1Cn-1(Δn-1) which is a -linear combination of linear simplices. For i=0,,n we have an affine map δi:Δn-1Δn and thus a chain (δi)#(θn-1)Cn-1(Δn), which is again a -linear combination of linear simplices. We put

θn =i=0n(-1)i(δi)#(θn-1)Cn-1(Δn)
θn =β(θn)=i=0n(-1)iβ((δi)#(θn-1))Cn(Δn).

This defines θn for all n by recursion. Next, suppose we have a space X and a map u:ΔnX, so uSn(X)Cn(X). The map u:ΔnX then gives a map u#:Cn(Δn)Cn(X), and we define sd(u)=u#(θn).

Example 18.5.

The map δ0:Δ1Δ2 sends e0, e1 and e01 to e1, e2 and e12 respectively. It follows that

(δ0)#(θ1) =(δ0)#(e01,e1-e01,e0)=e12,e2-e12,e1
β((δ0)#(θ1)) =e012,e12,e2-e012,e12,e1.

After expressing β((δ1)#(θ1)) and β((δ2)#(θ1)) in the same way, we obtain the advertised formula for θ2:

Example 18.6.

Suppose we have a path u:Δ1X. We identify Δ1 with [0,1] as usual, so the points e0, e1 and e01 become 0, 1 and 12 respectively. The map e01,e1:Δ1Δ1 is thus t(1+t)/2, and the map e01,e0:Δ1Δ1 is t(1-t)/2. This means that sd(u)=v-w, where v(t)=u((1+t)/2) and w(t)=((1-t)/2). In other words, v is the second half of u and w is the reverse of the first half of u, so u=w¯*v.

Remark 18.7.

An alternative approach is as follows. Let π be a permutation of {0,,n}. For 0in we put eiπ=(n-i+1)-1j=ineπ(j)Δn. This gives a linear n-simplex uπ=e0π,,enπSnΔn. It can be shown that θn=πsgn(π)uπCn(Δn). We could instead have taken this formula as the definition of θn; that would make some things easier and some other things harder.

Lemma 18.8.

For any f:XY and any uCn(X) we have f#(sd(u))=sd(f#(u)) in Cn(Y).


We can easily reduce to the case where uSn(X), or in other words u:ΔnX. We then have


Lemma 18.9.

If we let ιn denote the identity map ΔnΔn considered as an element of Cn(Δn), then we have θn=sd((ιn)), and therefore θn=β(sd((ιn))).


We have (ιn)=i=0n(-1)i(ιnδi)=i=0n(-1)iδi. By definition sd is linear and has sd(δi)=(δi)#(θn-1). It follows that sd((ιn))=i=0n(-1)isd(δi)=i=0n(δi)#(θn-1)=θn as claimed. ∎

Proposition 18.10.

The map sd:C*(X)C*(X) is a chain map.


We must show that for all n0 and all uCn(X) we have (sd(u))=sd((u)) in Cn-1(X). If n=0 then Cn-1(X)=0 and so the claim is clear. For n>0 we will argue by induction. We can easily reduce to the case where uSn(X)Cn(X), or in other words u:ΔnX. We then have sd(u)=u#(θn), and u# is a chain map, so (sd(u))=(u#(θn))=u#((θn)). On the other hand, we have


It will therefore be enough to prove that (θn)=θn.

We are assuming inductively that (sd(v))=sd((v)) for all vCn-1(X). We can take v=(ιn), so (v)=2(ιn)=0; it follows that (sd((ιn)))=0, or in other words (θn)=0. We also know from Lemma 18.3 that (β(θn))+β((θn))=θn. As β(θn)=θn and (θn)=0 this can be rewritten as (θn)=θn, as required. ∎

Proposition 18.11.

The chain map sd:C*(X)C*(X) is chain-homotopic to the identity.


We define chains κnCn+1(Δn) recursively as follows. We start with κ0=0. Now suppose that n>0 and we have already defined κn-1Cn(Δn-1). For 0in we have a face inclusion δi:Δn-1Δn, using which we can form (δi)#(κn-1)Cn(Δn). We put


and κn=β(κn)Cn+1(Δn); this completes the recursion step.

Next, given uSn(X) we note that u:ΔnX and κnCn+1(Δn) so u#(κn)Cn+1(X). We define σ(u)=u#(κn), and extend this linearly to get σ:Cn(X)Cn+1(X). We will prove that this gives the required chain homotopy.

As a first step, we will reformulate the definition of κn. We have (δi)#(κn-1)=σ(δi) so


It follows that


We will now prove by induction that (σ(u))+σ((u))=u-sd(u) for all spaces X and all uCn(X). When n=0 the claim is just that 0=0, which is true. Suppose we have proved the claim for n-1. We can then apply it to the element (ιn)Cn-1(Δn); we find that


Using 2=0 and sd=sd we can rewrite this as


or in other words (κn)=0. We can therefore take u=κn in Lemma 18.3 to get (β(κn))=κn. After recalling our formula above for κn and the definition σ(ιn)=κn=β(κn) we get


Now suppose we have a map u:ΔnX. We apply u# to the above equation, noting that u#=u# and u#sd=sdu# and u#σ=σu# and u#(ιn)=u. We get


or equivalently (σ(u))+σ((u))=u-sd(u). We have proved this for uSn(X), but it follows by linearity for all uCn(X), as required. ∎

Video (Definition 18.12 and Lemma 18.13)

Definition 18.12.

Let u:ΔnN be a linear simplex. We define

diam(u)=max{u(s)-u(t)|s,tΔn}=max{u(ei)-u(ej)| 0i,jn},

and we call this the diameter of u. More generally, given a chain u=m1u1++mrurCn(N) we put diam(u)=max(diam(u1),,diam(ur)).

Lemma 18.13.

If uCn(N) is a -linear combination of linear simplices then we have diam(sd(u))nn+1diam(u).


In the case n=0 all diameters are zero so the claim is clear. We can therefore assume that n>0 and argue by induction. The claim involves the number cn=n/(n+1)=1-(n+1)-1; from the second form it is clear that 0cn<cn+1<1.

We can easily reduce to the case where u is a single linear simplex, say u=a0,,an. Put d=diam(u), so ai-ajd for all i,j. Put b=(a0++an)/(n+1), which is the barycentre of u. For any i we can write ai as (n+1)-1j=0nai. Using this, we get ai-b=(n+1)-1j=0n(ai-aj). In the sum on right hand side, the term for j=i is zero and the other n terms have norm at most d; it follows that ai-bnn+1d=cnd. More generally, consider a point xu(Δn), say x=i=0ntiai with ti0 and iti=1. We can write b as itib, so


Now let v be a simplex occuring in sd(u). Then v=β(w) for some w occuring in sd(uδi) for some i, so the vertices of v are the vertices of w together with b. It is clear that diam(uδi)d so by induction we have diam(w)cn-1dcnd. Also, from the discussion above, any vertex in w has distance at most cnd from b. It follows that v has diameter at most cnd as required. ∎