1. Barycentric Subdivision

Michaelmas Term 2016 O. Randal-Williams

Part III Algebraic Topology // The small simplices theorem

U Let U = {Uα}α∈I be a collection of subsets of X whose interiors cover X, and C• (X) ⊂ C•(X) be the sub-chain complex generated by those singular simplices σ : ∆n → X whose image lies entirely within some Uα. The goal of this note is to prove the following theorem.

U U Theorem 1. The map H∗ (X) := H∗(C• (X)) → H∗(X) is an isomorphism.

1. Barycentric subdivision.

N Deﬁnition 2. If x = {x0, . . . , xn} is a collection of points in R which span an n-simplex, we 1 P n n+1 write bx = n+1 xi for the barycentre of the simplex x. In particular, we write bn ∈ ∆ ⊂ R for the barycentre of the standard n-simplex.

n n Let us write ιn : ∆ → ∆ for the identity map considered as a singular n-simplex, so as an n i n element of Cn(∆ ). If σ : ∆ → ∆ is a singular i-simplex, let

∆n i+1 n Conei (σ) : ∆ −→ ∆ (t1, . . . , ti+1) (t0, t1, . . . , ti+1) 7−→ t0 · bn + (1 − t0) · σ , 1 − t0

n n where we have used that ∆ is convex to linearly interpolate between bn, σ(t1, . . . , ti+1) ∈ ∆ . ∆n n n This construction extended linearly gives a homomorphism Conei : Ci(∆ ) → Ci+1(∆ ), which satisﬁes ( ∆n ∆n σ − Conei−1(dσ) i > 0 d(Conei (σ)) = σ − (σ) · bn i = 0. n n Therefore, if we let c• : C•(∆ ) → C•(∆ ) be the chain map given by c0(σ) = (σ) · bn on a 0-simplex σ, and by ci(σ) = 0 on simplices of higher dimension, then

∆n ∆n n dCone + Cone d = idC•(∆ ) − c•.

n Remark 3. This in particular shows that Hi(∆ ) = 0 for i > 0.

X Definition 4. If p• : C•(X) → C•(X) is a collection of chain maps, one for each space X, we say X Y they are natural if for each map f : X → Y of spaces we have fn ◦ pn = pn ◦ fn. We make the X analogous definition for a collection of chain homotopies F• : C•(X) → C•+1(X). X Definition 5. Define homomorphisms ρn : Cn(X) → Cn(X) inductively by: X (i) Let ρ0 = idC0(X) for all spaces X.

X (ii) If ρn−1 has been deﬁned for all spaces X, let

X ρn : Cn(X) −→ Cn(X) ∆n ∆n σ 7−→ σ#(Conen−1(ρn−1(dιn))).

X Lemma 6. ρ• : C•(X) → C•(X) is a natural chain map.

1 Proof. If f : X → Y then

X ∆n ∆n ∆n ∆n f#(ρn (σ)) = f#σ#(Conen−1(ρn−1(dιn))) = (f ◦ σ)#(Conen−1(ρn−1(dιn)))

Y Y which is ρn (f ◦ σ) = ρn (f#(σ)), so this is natural. X X Let us suppose for an induction that dρn−1 = ρn−2d for all spaces X, which is certainly satisﬁed when n − 1 = 0. Then for n ≥ 1 calculate

X ∆n ∆n dρn (σ) = σ#(d(Conen−1(ρn−1(dιn)))) ∆n ∆n ∆n = σ#(ρn−1(dιn) − Conen−2(dρn−1(dιn))) ∆n ∆n ∆n = σ#(ρn−1(dιn) − Conen−2(ρn−1(ddιn))) ∆n X = σ#(ρn−1(dιn)) = ρn−1(dσ)

∆n X as required, where at the end we have used the naturality property σ# ◦ ρn−1 = ρn−1 ◦ σ#. X We now wish to show that ρ• is naturally chain homotopic to the identity. X Definition 7. Define homomorphisms Tn : Cn(X) → Cn+1(X) inductively by: X (i) Let T0 = 0 for all spaces X. X (ii) If Tn−1 has been defined for all spaces X, let

X Tn : Cn(X) −→ Cn+1(X) ∆n ∆n ∆n σ 7−→ σ#(Conen (ρn (ιn) − ιn − Tn−1(dιn))).

X X Lemma 8. T• : C•(X) → C•+1(X) is a natural chain homotopy from ρ• to the identity. X X Proof. It is natural for the same reason ρn was: Tn (σ) is obtained by applying σ# to an element n of Cn+1(∆ ). X X X Suppose for an induction that dTn−1 + Tn−2d = ρn−1 − idCn−1(X) for all spaces X, which is certainly satisﬁed for n − 1 = 0. Then for n ≥ 1 calculate

X ∆n ∆n ∆n dTn (σ) = σ#(dConen (ρn (ιn) − ιn − Tn−1(dιn))) ∆n ∆n ∆n n = σ#((idCn(∆ ) − Conen−1d)(ρn (ιn) − ιn − Tn−1(dιn))). Now by the inductive assumption we have

∆n ∆n ∆n ∆n ∆n d(ρn (ιn) − ιn − Tn−1(dιn)) = ρn−1(dιn) − dιn − dTn−1(dιn) = Tn−2(ddιn) = 0, so the expression simpliﬁes to

X ∆n ∆n X X dTn (σ) = σ#(ρn (ιn) − ιn − Tn−1(dιn)) = ρn (σ) − σ − Tn−1(dσ) as required, where we have again used naturality of ρX and T X .

2. Some geometry of simplices. n n+1 The standard simplex ∆ is a metric space via the metric inherited from R , so we may talk n n about the diameter of a subset of ∆ . For points v0, v1, . . . , vn ∈ ∆ , we write [v0, v1, . . . , vn]: n n P ∆ → ∆ for the map (t0, t1, . . . , tn) 7→ i tivi; when it is injective, let us also write [v0, v1, . . . , vn] for the image of this map, which is the convex hull of the vi.

Lemma 9. diam([v0, v1, . . . , vn]) = maxi,j{|vi − vj|}

2 Proof. For v ∈ [v0, v1, . . . , vn] we have

X X X v − tivi = tiv − tivi i i i X ≤ ti|vi − v| i

≤ max{|v − vj|} j and by convexity the latter is maximised when v is a vertex.

∆n n Lemma 10. Each simplex of ρn ([v0, . . . , vn]) has diameter ≤ n+1 diam([v0, v1, . . . , vn]). Proof. Let us prove this by induction on dimension; it clearly holds for 0-simplices. ∆n Now ρn ([v0, . . . , vn]) is a signed sum of n-simplices [bv, x1, . . . , xn] where bv is the barycentre of [v0, . . . , vn] and the xi lie in the boundary of [v0, . . . , vn]. If the maximal distance between two vertices of such a simplex is between two xi’s, then this takes place in a face of [v0, . . . , vn], n which has dimension < n so the distance between them is ≤ n+1 diam([v0, v1, . . . , vn]) by inductive assumption. If the maximal distance is between bv and some xi, then xi lies in some face [v0,..., vbj, . . . , vn] so |bv − xi| ≤ |bv − vk| for some k. But

X |b − v | = 1 v − n+1 v v k n+1 i n+1 k i

X = 1 v − v n+1 i k i X 1 ≤ n+1 |vi − vk| i

Proposition 11. Let U = {Uα}α∈I be a collection of subsets of X whose interiors cover.

U X U (i) If c ∈ Cn (X) then ρn (c) ∈ Cn (X) too. X k U (ii) If c ∈ Cn(X) then there is a k 0 such that (ρn ) (c) ∈ Cn (X). X n Proof. The ﬁrst part follows from naturality of ρn : if σ : ∆ → Uα and iα : Uα → X is the X Uα inclusion of spaces, then ρn ((iα)#(σ)) = (iα)#(ρn (σ)) is a sum of simplices in Uα. For the second part, by (i) and the fact that an n-chain is a ﬁnite sum of singular n-simplices, n −1 we may suppose that c is a single singular n-simplex σ : ∆ → X. Then V = {σ U˚α}α∈I is an open cover of ∆n, which is a compact metric space. By the Lesbegue Number Lemma there is n −1 an > 0 such that each -ball in ∆ is contained in some σ U˚α. By iterating Lemma 10, each ∆n k n k n simplex of (ρn ) (ιn) has diameter ≤ ( n+1 ) diam(∆ ), so by choosing k 0 we may suppose that ∆n k −1˚ each simplex of (ρn ) (ιn) has diameter less than , and so lies in some σ Uα. ∆n k V n X k ∆n k U Hence (ρn ) (ιn) ∈ Cn (∆ ), and so (ρn ) (σ) = σ#((ρn ) (ιn)) ∈ Cn (X), as required.

3 3. Proof of Theorem 1. U We consider the chain map C• (X) → C•(X) given by inclusion, and the induced map U : U H∗ (X) → H∗(X) on homology. X k U X Let [c] ∈ Hn(X). By Proposition 11 there is a k 0 such that (ρn ) (c) ∈ Cn (X). As ρ• is X k naturally chain homotopic to the identity, so is the composition (ρ• ) . One could ﬁnd a formula X for such a chain homotopy in terms of T• , but the formula does not matter so let us simply write k k k X k F• for such a chain homotopy, satisfying dFn + Fn−1d = (ρn ) − id. Then

X k k k (ρn ) (c) − c = dFn (c) + Fn−1d(c)

X k but the last term vanishes as c is a cycle (because it represents a homology class). Thus (ρn ) (c) U is equivalent to c modulo boundaries, so U : Hn (X) → Hn(X) is surjective. U Now let [c] ∈ Hn (X) be such that U([c]) = 0 ∈ Hn(X). Thus there is a z ∈ Cn+1(X) such that X k U d(z) = c ∈ Cn(X). By Proposition 11 there is a k 0 such that (ρn+1) (z) ∈ Cn+1(X), and we have X k k k (ρn+1) (z) − z = dFn+1(z) + Fn d(z) and so applying d we get X k k d((ρn+1) (z) − Fn d(z)) = d(z) = c. X k U U Now (ρn+1) (z) ∈ Cn+1(X) by our choice of k, and as d(z) = c ∈ Cn (X) and the chain homotopy k k U U U F• is natural, Fn d(z) ∈ Cn+1(X) too. Thus c is a boundary in C• (X), so [c] = 0 ∈ Hn (X), and U hence U : Hn (X) → Hn(X) is injective.