Lecture 14: October 13 Embeddings of Manifolds. Recall from Last Time the Deﬁnition of a Partition of Unity

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Lecture 14: October 13 Embeddings of manifolds. Recall from last time the deﬁnition of a partition of unity. Given a ﬁnite open covering X = U U of a topological space, a 1 ∪···∪ n partition of unity is a collection of continuous functions φ1,...,φn : X [0, 1] with the following two properties: →

(1) The support of φk is contained in the open set Uk. (2) We have n φ (x) = 1 for every x X. k=1 k ∈ We can use Urysohn’s lemma to construct a partition of unity for every ﬁnite open covering of a normal space.

Lemma 14.1. If X is normal, then every finite open covering X = U1 Un admits a partition of unity. ∪···∪ Proof. We first use normality to find an open covering by slightly smaller open sets, and then apply Urysohn’s lemma to produce the desired functions. Step 1. We can find an open covering X = V1 Vn such that Vk Uk for every k =1,...,n. To get started, consider the set∪···∪ ⊆ A = X U U . \ 2 ∪···∪ n It is closed and contained in the open set U1, because X = U1 Un.Since X is normal, we can find an open set V with A V V U ;∪ by··· construction,∪ 1 ⊆ 1 ⊆ 1 ⊆ 1 V1,U2,...,Un is still an open covering of X. To get the remaining open sets, we proceed by induction. Suppose that we already have V1,...,Vk 1 with X = V1 Vk 1 Uk Un.Then − ∪···∪ − ∪ ∪···∪ B = X V1 Vk 1 Uk+1 Un \ ∪···∪ − ∪ ∪···∪ is closed and contained in the open set Uk.SinceX is normal, we can find an open set Vk with B Vk Vk Uk;nowV1,...,Vk,Uk+1,...,Un still cover X, and so we can continue⊆ the⊆ process⊆ until we reach k = n. Step 2. We construct the desired partition of unity φ + + φ = 1. Repeating 1 ··· n the argument in Step 1 for the open covering X = V1 Vn, we can find an open covering X = W W by even smaller open∪ sets··· with∪ 1 ∪···∪ n W V V U . k ⊆ k ⊆ k ⊆ k [Draw a picture.] Now the closed sets Wk and X Vk are disjoint, and so Urysohn’s lemma tells us that there is a continuous function\ ψ : X [0, 1] with k → 1ifx Wk, ψk(x)= ∈ 0ifx X V . ∈ \ k Because ψk is identically zero on X Vk, the support of ψk is contained in Vk Uk. To turn these functions into a partition\ of unity, we now consider their sum ⊆ ® ψ + + ψ . 1 ··· n At every x X, the value of the sum is at least 1: indeed, we have ψk(x) = 1 for x W , and∈ the open sets W ,...,W cover X. Therefore ∈ k 1 n ψ φ = k k ψ + + ψ 1 ··· n 2 is a continuous function from R into [0, 1], with the property that Supp φk Uk. Since φ + + φ = 1 is clear, we have found the desired partition of unity.⊆ 1 ··· n We can use the existence of partitions of unity to prove the following embedding theorem for compact manifolds. Theorem 14.2. Every compact manifold can be embedded into Euclidean space. Proof. Let X be a compact manifold of dimension m; we will construct an embedding F : X RN for some large integer N. By definition, every point of X → has a neighborhood homeomorphic to an open set in Rm;sinceX is compact, finitely many of these neighborhoods will cover X. We thus get an open covering X = U U as well as embeddings 1 ∪···∪ n m gk : Uk R . → Let φ1 + + φn = 1 be a partition of unity; it exists by Lemma 14.1 because X is compact··· Hausdorff, hence normal. For every k =1,...,n, consider the function

m φk(x)gk(x)ifx Uk, fk : X R ,fk(x)= ∈ → 0ifx X Supp φ . ∈ \ k The two definitions are compatible on the intersections of the open sets Uk and X Supp φk, and so fk is continuous. \Now it is an easy matter to obtain® the desired embedding. Set N = n + mn, and consider the function N n m n F : X R = R (R ) ,F(x)= φ1(x),...,φn(x),f1(x),...,fn(x) . → × Clearly, F is continuous; we want to show that it defines an embedding of X into RN . Because X is compact and RN is Hausdorff, all we have to do is prove that F is injective: a continuous bijection between a compact space and a Hausdorffspace is automatically a homeomorphism! So suppose that we have two points x, y X with F (x)=F (y). This means that φ (x)=φ (y) and f (x)=f (y) for every∈k =1,...,n.Sinceφ + +φ = 1, we k k k k 1 ··· n can find some index k for which φk(x)=φk(y) > 0. This forces x, y Uk (because Supp φ U ); but then ∈ k ⊆ k φk(x)gk(x)=fk(x)=fk(y)=φk(y)gk(y). m After dividing by φk(x)=φk(y), we see that gk(x)=gk(y); but gk : Uk R was → injective, and so x = y. The choice of N is far from optimal: with some additional tricks, one can show that every compact manifold of dimension m can actually be embedded into R2m.