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Framed points in one component of the frame bundle are framed cobordant

Statement

Assume ACω. Let M be a closed smooth m-manifold, m≥1, and let c:[0,1]→B(M) be a smooth path in the frame bundle from (x0,b0) to (x1,b1). Then the framed points (x0,b0) and (x1,b1), regarded as closed framed 0-dimensional submanifolds of M of codimension m with framings bi:Rm→TxiM, are framed cobordant in M.

Facts & Assumptions

Given: A closed smooth m-manifold M, m≥1, points x0,x1∈M, linear isomorphisms bi:Rm→TxiM, and a smooth path c:[0,1]→B(M) with c(0)=(x0,b0), c(1)=(x1,b1) (The frame bundle of a smooth manifold).

[F1]

The frame bundle B(M) is a smooth manifold with smooth projection π and smooth right action; writing c(s)=(x(s),b(s)), both s↦x(s) and s↦b(s) are smooth, and composing c with a smooth nondecreasing reparametrization λ:[0,1]→[0,1] with λ=0 near 0 and λ=1 near 1 gives a smooth path with the same endpoints that is constant near the ends (The frame bundle of a smooth manifold, Smooth embeddings, The chain rule for differentials of smooth maps). The interval is compact and its graph image is compact (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism).

[F2]

A framed cobordism from a closed framed codimension-k submanifold (N0,φ0) to (N1,φ1) in a closed X is data (W,ε,Ψ): a compact neat embedded W⊆X×I with ∂W=N0×{0}⊔N1×{1}, product ends W∩(X×[0,ε))=N0×[0,ε) and W∩(X×(1−ε,1])=N1×(1−ε,1], and a framing Ψ of ν(W⊆X×I) the pullback of φi over each end collar, along which the I-direction is tangent to W (Framed cobordism of framed submanifolds, Framings of a normal bundle, Normal and conormal bundles of an embedded submanifold, Neat submanifolds of a manifold with boundary, The Axiom of Countable Choice (ACω)).

Proof

technique · constructive
1.1F1givenconstruct

Choose a smooth nondecreasing λ:[0,1]→[0,1] given explicitly by λ(t)=σ(3t−1), with σ the standard smooth step function (The standard smooth step function) and ε=1/8, and replace c by the reparametrized smooth path c∘λ with the same endpoints, so that x(λ(t))=x0 and b(λ(t))=b0 for t≤ε and x(λ(t))=x1, b(λ(t))=b1 for t≥1−ε.

2.1F1step 1.1given

Define W:={(x(λ(t)),t):t∈[0,1]}⊆M×I. The map t↦(x(λ(t)),t) is smooth and injective (the second coordinate separates points) with derivative having second component 1≠0, so W is a compact embedded 1-submanifold with boundary the two endpoints (x0,0) and (x1,1); it is neat in M×[0,1], and by step 1.1 its ends are exactly W∩(M×[0,ε))={x0}×[0,ε) and W∩(M×(1−ε,1])={x1}×(1−ε,1].

3.1F2step 1.1step 2.1

Write γ(t)=x(λ(t)). At (γ(t),t) define Qt:Tγ(t)M⊕R→Tγ(t)M by Qt(v,a)=v−aγ˙(t). Its kernel is precisely R(γ˙(t),1)=T(γ(t),t)W, and it is surjective since Qt(v,0)=v; hence it induces a smooth isomorphism of the normal quotient with Tγ(t)M. The framing is Ψt=b(λ(t))−1∘Qt on that quotient. On each end collar γ˙=0 and b(λ(t))=bi, so Ψ is exactly the product pullback of φi=bi−1. This supplies the required quotient map and the framing in the trivialization direction of [F2].

4.1F2step 3.1discharge-construct∎

Therefore (W,ε,Ψ) satisfies all the data of a framed cobordism from the framed point (x0,φ0) to (x1,φ1) in the sense of [F2], and reading the framings through the frame-bundle dictionary the two framed points (x0,b0) and (x1,b1) are framed cobordant. No choice beyond the inherited countable choice and the finite choice of λ and ε is used.

Depends on

Used by

Dependency tree · two levels

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