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Disk bundles over compact bases are compact manifolds with boundary

Statement

Let E→M be a smooth rank-q vector bundle over a boundaryless smooth manifold, with a supplied smooth bundle metric h. The closed disk bundle Dh(E)={∥v∥h≤1} is a smooth manifold with boundary of dimension dim⁡M+q, with boundary Sh(E)={∥v∥h=1} and interior {∥v∥h<1}. The projection and zero section are smooth. If M is compact, the disk bundle and its boundary are compact. For q=0 the disk bundle is M and the boundary is empty; when dim⁡M+q≥1 the boundary is a closed embedded smooth manifold of dimension dim⁡M+q−1.

Facts & Assumptions

Given: A smooth vector bundle E→M over a boundaryless smooth manifold, with smooth metric h and rank q≥0.

[F1]

In a bundle chart the metric squared is H(u,v)=vTA(u)v, with A smooth positive definite (Smooth vector bundles, rank, fibres, and trivial bundles); disk and sphere bundles have their indicated inequalities (Disk, sphere, and Thom spaces of a metric vector bundle).

[L1]

At a regular level a smooth real-valued function has coordinate normal form, giving half-space charts for its sublevel (Local normal form for submersions, Regular sublevels are compact manifolds with boundary). Manifold boundaries are closed embedded submanifolds (The boundary of a positive-dimensional manifold is a closed embedded smooth (n-1)-manifold).

Proof

technique · direct
1.1F1L1givenalgebra

The smooth function H:E→R, H(v)=∥v∥h2, has vertical derivative w↦2h(v,w). At H(v)=1, evaluating on w=v gives 2, so 1 is regular. The local normal-form argument of [L1] supplies the subspace smooth half-space charts on H≤1, without needing compactness. Its boundary is H=1 and its interior is H<1. For q=0, H=0 and the disk bundle is simply M.

2.1L1step 1.1

These charts are restrictions of smooth ambient charts, so projection and zero section remain smooth. The boundary is closed and embedded by [L1], with the asserted dimension when the total dimension is positive.

3.1F1L2step 1.1algebra∎

If M is compact, cover it by finitely many compact coordinate pieces Ki lying inside bundle-trivialization domains. For q>0, on the compact set Ki×Sq−1 the function (u,z)↦zTA(u)z has a positive minimum ci by [L2]. Thus vTA(u)v≤1 implies ∣v∣≤ci−1/2. The disk bundle over Ki is a closed bounded subset of a Euclidean coordinate product, hence compact by [L2]. Their finite union is Dh(E), which is therefore compact; its closed boundary is compact too. Empty pieces are omitted. For q=0 compactness is just compactness of M.

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