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Dual elimination of top-index handles

Statement

Assume ACω. Let (W;M0,M1) be a compact connected triad with M1≠∅ and dim⁡W=n. Then W admits a handle decomposition relative to M0 with no n-handles; if M1 is disconnected the dual presentation ends with the corresponding dual (n−1)-handles coming from the connecting 1-handles in the reversed triad, and still has no n-handles. Equivalently, the n-handles of a presentation relative to M0 are the duals of the 0-handles in the reversed triad, so eliminating those 0-handles eliminates these n-handles.

Facts & Assumptions

[F1]

Smooth cobordism triad for Morse theory: the reversed triad of (W;M0,M1) is (W;M1,M0), a compact triad with the same collars and the faces exchanged; no orientation is used.

[F2]

Connected cobordisms admit presentations without superfluous zero handles: Assume ACω. A compact connected triad whose incoming boundary is nonempty admits a handle decomposition relative to that boundary with no 0-handles; if that incoming boundary has k components the presentation begins with k−1 connecting 1-handles; if the incoming boundary is empty, exactly the 0-handles needed to create the components remain.

[F3]

Morse functions and handle decompositions correspond: Assume ACω. Every finite handle decomposition of a compact triad relative to its incoming face is induced by an adapted excellent Morse function with one critical point per handle, of the same index.

[F4]

Handle duality from negating a Morse function: Assume ACω. If f is adapted excellent on a compact triad then 1−f is adapted excellent on the reversed triad with indices n−ind⁡(p) at the same critical points, and its handle decomposition relative to the opposite face is the dual of the decomposition of f.

[F5]

Dual handle decomposition: in the dual presentation a k-handle becomes an (n−k)-handle, the order is reversed, and attaching and belt spheres are interchanged.

[F6]

Index n handles cap boundary spheres: an n-handle attaches along its whole boundary sphere Sn−1 and caps it; a 0-handle attaches along the empty set and creates a component.

[F7]

The Axiom of Countable Choice (ACω): ACω: every at most countable family of nonempty sets has a choice function.

[F8]

Under ACω, Every smooth manifold admits a riemannian metric supplies a background metric, Morse lemma supplies the quadratic critical charts, A manifold bump for a compact set inside an open set supplies finite chart cutoffs, and Compactly supported smooth vector fields are complete makes a compactly supported smooth field on a boundaryless carrier complete.

Proof

Given: The compact connected triad (W;M0,M1) with dim⁡W=n and M1≠∅.

1.1F1F2givenconstruct

The reversed triad (W;M1,M0) of [F1] is compact and connected, its incoming face is M1≠∅, and its outgoing face is M0, which may be empty. By [F2] applied to the reversed triad, W admits a handle decomposition relative to M1 with no 0-handles; if M1 has k components, that presentation begins with k−1 connecting 1-handles. Fix this chosen presentation for the dual construction.

2.1F3F4F5F7F8step 1.1construct

Realize the presentation by a Morse function: by [F3] the decomposition of step 1.1 is induced by an adapted excellent Morse function f on the reversed triad (W;M1,M0). To supply the field required by duality, patch the background metric of [F8] to Euclidean metrics in smaller disjoint Morse charts and to product metrics on the realizing function's regular face collars, using finite cutoffs. Its negative gradient has the exact model (2u,−2v) and the required boundary signs. Extend the product collar field across signed face collars, with a cutoff vanishing before their outer ends. The resulting ambient field is compactly supported, hence complete by [F8], and restricts to an adapted field for f. Applying [F4] to that pair, the function 1−f is adapted excellent on the original triad (W;M0,M1), with the same critical points and with indices transformed by k↦n−k, and its handle decomposition relative to M0 is the dual of the decomposition of f, in the sense of [F5].

3.1F5step 1.1step 2.1algebra

Since the presentation of step 1.1 has no 0-handles, its dual presentation has no n-handles, because a 0-handle becomes an n-handle under k↦n−k by [F5]. The connecting 1-handles of step 1.1, which join the k components of M1 when M1 is disconnected, become handles of index n−1 in the dual presentation, again by [F5], and they come last because the order is reversed. Duality bijects the handles and complements their indices, so the resulting presentation of W relative to M0 has the same number of n-handles as the reversed presentation has 0-handles, namely zero.

4.1F2F5F6step 3.1algebra∎

Equivalently, the n-handles of any presentation relative to M0 are the duals of the 0-handles of the reversed presentation: a 0-handle is an n-disk attached along the empty set [F6], and its dual is an n-handle attached along the whole boundary sphere [F6], so eliminating the 0-handles of the reversed presentation by [F2] eliminates exactly the n-handles of the dual presentation relative to M0. This is the dual endpoint elimination; the argument uses the duality, correspondence and elimination suppliers, and through them ACω.

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Sources