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Geometric cancellation is a unit entry in the handle matrix

Statement

Assume ${A C}_{ω}$ . In the situation of the matrix definition, suppose the attaching sphere $A_{i}$ of the $(k + 1)$ -handle $g_{i}$ meets the belt sphere $B_{j}$ of the $k$ -handle $e_{j}$ transversely in exactly one point. Then the oriented entry is $M_{i j} = \pm 1$ , equal to the local intersection sign of that point; in particular a geometrically cancelling pair has a unit entry in $Z$ . Without orientations the mod-2 entry is $1 \in Z_{2}$ , also a unit.

Facts & Assumptions

Given: An index-ordered presentation with $1 \leq k \leq n - 2$ and transverse attaching and belt spheres, a $(k + 1)$ -handle $g_{i}$ whose attaching sphere $A_{i}$ meets the belt sphere $B_{j}$ of the $k$ -handle $e_{j}$ transversely in exactly one point.

[F1]

Attaching-belt intersection matrix of adjacent-index handles: the entry $M_{i j}$ is the oriented intersection number $I (A_{i}, B_{j})$ when the spheres carry the orientations induced by the framings and the boundary orientation, and the mod-2 number $I_{2} (A_{i}, B_{j})$ otherwise.

[F2]

The oriented intersection number and The local oriented intersection sign: for transverse complementary-dimensional submanifolds the oriented intersection number is the finite sum $\sum_{p \in A \cap B} ε (p)$ of local signs, each of which lies in ${+ 1, - 1}$ ; the empty intersection contributes $0$ .

[F3]

The mod 2 intersection number and The oriented intersection number reduces to the mod 2 number: the mod-2 intersection number is $# (A \cap B) m o d 2$ , and in the common oriented setting it is the reduction modulo two of the oriented number.

[F4]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ is assumed, as in the matrix definition; the single-entry computation below is choice-free.

Proof

technique · direct

1.1F1F2F4given

By [F1] the entry is the oriented intersection number of $A_{i}$ and $B_{j}$ in the middle boundary. By hypothesis the transverse intersection is exactly one point $P$ , so by [F2] the finite sum has the single term $ε (P)$ , and $ε (P) \in {+ 1, - 1}$ by definition of the local sign. Hence $M_{i j} = ε (P) = \pm 1$ , a unit of $Z$ . No other entry is involved.

2.1F3step 1.1

For the mod-2 version, [F3] gives $I_{2} (A_{i}, B_{j}) = # (A_{i} \cap B_{j}) m o d 2 = 1 m o d 2 = 1 \in Z_{2}$ , a unit of $Z_{2}$ ; and in the oriented setting this agrees with the reduction of the oriented entry by [F3].

3.1step 1.1step 2.1given∎

Therefore a geometrically cancelling pair — one transverse intersection point of the attaching sphere with the belt sphere — has a unit entry in the attaching-belt intersection matrix, over $Z$ with value the local sign of that point and over $Z_{2}$ with value $1$ .

Depends on

Used by

Adjacent-index handles with zero intersection do not cancel Counterexample
Algebraic intersection one with three geometric points Counterexample
An elementary cancelling handle pair gives a product cobordism Example
Algebraic cancellation does not yet give geometric cancellation Lemma

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John Milnor, Lectures on the h-Cobordism Theorem (notes by L. Siebenmann and J. Sondow; scanned edition with text layer) (standard reference, not scraped)
C. T. C. Wall, Differential Topology (Cambridge Studies in Advanced Mathematics 156; complete PDF) (standard reference, not scraped)