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The Milnor homotopy seven-spheres are homeomorphic to $S^{7}$

Statement

Assume the Axiom of Choice and countable choice. For $h + j = \pm 1$ , the Milnor homotopy seven-sphere $M_{h, j}$ is homeomorphic to $S^{7}$ .

Facts & Assumptions

Given: Integers $h, j$ with $h + j = \pm 1$ and the smooth homotopy seven-sphere $M_{h, j}$ of Euler number $\pm 1$ makes the Milnor sphere bundle a homotopy seven-sphere.

[A1]

AC and ${A C}_{ω}$ are assumed; AC implies ${A C}_{ω}$ (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), AC implies DC implies countable choice).

[L1]

Removing two disjoint disks from a smooth homotopy $d$ -sphere with $d \geq 6$ gives a compact simply connected h-cobordism between two standard $S^{d - 1}$ faces (The two-disk complement of a homotopy sphere is an h-cobordism).

[L2]

Assume ${A C}_{ω}$ . The smooth simply connected h-cobordism theorem: a compact connected smooth h-cobordism of dimension at least six between closed simply connected manifolds is diffeomorphic to a product relative to one face (The smooth simply connected h-cobordism theorem).

[L3]

Every homeomorphism $S^{d - 1} \to S^{d - 1}$ extends radially to a homeomorphism $D^{d} \to D^{d}$ (Alexander trick: a sphere homeomorphism extends radially over the disk).

Proof

technique · direct

1.1L1A1given

By [L1] with $d = 7$ the complement $W$ of the interiors of two disjoint disks in $M_{h, j}$ is a compact simply connected h-cobordism of dimension seven between two standard $S^{6}$ boundary faces; the dimension meets the threshold $7 \geq 6$ .

2.1step 1.1L2A1

By [L2] and [A1] the cobordism $W$ is diffeomorphic to the product $S^{6} \times [0, 1]$ relative to one face.

3.1step 2.1L3

Reattach the two seven-disks: one attaching boundary diffeomorphism can be taken to be the standard one, and the other is a homeomorphism of $S^{6}$ which by [L3] extends radially to a homeomorphism of the disk; hence the reattached space $D^{7} \cup_{S^{6}} (S^{6} \times [0, 1]) \cup_{S^{6}} D^{7}$ is homeomorphic to $S^{7}$ .

4.1step 3.1∎

Therefore $M_{h, j}$ is homeomorphic to $S^{7}$ , as asserted; the argument is topological at the gluing step and does not claim smoothness of the radial extension at the origin.

Depends on

Used by

Dependency tree · two levels

54 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, section 9, printed pp. 109-110 (standard reference, not scraped)
John Milnor, On Manifolds Homeomorphic to the 7-Sphere, Annals of Mathematics 64 (1956), 399-405 (standard reference, not scraped)

The Milnor homotopy seven-spheres are homeomorphic to S7S7