Math-1449 · mathematics · Topology

Math-1449

Kirby Problem 4.26

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Statement

Definitions and mathematical statement

An integral homology $3$-sphere is a closed connected oriented $3$-manifold $Y$ such that $H_i(Y;\mathbb Z)\cong H_i(S^3;\mathbb Z)$ for all $i$. Equip it with its smooth structure and choose a smoothly embedded closed ball $B^3\subset Y$. Set \[ Y_0=Y\setminus\operatorname{int}(B^3). \] This is a compact smooth $3$-manifold with boundary $S^2$; puncturing means removing the interior of a ball, not removing only a point.

Does every such $Y_0$ admit a smooth embedding into the standard smooth $S^4$? Smoothness is required also along the boundary, so near boundary points the embedding has the usual smooth half-space local model. The embedding need not send $\partial Y_0$ to a preassigned sphere, nor need that embedded $2$-sphere be unknotted.

No simple-connectivity assumption on $Y$ is made: integral homology alone does not determine its fundamental group. The question asks for all integral homology spheres, with no surgery presentation or covering description supplied. Removing a different smooth ball does not change the diffeomorphism type of the punctured manifold.

The desired embedding is into the standard sphere itself. An embedding into a homotopy $4$-sphere or an integral homology $4$-sphere would need an additional standardness argument. Nor is the problem equivalent to embedding the closed manifold $Y$: a missing $3$-ball might not be attachable within the ambient $S^4$ without intersections.

Short English statement

Does removing a ball from any integral homology three-sphere produce a manifold that smoothly embeds in the standard four-sphere?

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