Kirby Problem 4.40

Definitions and mathematical statement

A smooth link of $2$-spheres in $S^4$ is a finite disjoint union $L=K_1\sqcup\cdots\sqcup K_r$ of smoothly embedded copies of $S^2$, where $r\ge1$. Regard $S^4$ as the boundary of the standard smooth $5$-ball $B^5$. The link is smoothly slice if there exist disjoint proper smooth embeddings \[ D_j:B^3\hookrightarrow B^5\quad(1\le j\le r) \] whose boundary images are $K_j$. Proper means that the inverse image of $\partial B^5$ is exactly $\partial B^3$, with the usual smooth boundary-collar condition.

Is every smooth link of $2$-spheres in $S^4$ smoothly slice? Each component must bound its own $3$-ball, and these balls must be simultaneously disjoint in the same $B^5$. Individual sliceness of every component therefore does not alone prove sliceness of the link.

The analogous locally flat question replaces smooth embeddings by topological embeddings with local Euclidean pair charts, including half-space charts at the boundary. It asks for the corresponding locally flat disjoint balls. The source permits studying either category; the displayed main formulation fixes the smooth one.

The number of components is arbitrary and no unlinking, boundary-link, or complement-group hypothesis is imposed. The filling dimension is five: bounding balls inside the original $S^4$ would be a substantially stronger condition than the sliceness requested here.

Short English statement

Does every finite link of two-spheres in the four-sphere bound disjoint properly embedded three-balls in the five-ball?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2916 (KP-4.40), Kirby's Problems in Low-Dimensional Topology. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 K3: A New Problem List in Low-Dimensional Topology, Problem 4.40. Expanded formulation of the supplied catalogue statement. https://math.berkeley.edu/sites/default/files/surv-295-ruberman-watermarked-author-pdf.pdf.