Math-3243 · mathematics · Topology

Math-3243

Kirby Problem 4.47

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Statement

Definitions and mathematical statement

Shared notation.
$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Let $L=L_1\cup\cdots\cup L_m\subset S^3$ be a smooth oriented link with zero pairwise linking numbers: $\operatorname{lk}(L_i,L_j)=0$ for $i\ne j$. Linking number is the signed intersection with a Seifert surface for the other component. For each component choose its zero-framed longitude $\lambda_i$ and a meridian $\mu_i$ in the boundary diagram, placed so that $\operatorname{lk}(\lambda_i,\mu_i)=0$, as in the source construction. Both auxiliary curves carry zero framing.

A four-dimensional round one-handle is $[-1,1]\times S^1\times D^2$, attached along the two solid tori $\{-1,1\}\times S^1\times D^2$. Let $R(L)$ be obtained from $B^4$ by attaching one such handle at the neighborhoods of $\lambda_i$ and $\mu_i$, with their specified framings, for each $i$. The original link components remain as curves in the new boundary. Call $L$ round-handle slice if its components bound pairwise disjoint proper locally flat disks in $R(L)$; locally flat means the disk embedding has the standard local topological plane or half-plane model.

Statement recorded in the supplied source.
Is every link $L\subset S^3$ with zero pairwise linking numbers round-handle slice? Equivalently, must there exist a disjoint family of locally flat disks $D_1,\ldots,D_m\subset R(L)$ with $\partial D_i=L_i$, or is there a zero-pairwise-linking counterexample? Smooth slicing disks are not required. [S2]

Short English statement

Do all links with zero pairwise linking numbers become topologically slice after the specified round-one-handle attachments?

Sources

Catalogue entry

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