Math-1452 · mathematics · Topology

Math-1452

Kirby Problem 4.31

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Statement

Definitions and mathematical statement

Fix a smooth knot $K\subset S^3=\partial B^4$, meaning the image of an embedding of a circle. A spanning surface is a compact connected orientable smooth surface $F$ properly embedded in $B^4$ with $F\cap\partial B^4=\partial F=K$. Proper smooth embedding includes the standard collar condition at the boundary. Its genus is unrestricted.

Two such surfaces $F_0,F_1$ form an exotic pair relative to the boundary if they are ambiently topologically isotopic fixing $\partial B^4$ pointwise, but are not ambiently smoothly isotopic under that same restriction. Explicitly, the topological condition is a path $H_t\in\operatorname{Homeo}(B^4)$ with $H_0=\mathrm{id}$, $H_t|_{S^3}=\mathrm{id}$ and $H_1(F_0)=F_1$; the corresponding smooth family of diffeomorphisms must not exist.

Does every knot $K$ bound such an exotic pair? The surfaces may have positive genus, and that genus may depend on the knot. The question does not assert that every knot bounds exotic disks. The topological isotopy ensures the two surfaces have the same genus and the same boundary placement.

The universal quantifier includes the unknot as well as nontrivial knots. Producing pairs for infinitely many knots does not by itself establish the statement for an arbitrary fixed boundary knot. Both surfaces lie in the standard smooth $4$-ball; changing the ambient smooth structure is not part of the proposed construction.

Short English statement

Does every knot bound two orientable surfaces in the standard four-ball that are topologically but not smoothly isotopic relative to the boundary?

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