Math-1614 · mathematics · Topology

Math-1614

Kirby Problem 4.36

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Statement

Definitions and mathematical statement

Let $k:S^2\hookrightarrow S^4$ be a smooth or locally flat two-knot and let $E_k$ be its exterior, the complement of an open tubular neighborhood. Set $G_k=\pi_1(E_k)$ and $A_k=\pi_2(E_k)$. Deck transformations of the universal cover make $A_k$ a $\mathbb Z[G_k]$-module.

The homotopy two-type of $E_k$ is its second Postnikov stage: a space with the same fundamental and second homotopy groups as $E_k$ and with higher homotopy groups zero. It is specified by \[ (G_k,A_k,\kappa_k),\qquad \kappa_k\in H^3(G_k;A_k), \] where $\kappa_k$ is the first Postnikov invariant, classifying the fibration with fiber $K(A_k,2)$ over $K(G_k,1)$.

If two two-knot exteriors have isomorphic homotopy two-types, must the exteriors be homotopy equivalent? Isomorphism here means a group isomorphism, a compatible module isomorphism, and preservation of the cohomology class $\kappa$ under the induced identifications. The question ranges over arbitrary two-knots in the stated category.

The conclusion concerns ordinary homotopy type of the exterior. It does not ask for a homeomorphism, equivalence of the knots, or preservation of a peripheral marking. Equality of $G$ and $A$ without the Postnikov invariant is a weaker hypothesis. The issue is whether higher homotopy information in this particular class of four-dimensional exteriors is already forced by the full two-type.

Short English statement

Does the fundamental group, second homotopy module, and first Postnikov invariant determine a two-knot exterior up to homotopy?

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