Suppose I have a compact 3-dimensional submanifold N of S X (0,1) which has one boundary component, where S is a closed surface. Must N be a handlebody?
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2 Answers
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No. N could be homeomorphic to the exterior of any knot in the 3-sphere (i.e. a cube with a knotted hole).
answered Mar 1, 2010 at 21:17
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There is a class of three-manifolds which are called Handlebodies with Wormholes. M is a handlebody with wormholes if it is homeomorphic to the result of removing the regular neighborhood of a system of properly embedded arcs from a handlebody, that includes the example that Richard gave above. There is a Topology paper by Menasco and Thompson "Compressing Handlebodies with holes" that studies this class of manifolds.
If you also allowed the removal of the regular neighborhoods of graphs that did not touch the boundary of $S$, that class of manifolds would include every compact three-manifold with boundary that can be embedded in $S\times [0,1]$.
More simply, suppose you decided to remove the neighborhood of one properly embedded arc in $S\times [0,1]$ , whose boundary consisted of a point in $S\times \{0\}$ and one in $S\times \{1\}$, I proved that the complement is a handlebody if and only if the arc is isotopic to a vertical arc $\{pt\}\times [0,1]$. That appeared in,
Minimal surfaces and Heegaard splittings of the three-torus. Pacific J. Math. 124 (1986), no. 1, 119--130
I then developed an unknotting lemma for systems of arcs
An unknotting lemma for systems of arcs in $F\times I$. Pacific J. Math. 139 (1989), no. 1, 59--66.
There is related work of Gordon, that was done about the same time, but the referee lost my paper so that there was a two year lag publication.
On primitive sets of loops in the boundary of a handlebody. Topology Appl. 27 (1987), no. 3, 285--299.
appeared.
