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Due to Mazur, Akbulut and Kirby and many others, there are many examples of integer homology 3-spheres which bound contractible 4-manifolds given by attaching a single 2-handle to $S^1 \times D^3$ which algebraically intersects the one-handle once but maybe geometrically many times. Many of these can be distinguished from the 3-sphere by things like the Casson invariant and other invariants specifically developed for integral homology $S^3$'s for instance in https://arxiv.org/pdf/1508.01491.pdf

I would like to know how to distinguish the analogous constructions (which don't reduce to the above case) for homology $S^1 \times S^2$'s bounding homotopy $S^1 \times D^3$'s (homotopy equivalent not homeomorphic). Specifically are there any examples of knots in $S^1\times S^2 \# S^1 \times S^2$ which algebraically intersect both one handles once but not geometrically, that surger to (hopefully many or even infinitely for the application I have in mind) non-trivial examples of homology $S^1 \times S^2$'s? And if so, how does one distinguish these?

Note by construction these all bound homotopy $S^1 \times D^3$'s. It's very easy to draw such candidates knots, but they end up having a decent amount of crossings, and even then I don't know what tools I would use to distinguish their surgeries from $S^1 \times S^2$ or each other.

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    $\begingroup$ There's two aspects (in my mind) to your question: distinguishing homology $S^1\times S^2$'s, and distinguishing the 4-manifolds (homotopy $S^1$) that a homology $S^1\times S^2$ bounds. Which one are you having issues with? $\endgroup$
    – Ian Agol
    Commented Sep 1, 2016 at 4:26
  • $\begingroup$ @IanAgol I would like to do the first. I want to have examples analogous to Mazur manifolds (homotopy equivalent 4-manifolds which are distinguished by their boundary) that hopefully can be described in the simplest possible way (surgery on a single knot in $S^1 \times S^2 \# S^1 \times S^2$). $\endgroup$
    – PVAL
    Commented Sep 1, 2016 at 12:13
  • $\begingroup$ There are plenty of tools for distinguishing 3-manifolds in practice. See the website math.uiuc.edu/~nmd/computop Snappea is particularly effective. $\endgroup$
    – Ian Agol
    Commented Sep 1, 2016 at 15:09

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A standard construction would be to take a 3-manifold $Y$ given by 0-surgery on a knot $K$. If $K$ is the boundary of a slice disk $D \subset B^4$, then the complement, say $W$ of a neighborhood of $D$ is a homology $S^1 \times B^3$ with boundary $Y$. But you are asking for more, since $W$ should be a homotopy $S^1 \times B^3$; this is equivalent to its fundamental group being ${\bf Z}$. This would require that the Alexander polynomial of $K$ be equal to $1$.

There are plenty of examples of Alexander polynomial 1 knots bounding a disk with $\pi_1(B^4-D) = \bf{Z}$. Three nice ones are given in the paper Doubly slice knots with low crossing number (New York J. Math. 21 (2015) 1007–1026) of Livingston and Meier. See section 4.2.

If you are willing to accept a topological (rather than smooth) $W$, then Freedman has shown any homology $S^1 \times S^2$ with Alexander polynomial equal to 1 bounds such a $W$. Lots of $S^1 \times S^2$s with Alexander polynomial equal to 1 do not bound a smooth homotopy $S^1 \times B^3$.

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  • $\begingroup$ Thanks a lot. I think the three from the Livingston and Meier paper are of the type I want (the slice disk exterior is diffeomorphic to a boundary sum of two $S^1 \times B^3$'s with a single 2-handle attached). $\endgroup$
    – PVAL
    Commented Sep 1, 2016 at 15:11
  • $\begingroup$ I second @IanAgol's suggestion to use Snappea to distinguish the resulting 3-manifolds. $\endgroup$
    – Danny Ruberman
    Commented Sep 1, 2016 at 15:37
  • $\begingroup$ I think you likely already realized this, but in case anyone else is reading this the examples you gave are already distinguished from $S^1 \times S^2$ by property R. $\endgroup$
    – PVAL
    Commented Sep 1, 2016 at 19:05
  • $\begingroup$ I read the question as asking about distinguishing these from each other. You're quite right that Property R implies that none of them is $S^1\times S^2$. $\endgroup$
    – Danny Ruberman
    Commented Sep 1, 2016 at 19:44

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