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.



Distinguishing homology $S^1 \times S^2$'s which bound homotopy $S^1$'s
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](http://nyjm.albany.edu/j/2015/21-47v.pdf) (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 $S^1 \times B^3$.


