Consider a smooth torus endowed with the non-bounding spin structure. Pick a basis of its first homology and a diffeomorphism inducing the S-transformation

$\left(\begin{array}{cc} 0 & 1 \\-1 & 0\end{array}\right)$

in the chosen basis. Construct the mapping torus $U$ of this diffeomorphism. Take the non-bounding spin structure on the $S^1$ base. This induces a spin structure on $U$. As the 3rd spin cobordism group vanishes, there is a smooth spin 4-manifold $W$ whose boundary is $U$.

The question is: How can one concretely construct such a $W$?

More generally, when given a manifold that is known to bound, what are the tools available to construct a concrete bordism to the empty manifold?

**Update**: I explain why Marco's answer below does not provide 4-manifolds bounding the mapping torus above with the required spin structure. It is not difficult to see that $H^1(U;\mathbb{Z}_2) = \mathbb{Z}_2^2$, with one generator associated to the base circle and one generator associated to the cycle on the torus fiber left invariant under the S transformation. The required spin structure restricts to the non-bounding spin structure on representative loops in both classes. This means that these loops cannot become homotopically trivial in the 4-manifold. As the 4-manifolds Marco constructs are all simply connected, they cannot carry a spin structure restricting to the one I am interested in on the boundary.

simply connectedspin 4-manifold that it bounds. This can also be seen more abstractly; take any spin 4-manifold with given spin boundary, and do (spin) surgery on generators of the fundamental group. to make it simply connected. What your argument might show is that there is no single 4-manifold over which every spin structure on U extends. But there's no reason to expect that there would be such a 4-manifold. $\endgroup$ – Danny Ruberman Sep 28 '16 at 12:19