Some thoughts on mathematics

Back for more!  Recall that we are interested in proving the following proposition:  Proposition:  There exists an irreducible, i.e. prime, homology 3-sphere which bounds a contractible 4-manifold and no bounding contractible four-manifold can be built with fewer than five handles.   In the last post, we showed that it suffices to show that there is a prime homology 3-sphere that bounds a contractible 4-manifold that is *not* surgery on a knot in $S^2 \times S^1$.  That was because being surgery on a knot in $S^2 \times S^1$ is equivalent to bounding a Mazur manifold, and those are the only contractible 4-manifolds with a handle decomposition with fewer than five handles.  That's our goal.  If you don't care about a proof of the proposition, here's a surgery picture for an example (and if you do care, use this picture as reference for the constructive proof below):     We'll start the proof by giving an obstruction to...

This sequence of posts was inspired by some conversations with Mike Miller Eismeier In this post, I want to talk about some contractible four-manifolds and connections with Dehn surgery .   Recall that lots of homology 3-spheres bound contractible four-manifolds.  For example, $S^3_{1/n}(K)$ bounds a contractible four-manifold if $K$ is a smoothly slice knot, as do the Brieskorn spheres $\Sigma(3,4,5)$ and $\Sigma(2,3,13)$.  By a quick Euler characteristic computation, it is easy to see that every contractible four-manifold other than the four-ball requires at least three handles.  In this sequence of posts I want to give a proof of the following goofy result (and give it some context).  Proposition:  There exists an irreducible, i.e. prime, homology 3-sphere which bounds a contractible 4-manifold and no bounding contractible four-manifold can be built with fewer than five handles.  Alternatively, there are prime homology 3-spheres that bound con...